A tricky integral to evaluate I came across this integral in some work. So, I would like to ask:

QUESTION. Can you evaluate this integral with proofs?
$$\int_0^1\frac{\log x\cdot\log(x+2)}{x+1}\,dx.$$

 A: $$\int_0^1\frac{\ln x\cdot\ln(x+2)}{x+1}\,dx=$$
$$=\text{Li}_3\left(-\tfrac{1}{3}\right)-2 \,\text{Li}_3\left(\tfrac{1}{3}\right)+\tfrac{1}{2} \ln 3\left[ \text{Li}_2\left(\tfrac{1}{9}\right)-6\, \text{Li}_2\left(\tfrac{1}{3}\right)  -\tfrac{2}{3} \ln ^2 3\right]+\tfrac{13}{8} \zeta (3).$$
I checked that this combination of polylog's evaluates to $-0.651114$, equal to a numerical evaluation of the integral.
Update:
As Timothy Budd pointed out, that this combination of polylog's simplifies to $-\frac{13}{24}\zeta(3)$ is proven by Przemo at MSE.
The identities that enable this simplification are
$$\text{Li}_3\left(-\tfrac{1}{3}\right)-2 \,\text{Li}_3\left(\tfrac{1}{3}\right) = -\tfrac{1}{6} \ln^3 3 + \tfrac{1}{6}\pi^2 \ln 3 - \tfrac{13}{6} \zeta(3),$$
$$\text{Li}_2(\tfrac{1}{9})=2\,\text{Li}_2(-\tfrac{1}{3})+2\,\text{Li}_2(\tfrac{1}{3}),$$
$$2\text{Li}_2\left(-\tfrac{1}{3}\right)-4 \,\text{Li}_2\left(\tfrac{1}{3}\right) =
\ln^2 3 -\tfrac{1}{3}\pi^2 .$$
A: For $t\in(0,1]$, let
\begin{equation}
    I(t):=\int_0^1\frac{\log(x)\,\log(1+t(x+1))}{x+1}\,dx 
\end{equation}
(so that the integral in question is $I(1)$),
\begin{equation}
\begin{aligned}
    J(t)&:=-\text{Li}_3(2+1/t)+\text{Li}_3(-2 t-1)+\text{Li}_3(t+1) \\ 
    &-\text{Li}_3(2t+1) 
   +\text{Li}_2(2+1/t) (\log (2 t+1)+i \pi ) \\ 
   &+\text{Li}_2(-2 t-1)(-\log (2 t+1)-i \pi ) \\ 
   &+\text{Li}_2(t+1) (-\log (t+1)-i \pi )+\text{Li}_2(2 t+1) (\log (2
   t+1)+i \pi ) \\ 
   &+\frac{1}{6} (\log ^3(t)+(-3 \log ^2(t+1)-6 i \pi  \log (t+1)+4 \pi
   ^2) \log (t) \\ 
   &+3 \pi  (-i \log ^2(t+1)+2 i \log ^2(2 t+1)+2 \pi  \log (t+1)-4
   \pi  \log (2 t+1))) \\ 
   &+\frac{3 \zeta (3)}{4}-\frac{5 i \pi ^3}{12}, 
\end{aligned}   
\end{equation}
\begin{equation}
\begin{aligned}
    I_1(t)&:=6tI'(t)=6t\int_0^1\frac{\log(x)}{1+t(x+1)}\,dx \\ 
&   =6 \text{Li}_2\left(\frac{t+1}{2 t+1}\right)-3 \log ^2(t+1)+3 \log ^2(2 t+1) \\ 
&+6 \log (t) \log
   \left(\frac{t+1}{2 t+1}\right)-\pi ^2,
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
    J_1(t)&:=6tJ'(t)=6 \text{Li}_2\left(2+1/t\right)+3 \log ^2(t)-3 \log ^2(t+1) \\ 
    &-6 (\log (2 t+1)+i \pi )
   \log (t)+6 \log (t+1) \log (2 t+1) \\ 
   &+6 i \pi  \log (2 t+1)-2 \pi ^2. 
\end{aligned}
\end{equation}
Then $I'_1=J'_1$ and $I_1(0+)=J_1(0+)$, so that $I_1=J_1$, and hence $I'=J'$. Also, $I(0)=I(0+)=0=J(0+)$, so that $I=J$, and the integral in question is
\begin{equation}
\begin{aligned}
    I(1)&=J(1)=\text{Li}_3(-3)-2 \text{Li}_3(3)+i \pi  \left(-\text{Li}_2(-3)+2 \text{Li}_2(3)+\log
   ^2(3)\right) \\ 
   &-\text{Li}_2(-3) \log (3)+\text{Li}_2(3) \log (9)+\frac{13 \zeta
   (3)}{8}-\frac{2 i \pi ^3}{3}-\pi ^2 \log (9) \\
   &=-0.651114\dots.  
\end{aligned}   
\end{equation}
