<s>[INST] An integral involving $\ln(3)$ 
$$\int_{-\infty}^{\infty}{x^2\over 1+3\cosh^2\left({x\over \alpha}\right)}\mathrm dx={\alpha^3\over 4!}[\pi^2\ln(3)+\ln^3(3)]\tag1$$
  $\alpha\ge 1$
How to show that $(1)={\alpha^3\over 4!}[\pi^2\ln(3)+\ln^3(3)]?$

$u=3\cosh^2({x\over \alpha})$ then $\mathrm du ={3\over \alpha}\sinh\left({2x\over \alpha}\right)\mathrm dx$
$x=\alpha\cosh^{-1}\left(\sqrt{u\over 3}\right)$
$${\alpha\over 3}\int{x^2\over 1+u}{1\over \sinh({2x\over \alpha})}\mathrm du\tag2$$
$${\alpha^3\over 3}\int{\left(\cosh^{-1}\left(\sqrt{u\over 3}\right)\right)^2\over 1+u}{\mathrm du\over \sinh\left(2\cosh^{-1}\left(\sqrt{u\over 3}\right)\right)}\tag3$$
over complicate...
 [/INST] The $\alpha$ parameter is irrelevant, it can be removed through a substitution. The problem boils down to computing
$$ \int_{0}^{+\infty}\frac{x^2}{1+3\cosh^2 x}\,dx = \int_{1}^{+\infty}\frac{u\log^2(u)}{u^2+\frac{3}{4}(1+u^2)^2}\,du=\int_{0}^{1}\frac{u\log^2(u)}{u^2+\frac{3}{4}(1+u^2)^2}\,du$$
which can be done through partial fraction decomposition. We have
$$ \frac{4u}{u^2+\frac{3}{4}(1+u^2)^2} =t\frac{1}{u+\frac{i}{\sqrt{3}}}+\frac{1}{u-\frac{i}{\sqrt{3}}}-\frac{1}{u-i\sqrt{3}}-\frac{1}{u+i\sqrt{3}}$$
and
$$ \int_{0}^{1}\frac{u\log^2(u)}{u^2+a^2}\,du = -\frac{1}{4}\,\text{Li}_3\left(-\frac{1}{a^2}\right),$$
so the claim follows from the functional identities for the trilogarithm.
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