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Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that $$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2 \frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$ Integrate term-wise to get (after some regrouping) $$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$ Perhaps there is some hope in view of what I see as $$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$ which is a Wallis-type formula $$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$

UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate $$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$ Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and $$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$ Next, put all these together and simplify \begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k &=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\ &=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}. \end{align}

Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that $$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2 \frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$ Integrate term-wise to get (after some regrouping) $$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$ Perhaps there is some hope in view of what I see as $$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$ which is a Wallis-type formula $$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$

UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate $$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$ Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and $$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$ Next, put all these together and simplify \begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k &=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\ &=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}. \end{align}

I do not know if this helps but $$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$ It's not clear what the closed form would be for this value. Perhaps there is some hope in view of what I see as $$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$ which is a Wallis-type formula $$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$

UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate $$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$ Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and $$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$ Next, put all these together and simplify \begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k &=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\ &=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}. \end{align}

Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that $$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2 \frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$ Integrate term-wise to get (after some regrouping) $$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$ Perhaps there is some hope in view of what I see as $$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$ which is a Wallis-type formula $$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$

UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate $$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$ Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and $$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$ Next, put all these together and simplify \begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k &=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\ &=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}. \end{align}

I do not know if this helps but $$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$ It's not clear what the closed form would be for this value. Perhaps there is some hope in view of what I see as $$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$ which is a Wallis-type formula $$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$

UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate $$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$ Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and $$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$ Next, put all these together and simplify \begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k &=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\ &=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}, \end{align} which is in agreement with Igor's answer.

I do not know if this helps but $$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$ It's not clear what the closed form would be for this value. Perhaps there is some hope in view of what I see as $$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$ which is a Wallis-type formula $$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$

UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate $$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$ Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and $$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$ Next, put all these together and simplify \begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k &=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\ &=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}. \end{align}