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$$ \begin{array}{|c|c|} \hline \text{Words in few symbols} & \text{Tuples of few functions}\\\hline A &\frac{dt}t\\\hline B=B_0 &\frac{dt}{1-t}\\\hline B'=B_1 &-\frac{dt}{1+t}\\\hline X=\text{Alphabet }\{A,B,C\}^* & \Omega=\text{Alphabet in }\frac{dt}t,\ \frac{dt}{1-t},\ -\frac{dt}{1+t}\\\hline X\text{-word} & \Omega\text{-word}\\\hline X\text{-word }w & \Omega\text{-word often also written }w\\\hline \int_0^1 w\in\Bbb R & \operatorname{It}(w)=\operatorname{It}_{[0,1]}(w)\in\Bbb R\\\hline \operatorname{Li}_w(t)=\operatorname{Li}(w,t)= \int_0^t w & \operatorname{It}_{[0,t]}(w)\\\hline \int_\gamma w & \operatorname{It}_\gamma(w)\\\hline \bbox[lightyellow]{\qquad \int_\gamma w\cdot \int_\gamma w' = \int_\gamma w\ ш\ w'\qquad} & \operatorname{It}_\gamma(w)\operatorname{It}_\gamma(w') = \operatorname{It}_\gamma(w \ ш\ w')\\\hline ш\text{ is the shuffle product of words} &\\\hline \int_t^1 A^n & \frac 1{n!}(-1)^n\log^n t\\\hline \int_0^t B^n & \frac 1{n!}(-1)^n\log^n (1-t)=\frac 1{n!}\operatorname{Li}_1(t)^n\\\hline \int_0^t C^n & \frac 1{n!}(-1)^n\log^n (1+t)=\frac 1{n!}\operatorname{Li}_1(-t)^n\\\hline \operatorname{Li}(B,t) & \operatorname{Li}_1(t) =-\log(1-t)=\int_0^tB=\int_0^t\frac{dt}{1-t}\\\hline \operatorname{Li}(AB,t) & \operatorname{Li}_2(t) =\int_0^t\frac {\operatorname{Li}_1(t)}{t}\;dt =\int_0^t\frac {dt}t\int_0^t\frac {du}{1-u} \operatorname{It}_{[0,t]}\left(\frac 1t,\frac1{1-t}\right) \\\hline \operatorname{Li}(AAB,t) & \operatorname{Li}_3(t) =\int_0^t\frac {\operatorname{Li}_2(t)}{t}\;dt \operatorname{It}_{[0,t]}\left(\frac 1t,\frac 1t,\frac1{1-t}\right) \\\hline \operatorname{Li}(\underbrace{AA\dots AB}_{n\text{ letters}},t) & \operatorname{Li}_n(t) \\\hline Z(s) = Z_0(s)=\underbrace{AA\dots AB}_{s\text{ letters}} &\\ Z'(s) = Z_1(s) = \underbrace{AA\dots AB'}_{s\text{ letters}} &\\\hline Z_j(s)= \underbrace{AA\dots AB_j}_{s\text{ letters}} &\\\hline s =(s_1,s_2,\dots,s_d)\text{ poly-index} & s =(s_1,s_2,\dots,s_d)\in\Bbb N_{>0}^d\\\hline j =(j_1,j_2,\dots,j_d)\text{ poly-index} & j =(j_1,j_2,\dots,j_d)\in(\Bbb Z/N)^d\\\hline \Delta j=(j_1-0, j_2-j_1,\dots,j_d-j_{d-1}) & \\\hline x =(x_1,x_2,\dots,x_d)\text{ poly-argument} & x =(x_1,x_2,\dots,x_d)\\\hline \mu =(\mu,\mu,\dots,\mu)\text{ as poly-argument} & \mu =(\mu,\mu,\dots,\mu)\\\hline \prod Z(s,j) =\prod_j Z_j(s)\\ =Z_{j_1}(s_1)\dots Z_{j_d}(s_d) &\\\hline & \displaystyle \zeta(s) = L(s)=L(s,1)\\ & \displaystyle =\sum_{n_1>\dots >n_d\ge 1}\frac 1{n_1^{s_1}\dots n_d^{s_d}}\\\hline & \displaystyle L(s,x)=\sum_{n_1>\dots >n_d\ge 1}\frac {x_1^{n_1}\dots x_d^{n_d}}{n_1^{s_1}\dots n_d^{s_d}}\\\hline \int_0^1\prod Z(s,j) & L(s, \mu^{\Delta j})\\\hline \end{array} $$

The computation of $J_1$. The following part $J_{11}$ of $J_1$ is simpler, because only $y$ and $1-y$ appear, so the symbolic part involve only $A,B$: $$ \begin{aligned} J_{11} &= \int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1-y}\; dy \\ &\overset{\text{up to }\pm} = \int_0^1\frac {dy}{1-y} \left(\int_y^1\frac{du}{u}\right) \left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right) \\ &= \int_0^1 B\int_0^t A\ ш\ AB \\ &= \int_0^1 B( \color{blue}{A}\ ш\ AB ) \\ &= \int_0^1 ( B\color{blue}{A}AB + BA\color{blue}{A}B + BAB\color{blue}{A} ) =2\int_0^1\color{brown}{B}AAB+\int_0^1 \color{brown}{B}ABA \\ &\qquad\text{ move $\color{brown}{B}$ from the first place behind}\\ &= -2\int_0^1 (A\color{brown}{B}AB + AA\color{brown}{B}B + AAB\color{brown}{B}) -\int_0^1 (A\color{brown}{B}BA + AB\color{brown}{B}A+ ABA\color{brown}{B}) \\ &= -3\int_0^1ABAB -4\int_0^1AABB -2\int_0^1ABB\color{blue}{A} \\ &\qquad\text{ move $\color{blue}{A}$ from the last place to the left}\\ \\ &=-3\int_0^1ABAB -4\int_0^1AABB +2\int_0^1(AB\color{blue}{A}B + A\color{blue}{A}BB + \color{blue}{A}ABB) \\ &=-\int_0^1ABAB=-\int_0^1Z(2)Z(2) \\ &\overset{\text{up to }\pm}=\int_0^1 Z(2,2)=\zeta(2,2) \\ &\qquad\text{ recall $\zeta(k)\zeta(n)=\zeta(k,n)+\zeta(n,k)+\zeta(n+k)$, use $k=n=2$} \\ &=\frac 12(\zeta(2)^2-\zeta(4)) = \frac 1{120}\pi^4\ . \end{aligned} $$ Numerical check, pari/gp:

The computation of the remained part $J_{12}$ from $J_1$ follows, we have algebraically a similar situation, but together with $y$ and $1-y$ there appears also $1+y$, so the symbolic part involve $A,B$ and also the $2$-chromatic piece $B'$: $$ \begin{aligned} J_{12} &= \int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1+y}\; dy \\ &\overset{\text{up to }\pm} = \int_0^1\frac {dy}{1+y} \left(\int_y^1\frac{du}{u}\right) \left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right) \\ &= \int_0^1 B'\int_0^t A\ ш\ AB \\ &= \int_0^1 B'( \color{blue}{A}\ ш\ AB ) \\ &= \int_0^1 ( B'\color{blue}{A}AB + B'A\color{blue}{A}B + B'AB\color{blue}{A} ) =2\int_0^1\color{brown}{B'}AAB+\int_0^1 \color{brown}{B'}ABA \\ &\qquad\text{ move $\color{brown}{B'}$ from the first place behind}\\ &= -2\int_0^1 (A\color{brown}{B'}AB + AA\color{brown}{B'}B + AAB\color{brown}{B'}) -\int_0^1 (A\color{brown}{B'}BA + AB\color{brown}{B'}A+ ABA\color{brown}{B'}) \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) -\int_0^1(AB'B+ABB')\color{blue}{A} -\int_0^1ABAB' \\ &\qquad\text{ move $\color{blue}{A}$ from the last place to the left} \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) +\int_0^1\color{blue}{A}A(B'B+BB') \\ &\qquad +\int_0^1A\color{blue}{A}(B'B+BB') +\int_0^1AB'\color{blue}{A}B +\int_0^1AB\color{blue}{A}B' \\ &=-\int_0^1AB'AB=-\int_0^1Z_1(2)Z_0(2) \\ &\overset{\text{up to }\pm}=\int_0^1 Z_{(1,0)}(2,2)=L_{(1,1)}(2,2) \\ &=\zeta(\bar 2,\bar 2) \\ &\qquad\text{ (use $\zeta(\bar k)\zeta(\bar n)=\zeta(\bar k,\bar n)+\zeta(\bar n,\bar k)+\zeta(n+k)$, $n+k$ even)} \\ &\qquad\text{ recall $\zeta(\bar 2)=-\frac 1{1^2}+\frac 1{2^2}-\frac 1{3^2}+\frac 1{4^2}\pm \dots=-\left(1-\frac 2{2^2}\right)\zeta(2)=-\frac1{12}\pi^2$, } \\ &=\frac 12(\zeta(\bar 2)^2-\zeta(4)) = \frac 12\pi^4\left(\frac 1{12^2}-\frac 1{90}\right) = -\frac 1{480}\pi^4\ . \end{aligned} $$ Numerical check, pari/gp:

Later EDIT: The formula for $J_3$ can be obtained in a similar symbolic framework, and then checked in pari/gp, i will do this now in order to answer some comments. The fomula for $J_3$ is $$ J_3 = \int_0^1\log^2 y\;\log\frac{1-y}{1+y}\cdot \frac1{1-y^2}\; dy $$ so is a linear combination of integrals of $\frac 1{1\pm y}\log^2y\frac 1{1\pm y}$. Working symbolically, $\frac 1{2!}\log^2y$ corresponds to $A$, and $\frac 1{1-y}$, $-\frac 1{1+y}$ respectively to $B=B_0$, $B'=B_1$. We use the indices $j,k$ in the $B$-symbols to cover all four cases. We use $\mp_j$ for $(-1)^{j-1}$, so $\mp_0=\mp$. Then: $$ \begin{aligned} J_{3jk}&:=\int_0^1\frac1{1-\mp_j y}\log^2 y\;\log(1\mp_k y)\; dy \\ &\overset{\text{up to }pm}= \int_0^1 B_j(2A^2\ ш\ B_k) =2\int_0^1B_j(B_kAA +AB_kA+AAB_k)\qquad\text{(right-$A^2$-moves)} \\ &=-\int_0^1(AB_jB_kA+B_jAB_kA)+2\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k \\ &=-\int_0^1AB_jB_kA+\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k\qquad\text{(right-$A$-moves)} \\ &=+\int_0^1(2A^2B_jB_kA + AB_jAB_k) - \int_0^1(AB_jAB_k + 2B_jAAB_k) +2\int_0^1B_jAAB_k \\ &=+2\int_0^1A^2B_jB_k=2\int_0^1Z_j(3)Z_k(1) \\ &= 2L_{3, 1}(\ (-1)^j\ ,\ (-1)^{j+k}\ )\ , \\ &\qquad\qquad\text{ and we use the values:} \\ &\qquad L_{3,1}(1,1) = \zeta(3,1)=\frac 14\zeta(4)=\frac 1{360}\pi^4\ ,\\ &\qquad L_{3,1}(1,-1) = \zeta(3,\bar 1)\ ,\\ &\qquad L_{3,1}(-1,1) = \zeta(\bar 3,1)\ ,\\ &\qquad L_{3,1}(-1,-1) = \zeta(-3,-1)\ ,\\ &\qquad\qquad\text{ and we have:}\\ \frac 12J_{300} &=-\zeta(3,1)=-\frac 1{360}\pi^4\ , \\ \frac 12 J_{310} &= -\zeta(\bar 3,\bar 1) = \frac 1{180}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) - \frac 1{12}\log^4 2\ , \\ -\frac 12J_{301} &=\zeta(3,\bar 1)= \frac{19}{1440}\pi^4 -\frac 74\log 2\;\zeta(3) \ , \\ -\frac 12 J_{311} &= \zeta(\bar 3,1) = \frac 1{48}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) -\frac 74\log2\;\zeta(3) - \frac 1{12}\log^4 2\ , \end{aligned} $$ and this gives finally the claimed value for $J_3$, $$ J_3=\frac 12(J_{300} +J_{310}-J_{301}-J_{311})\\ =\frac {53}{1440}\pi^4 +\frac 16\pi^2\log^2 2 - 4\operatorname{Li}_4\left(\frac 12\right) -\frac 72\log 2\;\zeta(3) - \frac 16\log^4 2 \ . $$

$$ \begin{array}{|c|c|} \hline \text{Words in few symbols} & \text{Tuples of few functions}\\\hline A &\frac{dt}t\\\hline B=B_0 &\frac{dt}{1-t}\\\hline B'=B_1 &-\frac{dt}{1+t}\\\hline X=\text{Alphabet }\{A,B,C\}^* & \Omega=\text{Alphabet in }\frac{dt}t,\ \frac{dt}{1-t},\ -\frac{dt}{1+t}\\\hline X\text{-word} & \Omega\text{-word}\\\hline X\text{-word }w & \Omega\text{-word often also written }w\\\hline \int_0^1 w\in\Bbb R & \operatorname{It}(w)=\operatorname{It}_{[0,1]}(w)\in\Bbb R\\\hline \operatorname{Li}_w(t)=\operatorname{Li}(w,t)= \int_0^t w & \operatorname{It}_{[0,t]}(w)\\\hline \int_\gamma w & \operatorname{It}_\gamma(w)\\\hline \bbox[lightyellow]{\qquad \int_\gamma w\cdot \int_\gamma w' = \int_\gamma w\ ш\ w'\qquad} & \operatorname{It}_\gamma(w)\operatorname{It}_\gamma(w') = \operatorname{It}_\gamma(w \ ш\ w')\\\hline ш\text{ is the shuffle product of words} &\\\hline \int_t^1 A^n & \frac 1{n!}(-1)^n\log^n t\\\hline \int_0^t B^n & \frac 1{n!}(-1)^n\log^n (1-t)=\frac 1{n!}\operatorname{Li}_1(t)^n\\\hline \int_0^t C^n & \frac 1{n!}(-1)^n\log^n (1+t)=\frac 1{n!}\operatorname{Li}_1(-t)^n\\\hline \operatorname{Li}(B,t) & \operatorname{Li}_1(t) =-\log(1-t)=\int_0^tB=\int_0^t\frac{dt}{1-t}\\\hline \operatorname{Li}(AB,t) & \operatorname{Li}_2(t) =\int_0^t\frac {\operatorname{Li}_1(t)}{t}\;dt \\ &=\int_0^t\frac {dt}t\int_0^t\frac {du}{1-u} = \operatorname{It}_{[0,t]}\left(\frac 1t,\frac1{1-t}\right) \\\hline \operatorname{Li}(AAB,t) & \operatorname{Li}_3(t) =\int_0^t\frac {\operatorname{Li}_2(t)}{t}\;dt = \operatorname{It}_{[0,t]}\left(\frac 1t,\frac 1t,\frac1{1-t}\right) \\\hline \operatorname{Li}(\underbrace{AA\dots AB}_{n\text{ letters}},t) & \operatorname{Li}_n(t) \\\hline Z(s) = Z_0(s)=\underbrace{AA\dots AB}_{s\text{ letters}} &\\ Z'(s) = Z_1(s) = \underbrace{AA\dots AB'}_{s\text{ letters}} &\\\hline Z_j(s)= \underbrace{AA\dots AB_j}_{s\text{ letters}} &\\\hline s =(s_1,s_2,\dots,s_d)\text{ poly-index} & s =(s_1,s_2,\dots,s_d)\in\Bbb N_{>0}^d\\\hline j =(j_1,j_2,\dots,j_d)\text{ poly-index} & j =(j_1,j_2,\dots,j_d)\in(\Bbb Z/N)^d\\\hline \Delta j=(j_1-0, j_2-j_1,\dots,j_d-j_{d-1}) & \\\hline x =(x_1,x_2,\dots,x_d)\text{ poly-argument} & x =(x_1,x_2,\dots,x_d)\\\hline \mu =(\mu,\mu,\dots,\mu)\text{ as poly-argument} & \mu =(\mu,\mu,\dots,\mu)\\\hline \prod Z(s,j) =\prod_j Z_j(s)\\ =Z_{j_1}(s_1)\dots Z_{j_d}(s_d) &\\\hline & \displaystyle \zeta(s) = L(s)=L(s,1)\\ & \displaystyle =\sum_{n_1>\dots >n_d\ge 1}\frac 1{n_1^{s_1}\dots n_d^{s_d}}\\\hline & \displaystyle L(s,x)=\sum_{n_1>\dots >n_d\ge 1}\frac {x_1^{n_1}\dots x_d^{n_d}}{n_1^{s_1}\dots n_d^{s_d}}\\\hline \int_0^1\prod Z(s,j) & L(s, \mu^{\Delta j})\\\hline \end{array} $$

The computation of $J_1$. The following part $J_{11}$ of $J_1$ is simpler, because only $y$ and $1-y$ appear, so the symbolic part involve only $A,B$: $$ \begin{aligned} J_{11} &= \int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1-y}\; dy \\ &\overset{\text{up to }\pm} = \int_0^1\frac {dy}{1-y} \left(\int_y^1\frac{du}{u}\right) \left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right) \\ &= \int_0^1 B\int_0^t A\ ш\ AB \\ &= \int_0^1 B( \color{blue}{A}\ ш\ AB ) \\ &= \int_0^1 ( B\color{blue}{A}AB + BA\color{blue}{A}B + BAB\color{blue}{A} ) =2\int_0^1\color{brown}{B}AAB+\int_0^1 \color{brown}{B}ABA \\ &\qquad\text{ move $\color{brown}{B}$ from the first place behind}\\ &= -2\int_0^1 (A\color{brown}{B}AB + AA\color{brown}{B}B + AAB\color{brown}{B}) -\int_0^1 (A\color{brown}{B}BA + AB\color{brown}{B}A+ ABA\color{brown}{B}) \\ &= -3\int_0^1ABAB -4\int_0^1AABB -2\int_0^1ABB\color{blue}{A} \\ &\qquad\text{ move $\color{blue}{A}$ from the last place to the left}\\ \\ &=-3\int_0^1ABAB -4\int_0^1AABB +2\int_0^1(AB\color{blue}{A}B + A\color{blue}{A}BB + \color{blue}{A}ABB) \\ &=-\int_0^1ABAB=-\int_0^1Z(2)Z(2) \\ &\overset{\text{up to }\pm}=\int_0^1 Z(2,2)=\zeta(2,2) \\ &\qquad\text{ recall $\zeta(k)\zeta(n)=\zeta(k,n)+\zeta(n,k)+\zeta(n+k)$, use $k=n=2$} \\ &=\frac 12(\zeta(2)^2-\zeta(4)) = \frac 1{120}\pi^4\ . \end{aligned} $$ Numerical check, pari/gp:

The computation of the remained part $J_{12}$ from $J_1$ follows, we have algebraically a similar situation, but together with $y$ and $1-y$ there appears also $1+y$, so the symbolic part involve $A,B$ and also the $2$-chromatic piece $B'$: $$ \begin{aligned} J_{12} &= \int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1+y}\; dy \\ &\overset{\text{up to }\pm} = \int_0^1\frac {dy}{1+y} \left(\int_y^1\frac{du}{u}\right) \left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right) \\ &= \int_0^1 B'\int_0^t A\ ш\ AB \\ &= \int_0^1 B'( \color{blue}{A}\ ш\ AB ) \\ &= \int_0^1 ( B'\color{blue}{A}AB + B'A\color{blue}{A}B + B'AB\color{blue}{A} ) =2\int_0^1\color{brown}{B'}AAB+\int_0^1 \color{brown}{B'}ABA \\ &\qquad\text{ move $\color{brown}{B'}$ from the first place behind}\\ &= -2\int_0^1 (A\color{brown}{B'}AB + AA\color{brown}{B'}B + AAB\color{brown}{B'}) -\int_0^1 (A\color{brown}{B'}BA + AB\color{brown}{B'}A+ ABA\color{brown}{B'}) \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) -\int_0^1(AB'B+ABB')\color{blue}{A} -\int_0^1ABAB' \\ &\qquad\text{ move $\color{blue}{A}$ from the last place to the left} \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) +\int_0^1\color{blue}{A}A(B'B+BB') \\ &\qquad +\int_0^1A\color{blue}{A}(B'B+BB') +\int_0^1AB'\color{blue}{A}B +\int_0^1AB\color{blue}{A}B' \\ &=-\int_0^1AB'AB=-\int_0^1Z_1(2)Z_0(2) \\ &\overset{\text{up to }\pm}=\int_0^1 Z_{(1,0)}(2,2)=L_{(1,1)}(2,2) \\ &=\zeta(\bar 2,\bar 2) \\ &\qquad\text{ use $\zeta(\bar k)\zeta(\bar n)=\zeta(\bar k,\bar n)+\zeta(\bar n,\bar k)+\zeta(n+k)$, $n+k$ even, for $n=k=2$} \\ &\qquad\text{ recall $\zeta(\bar 2)=-\frac 1{1^2}+\frac 1{2^2}-\frac 1{3^2}+\frac 1{4^2}\pm \dots=-\left(1-\frac 2{2^2}\right)\zeta(2)=-\frac1{12}\pi^2$, } \\ &=\frac 12(\zeta(\bar 2)^2-\zeta(4)) = \frac 12\pi^4\left(\frac 1{12^2}-\frac 1{90}\right) = -\frac 1{480}\pi^4\ . \end{aligned} $$ Numerical check, pari/gp:

Later EDIT: The formula for $J_3$ can be obtained in a similar symbolic framework, and then checked in pari/gp, i will do this now in order to answer some comments. The fomula for $J_3$ is $$ J_3 = \int_0^1\log^2 y\;\log\frac{1-y}{1+y}\cdot \frac1{1-y^2}\; dy $$ so is a linear combination of integrals of $\frac 1{1\pm y}\log^2y\frac 1{1\pm y}$. Working symbolically, $\frac 1{2!}\log^2y$ corresponds to $A$, and $\frac 1{1-y}$, $-\frac 1{1+y}$ respectively to $B=B_0$, $B'=B_1$. We use the indices $j,k$ in the $B$-symbols to cover all four cases. We use $\mp_j$ for $(-1)^{j-1}$, so $\mp_0=\mp$. Then: $$ \begin{aligned} J_{3jk}&:=\int_0^1\frac1{1-\mp_j y}\log^2 y\;\log(1\mp_k y)\; dy \\ &\overset{\text{up to }\pm}= \int_0^1 B_j(2A^2\ ш\ B_k) =2\int_0^1B_j(B_kAA +AB_kA+AAB_k)\qquad\text{(right-$A^2$-moves)} \\ &=-\int_0^1(AB_jB_kA+B_jAB_kA)+2\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k \\ &=-\int_0^1AB_jB_kA+\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k\qquad\text{(right-$A$-moves)} \\ &=+\int_0^1(2A^2B_jB_kA + AB_jAB_k) - \int_0^1(AB_jAB_k + 2B_jAAB_k) +2\int_0^1B_jAAB_k \\ &=+2\int_0^1A^2B_jB_k=2\int_0^1Z_j(3)Z_k(1) \\ &= 2L_{3, 1}(\ (-1)^j\ ,\ (-1)^{j+k}\ )\ , \\ &\qquad\qquad\text{ and we use the values:} \\ &\qquad L_{3,1}(1,1) = \zeta(3,1)=\frac 14\zeta(4)=\frac 1{360}\pi^4\ ,\\ &\qquad L_{3,1}(1,-1) = \zeta(3,\bar 1)\ ,\\ &\qquad L_{3,1}(-1,1) = \zeta(\bar 3,1)\ ,\\ &\qquad L_{3,1}(-1,-1) = \zeta(\bar 3,\bar 1)\ ,\\ &\qquad\qquad\text{ and we have:}\\ \frac 12J_{300} &=-\zeta(3,1)=-\frac 1{360}\pi^4\ , \\ \frac 12 J_{310} &= +\zeta(\bar 3,\bar 1) = \frac 1{180}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) - \frac 1{12}\log^4 2\ , \\ -\frac 12J_{301} &=+\zeta(3,\bar 1)= \frac{19}{1440}\pi^4 -\frac 74\log 2\;\zeta(3) \ , \\ -\frac 12 J_{311} &= -\zeta(\bar 3,1) = \frac 1{48}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) -\frac 74\log2\;\zeta(3) - \frac 1{12}\log^4 2\ , \end{aligned} $$ and this gives finally the claimed value for $J_3$, $$ J_3=\frac 12(J_{300} +J_{310}-J_{301}-J_{311}) =\frac {53}{1440}\pi^4 +\frac 16\pi^2\log^2 2 - 4\operatorname{Li}_4\left(\frac 12\right) -\frac 72\log 2\;\zeta(3) - \frac 16\log^4 2 \ . $$

Note: The sum $\zeta(3,\bar 1)+\sum(\bar 3,\bar 1)$ can be also extracted from [Au2], Example 3.2, where $-\zeta(3,\bar 1)-\sum(\bar 3,\bar 1)$ is given (with no further reference). The value $\zeta(3,1)=\frac 14\zeta(4)$ can be obtained from $\zeta(2)^2=\int_0^1AB\cdot\int_0^1 AB=\int_0^1AB\ ш\ AB =2\int_0^1ABAB + 4\int_0^1AABB=2\int_0^1Z(2)Z(2)+4\int_0^1Z(3)Z(1)=2\zeta(2,2)+\zeta(3,1)$. The book [Z] has in Appendix, Euler sums for lower weights, page 537, also formulas relating these MZVs. All weight four alternating MZV are listed.

Later EDIT: The formula for $J_3$ can be obtained in a similar symbolic framework, and then checked in pari/gp, i will do this now in order to answer some comments. The fomula for $J_3$ is $$ J_3 = \int_0^1\log^2 y\;\log\frac{1-y}{1+y}}\cdot \frac1{1-y^2}\; dy $$ so is a linear combination of integrals of $\frac 1{1\pm y}\log^2y\frac 1{1\pm y}$. Working symbolically, $\frac 1{2!}\log^2y$ corresponds to $A$, and $\frac 1{1-y}$, $-\frac 1{1+y}$ respectively to $B=B_0$, $B'=B_1$. We use the indices $j,k$ in the $B$-symbols to cover all four cases. We use $\mp_j$ for $(-1)^{j-1}$, so $\mp_0=\mp$. Then: $$ \begin{aligned} J_{3jk}&:=\int_0^1\frac1{1-\mp_j y}\log^2 y\;\log(1\mp_k y)\; dy \\ &\overset{\text{up to }pm}= \int_0^1 B_j(2A^2\ ш\ B_k) =2\int_0^1B_j(B_kAA +AB_kA+AAB_k)\qquad\text{(right-$A^2$-moves)} \\ &=-\int_0^1(AB_jB_kA+B_jAB_kA)+2\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k \\ &=-\int_0^1AB_jB_kA+\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k\qquad\text{(right-$A$-moves)} \\ &=+\int_0^1(2A^2B_jB_kA + AB_jAB_k) - \int_0^1(AB_jAB_k + 2B_jAAB_k) +2\int_0^1B_jAAB_k \\ &=+2\int_0^1A^2B_jB_k=2\int_0^1Z_j(3)Z_k(1) \\ &= 2L_{3, 1}(\ (-1)^j\ ,\ (-1)^{j+k}\ )\ , \\ &\qquad\qquad\text{ and we use the values:} \\ &\qquad L_{3,1}(1,1) = \zeta(3,1)=\frac 14\zeta(4)=\frac 1{360}\pi^4\ ,\\ &\qquad L_{3,1}(1,-1) = \zeta(3,\bar 1)\ ,\\ &\qquad L_{3,1}(-1,1) = \zeta(\bar 3,1)\ ,\\ &\qquad L_{3,1}(-1,-1) = \zeta(-3,-1)\ ,\\ &\qquad\qquad\text{ and we have:}\\ \frac 12J_{300} &=-\zeta(3,1)=-\frac 1{360}\pi^4\ , \\ \frac 12 J_{310} &= -\zeta(\bar 3,\bar 1) = \frac 1{180}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) - \frac 1{12}\log^4 2\ , \\ % ==================================================================================================== -\frac 12J_{301} &=\zeta(3,\bar 1)= \frac{19}{1440}\pi^4 -\frac 74\log 2\;\zeta(3) \ , \\ -\frac 12 J_{311} &= \zeta(\bar 3,1) = \frac 1{48}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) -\frac 74\log2\;\zeta(3) - \frac 1{12}\log^4 2\ , \end{aligned} $$ and this gives finally the claimed value for $J_3$, $$ J_3=\frac 12(J_{300} +J_{310}-J_{301}-J_{311}) =\frac 53{1440}\pi^4 +\frac 16\pi^2\log^2 2 - 4\operatorname{Li}_4\left(\frac 12\right) -\frac 72\log 2\;\zeta(3) - \frac 16\log^4 2 \ . $$

Later EDIT: The formula for $J_3$ can be obtained in a similar symbolic framework, and then checked in pari/gp, i will do this now in order to answer some comments. The fomula for $J_3$ is $$ J_3 = \int_0^1\log^2 y\;\log\frac{1-y}{1+y}\cdot \frac1{1-y^2}\; dy $$ so is a linear combination of integrals of $\frac 1{1\pm y}\log^2y\frac 1{1\pm y}$. Working symbolically, $\frac 1{2!}\log^2y$ corresponds to $A$, and $\frac 1{1-y}$, $-\frac 1{1+y}$ respectively to $B=B_0$, $B'=B_1$. We use the indices $j,k$ in the $B$-symbols to cover all four cases. We use $\mp_j$ for $(-1)^{j-1}$, so $\mp_0=\mp$. Then: $$ \begin{aligned} J_{3jk}&:=\int_0^1\frac1{1-\mp_j y}\log^2 y\;\log(1\mp_k y)\; dy \\ &\overset{\text{up to }pm}= \int_0^1 B_j(2A^2\ ш\ B_k) =2\int_0^1B_j(B_kAA +AB_kA+AAB_k)\qquad\text{(right-$A^2$-moves)} \\ &=-\int_0^1(AB_jB_kA+B_jAB_kA)+2\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k \\ &=-\int_0^1AB_jB_kA+\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k\qquad\text{(right-$A$-moves)} \\ &=+\int_0^1(2A^2B_jB_kA + AB_jAB_k) - \int_0^1(AB_jAB_k + 2B_jAAB_k) +2\int_0^1B_jAAB_k \\ &=+2\int_0^1A^2B_jB_k=2\int_0^1Z_j(3)Z_k(1) \\ &= 2L_{3, 1}(\ (-1)^j\ ,\ (-1)^{j+k}\ )\ , \\ &\qquad\qquad\text{ and we use the values:} \\ &\qquad L_{3,1}(1,1) = \zeta(3,1)=\frac 14\zeta(4)=\frac 1{360}\pi^4\ ,\\ &\qquad L_{3,1}(1,-1) = \zeta(3,\bar 1)\ ,\\ &\qquad L_{3,1}(-1,1) = \zeta(\bar 3,1)\ ,\\ &\qquad L_{3,1}(-1,-1) = \zeta(-3,-1)\ ,\\ &\qquad\qquad\text{ and we have:}\\ \frac 12J_{300} &=-\zeta(3,1)=-\frac 1{360}\pi^4\ , \\ \frac 12 J_{310} &= -\zeta(\bar 3,\bar 1) = \frac 1{180}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) - \frac 1{12}\log^4 2\ , \\ -\frac 12J_{301} &=\zeta(3,\bar 1)= \frac{19}{1440}\pi^4 -\frac 74\log 2\;\zeta(3) \ , \\ -\frac 12 J_{311} &= \zeta(\bar 3,1) = \frac 1{48}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) -\frac 74\log2\;\zeta(3) - \frac 1{12}\log^4 2\ , \end{aligned} $$ and this gives finally the claimed value for $J_3$, $$ J_3=\frac 12(J_{300} +J_{310}-J_{301}-J_{311})\\ =\frac {53}{1440}\pi^4 +\frac 16\pi^2\log^2 2 - 4\operatorname{Li}_4\left(\frac 12\right) -\frac 72\log 2\;\zeta(3) - \frac 16\log^4 2 \ . $$

Then we have from $(\dagger)$ a linear dependence between the real part of the following integrals $J, J_1,J_2,J_3$: $$ \begin{aligned} J &= \int_0^1\log y\; \operatorname{Li}_2(-y)\;\frac1{1-y^2}\; dy\ ,\text{ the integral of interest,} \\ J_1 &= \int_0^1\log y\; \color{blue}{\operatorname{Li}_2(y)}\;\frac1{1-y^2}\; dy =\bbox[yellow]{\ -\frac 1{192}\pi^4\ }\ ,\text{ a clean value to be shown below,} \\[3mm] J_2 &= \Re \int_0^1\log y\; \color{brown}{\Big( \operatorname{Li}_2(1+y) - \operatorname{Li}_2(1-y) \Big)} \;\frac1{1-y^2}\; dy \\ &= \frac 12 \Re\int_0^1\log y\; \left( \frac{\operatorname{Li}_2(1+y)}{1+y} + \frac{\operatorname{Li}_2(1+y)}{1-y} - \frac{\operatorname{Li}_2(1-y)}{1+y} - \frac{\operatorname{Li}_2(1-y)}{1-y} \right)\; dy \\ &\equiv -\frac 12 \int_0^1 \Big(\Re\operatorname{Li}_2(1+y)\Big)'\Re\operatorname{Li}_2(1+y)\; dy - \frac 12 \int_0^1 \Big(\operatorname{Li}_2(1-y)\Big)'\operatorname{Li}_2(1-y)\; dy \\ &\qquad\qquad + \frac 12 \int_0^1 \underbrace{\frac{\log y}{1-y}} _{=(\ +\operatorname{Li}_2(1-y)\ )'} \Re\operatorname{Li}_2(1+y)\; dy - \frac 12 \underbrace{\int_0^1\frac{\log y}{1+y}} _{\sim (\ -\Re\operatorname{Li}_2(1+y)\ )'} \operatorname{Li}_2(1-y)\; dy \\ &= -\frac 14\Bigg[\ (\Re\operatorname{Li}_2(1+y))^2\ \Bigg]_0^1 -\frac 14\Bigg[\ \operatorname{Li}_2(1-y)^2\ \Bigg]_0^1 \\ &\qquad\qquad +\frac 12\Bigg[\ \operatorname{Li}_2(1-y)\Re\operatorname{Li}_2(1+y)\ \Bigg]_0^1 \\ &= -\frac 14\Big[\ (\Re \operatorname{Li}_2(2))^2 - \operatorname{Li}_2(1)^2\ \Big] -\frac 14\Big[\ 0-\operatorname{Li}_2(1)^2 \ \Big] +\frac 12\Big[ \ 0-\operatorname{Li}_2(1))^2 \ \Big] \\ &=-\frac 14\cdot (\Re \operatorname{Li}_2(2))^2 =\bbox[yellow]{\ -\frac {\pi^4}{64}\ }\ , \\[3mm] J_3 &=\int_0^1\log y\cdot \color{green}{\log y\log\frac{1-y}{1+y}}\cdot \frac1{1-y^2}\; dy \\ &=\frac 12\Bbb J_{120-021,100+001}\qquad\text{(with a linear split w.r.t. the symbolic indices)} \\ &= \frac{53}{1440}\pi^4 +\frac 16\pi^2\log^2 2 -4\operatorname{Li}_2\left(\frac 12\right) -\frac 72\log(2)\,\zeta(3) -\frac 16\log^42\ . \end{aligned} $$ Then we obtain a formula for $J$, from $J=J_1-J_2+J_3$, $$ \bbox[lightyellow]{\qquad J= \frac{17}{360}\pi^4 +\frac 16\pi^2\log^2 2 -4\operatorname{Li}_2\left(\frac 12\right) -\frac 72\log(2)\,\zeta(3) -\frac 16\log^42\ . \qquad} $$

The computation of the remained part $J_{12}$ from $J_1$ follows, we have algebraically a similar situation, but together with $y$ and $1-y$ there appears also $1+y$, so the symbolic part involve $A,B$ and also the $2$-chromatic piece $B'$: $$ \begin{aligned} J_{12} &= \int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1+y}\; dy \\ &\overset{\text{up to }\pm} = \int_0^1\frac {dy}{1+y} \left(\int_y^1\frac{du}{u}\right) \left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right) \\ &= \int_0^1 B'\int_0^t A\ ш\ AB \\ &= \int_0^1 B'( \color{blue}{A}\ ш\ AB ) \\ &= \int_0^1 ( B'\color{blue}{A}AB + B'A\color{blue}{A}B + B'AB\color{blue}{A} ) =2\int_0^1\color{brown}{B'}AAB+\int_0^1 \color{brown}{B'}ABA \\ &\qquad\text{ move $\color{brown}{B'}$ from the first place behind}\\ &= -2\int_0^1 (A\color{brown}{B'}AB + AA\color{brown}{B'}B + AAB\color{brown}{B'}) -\int_0^1 (A\color{brown}{B'}BA + AB\color{brown}{B'}A+ ABA\color{brown}{B'}) \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) -\int_0^1(AB'B+ABB')\color{blue}{A} -\int_0^1ABAB' \\ &\qquad\text{ move $\color{blue}{A}$ from the last place to the left} \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) +\int_0^1\color{blue}{A}A(B'B+BB') \\ &\qquad +\int_0^1A\color{blue}{A}(B'B+BB') +\int_0^1AB'\color{blue}{A}B +\int_0^1AB\color{blue}{A}B' \\ &=-\int_0^1AB'AB=-\int_0^1Z_1(2)Z_0(2) \\ &\overset{\text{up to }\pm}=\int_0^1 Z_{(1,0)}(2,2)=L_{(1,1)}(2,2) \\ &=\zeta(\bar 2,\bar 2) \\ &\qquad\text{ recall $\zeta(\bar k)\zeta(\bar n)=\zeta(\bar k,\bar n)+\zeta(\bar n,\bar k)+\zeta(n+k)$, use $k=n=2$,} \\ &\qquad\text{ recall $\zeta(\bar 2)=-\frac 1{1^2}+\frac 1{2^2}-\frac 1{3^2}+\frac 1{4^2}\pm \dots=-\left(1-\frac 2{2^2}\right)\zeta(2)=-\frac1{12}\pi^2$, } \\ &=\frac 12(\zeta(\bar 2)^2-\zeta(4)) = \frac 12\pi^4\left(\frac 1{12^2}-\frac 1{90}\right) = -\frac 1{480}\pi^4\ . \end{aligned} $$ Numerical check, pari/gp:

Then we have from $(\dagger)$ a linear dependence between the real part of the following integrals $J, J_1,J_2,J_3$: $$ \begin{aligned} J &= \int_0^1\log y\; \operatorname{Li}_2(-y)\;\frac1{1-y^2}\; dy\ ,\text{ the integral of interest,} \\ J_1 &= \int_0^1\log y\; \color{blue}{\operatorname{Li}_2(y)}\;\frac1{1-y^2}\; dy =\bbox[yellow]{\ -\frac 1{192}\pi^4\ }\ ,\text{ a clean value to be shown below,} \\[3mm] J_2 &= \Re \int_0^1\log y\; \color{brown}{\Big( \operatorname{Li}_2(1+y) - \operatorname{Li}_2(1-y) \Big)} \;\frac1{1-y^2}\; dy \\ &= \frac 12 \Re\int_0^1\log y\; \left( \frac{\operatorname{Li}_2(1+y)}{1+y} + \frac{\operatorname{Li}_2(1+y)}{1-y} - \frac{\operatorname{Li}_2(1-y)}{1+y} - \frac{\operatorname{Li}_2(1-y)}{1-y} \right)\; dy \\ &\equiv -\frac 12 \int_0^1 \Big(\Re\operatorname{Li}_2(1+y)\Big)'\Re\operatorname{Li}_2(1+y)\; dy - \frac 12 \int_0^1 \Big(\operatorname{Li}_2(1-y)\Big)'\operatorname{Li}_2(1-y)\; dy \\ &\qquad\qquad + \frac 12 \int_0^1 \underbrace{\frac{\log y}{1-y}} _{=(\ +\operatorname{Li}_2(1-y)\ )'} \Re\operatorname{Li}_2(1+y)\; dy - \frac 12 \underbrace{\int_0^1\frac{\log y}{1+y}} _{\sim (\ -\Re\operatorname{Li}_2(1+y)\ )'} \operatorname{Li}_2(1-y)\; dy \\ &= -\frac 14\Bigg[\ (\Re\operatorname{Li}_2(1+y))^2\ \Bigg]_0^1 -\frac 14\Bigg[\ \operatorname{Li}_2(1-y)^2\ \Bigg]_0^1 \\ &\qquad\qquad +\frac 12\Bigg[\ \operatorname{Li}_2(1-y)\Re\operatorname{Li}_2(1+y)\ \Bigg]_0^1 \\ &= -\frac 14\Big[\ (\Re \operatorname{Li}_2(2))^2 - \operatorname{Li}_2(1)^2\ \Big] -\frac 14\Big[\ 0-\operatorname{Li}_2(1)^2 \ \Big] +\frac 12\Big[ \ 0-\operatorname{Li}_2(1))^2 \ \Big] \\ &=-\frac 14\cdot (\Re \operatorname{Li}_2(2))^2 =\bbox[yellow]{\ -\frac {\pi^4}{64}\ }\ , \\[3mm] J_3 &=\int_0^1\log y\cdot \color{green}{\log y\log\frac{1-y}{1+y}}\cdot \frac1{1-y^2}\; dy \\ &=\frac 12\Bbb J_{120-021,100+001}\qquad\text{(with a linear split w.r.t. the symbolic indices)} \\ &= \frac{53}{1440}\pi^4 +\frac 16\pi^2\log^2 2 -4\operatorname{Li}_{\color{red}4}\left(\frac 12\right) -\frac 72\log(2)\,\zeta(3) -\frac 16\log^42\ . \end{aligned} $$ Then we obtain a formula for $J$, from $J=J_1-J_2+J_3$, $$ \bbox[lightyellow]{\qquad J= \frac{17}{360}\pi^4 +\frac 16\pi^2\log^2 2 -4\operatorname{Li}_{\color{red}4}\left(\frac 12\right) -\frac 72\log(2)\,\zeta(3) -\frac 16\log^42\ . \qquad} $$

The computation of the remained part $J_{12}$ from $J_1$ follows, we have algebraically a similar situation, but together with $y$ and $1-y$ there appears also $1+y$, so the symbolic part involve $A,B$ and also the $2$-chromatic piece $B'$: $$ \begin{aligned} J_{12} &= \int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1+y}\; dy \\ &\overset{\text{up to }\pm} = \int_0^1\frac {dy}{1+y} \left(\int_y^1\frac{du}{u}\right) \left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right) \\ &= \int_0^1 B'\int_0^t A\ ш\ AB \\ &= \int_0^1 B'( \color{blue}{A}\ ш\ AB ) \\ &= \int_0^1 ( B'\color{blue}{A}AB + B'A\color{blue}{A}B + B'AB\color{blue}{A} ) =2\int_0^1\color{brown}{B'}AAB+\int_0^1 \color{brown}{B'}ABA \\ &\qquad\text{ move $\color{brown}{B'}$ from the first place behind}\\ &= -2\int_0^1 (A\color{brown}{B'}AB + AA\color{brown}{B'}B + AAB\color{brown}{B'}) -\int_0^1 (A\color{brown}{B'}BA + AB\color{brown}{B'}A+ ABA\color{brown}{B'}) \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) -\int_0^1(AB'B+ABB')\color{blue}{A} -\int_0^1ABAB' \\ &\qquad\text{ move $\color{blue}{A}$ from the last place to the left} \\ &= -2\int AB'AB-2\int_0^1AA(BB'+B'B) +\int_0^1\color{blue}{A}A(B'B+BB') \\ &\qquad +\int_0^1A\color{blue}{A}(B'B+BB') +\int_0^1AB'\color{blue}{A}B +\int_0^1AB\color{blue}{A}B' \\ &=-\int_0^1AB'AB=-\int_0^1Z_1(2)Z_0(2) \\ &\overset{\text{up to }\pm}=\int_0^1 Z_{(1,0)}(2,2)=L_{(1,1)}(2,2) \\ &=\zeta(\bar 2,\bar 2) \\ &\qquad\text{ (use $\zeta(\bar k)\zeta(\bar n)=\zeta(\bar k,\bar n)+\zeta(\bar n,\bar k)+\zeta(n+k)$, $n+k$ even)} \\ &\qquad\text{ recall $\zeta(\bar 2)=-\frac 1{1^2}+\frac 1{2^2}-\frac 1{3^2}+\frac 1{4^2}\pm \dots=-\left(1-\frac 2{2^2}\right)\zeta(2)=-\frac1{12}\pi^2$, } \\ &=\frac 12(\zeta(\bar 2)^2-\zeta(4)) = \frac 12\pi^4\left(\frac 1{12^2}-\frac 1{90}\right) = -\frac 1{480}\pi^4\ . \end{aligned} $$ Numerical check, pari/gp:

Later EDIT: The formula for $J_3$ can be obtained in a similar symbolic framework, and then checked in pari/gp, i will do this now in order to answer some comments. The fomula for $J_3$ is $$ J_3 = \int_0^1\log^2 y\;\log\frac{1-y}{1+y}}\cdot \frac1{1-y^2}\; dy $$ so is a linear combination of integrals of $\frac 1{1\pm y}\log^2y\frac 1{1\pm y}$. Working symbolically, $\frac 1{2!}\log^2y$ corresponds to $A$, and $\frac 1{1-y}$, $-\frac 1{1+y}$ respectively to $B=B_0$, $B'=B_1$. We use the indices $j,k$ in the $B$-symbols to cover all four cases. We use $\mp_j$ for $(-1)^{j-1}$, so $\mp_0=\mp$. Then: $$ \begin{aligned} J_{3jk}&:=\int_0^1\frac1{1-\mp_j y}\log^2 y\;\log(1\mp_k y)\; dy \\ &\overset{\text{up to }pm}= \int_0^1 B_j(2A^2\ ш\ B_k) =2\int_0^1B_j(B_kAA +AB_kA+AAB_k)\qquad\text{(right-$A^2$-moves)} \\ &=-\int_0^1(AB_jB_kA+B_jAB_kA)+2\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k \\ &=-\int_0^1AB_jB_kA+\int_0^1B_jAB_kA+2\int_0^1B_jAAB_k\qquad\text{(right-$A$-moves)} \\ &=+\int_0^1(2A^2B_jB_kA + AB_jAB_k) - \int_0^1(AB_jAB_k + 2B_jAAB_k) +2\int_0^1B_jAAB_k \\ &=+2\int_0^1A^2B_jB_k=2\int_0^1Z_j(3)Z_k(1) \\ &= 2L_{3, 1}(\ (-1)^j\ ,\ (-1)^{j+k}\ )\ , \\ &\qquad\qquad\text{ and we use the values:} \\ &\qquad L_{3,1}(1,1) = \zeta(3,1)=\frac 14\zeta(4)=\frac 1{360}\pi^4\ ,\\ &\qquad L_{3,1}(1,-1) = \zeta(3,\bar 1)\ ,\\ &\qquad L_{3,1}(-1,1) = \zeta(\bar 3,1)\ ,\\ &\qquad L_{3,1}(-1,-1) = \zeta(-3,-1)\ ,\\ &\qquad\qquad\text{ and we have:}\\ \frac 12J_{300} &=-\zeta(3,1)=-\frac 1{360}\pi^4\ , \\ \frac 12 J_{310} &= -\zeta(\bar 3,\bar 1) = \frac 1{180}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) - \frac 1{12}\log^4 2\ , \\ % ==================================================================================================== -\frac 12J_{301} &=\zeta(3,\bar 1)= \frac{19}{1440}\pi^4 -\frac 74\log 2\;\zeta(3) \ , \\ -\frac 12 J_{311} &= \zeta(\bar 3,1) = \frac 1{48}\pi^4 +\frac 1{12}\pi^2\log^2 2 - 2\operatorname{Li}_4\left(\frac 12\right) -\frac 74\log2\;\zeta(3) - \frac 1{12}\log^4 2\ , \end{aligned} $$ and this gives finally the claimed value for $J_3$, $$ J_3=\frac 12(J_{300} +J_{310}-J_{301}-J_{311}) =\frac 53{1440}\pi^4 +\frac 16\pi^2\log^2 2 - 4\operatorname{Li}_4\left(\frac 12\right) -\frac 72\log 2\;\zeta(3) - \frac 16\log^4 2 \ . $$