Is it possible to evaluate $-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx$? Few minutes ago I was playing with Wolfram Alpha about integrals like this 
$$-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx.$$
Previous integral is convergent, and Wolfram Alpha online calculator provide me a closed-form for the indefinite integral 
int -log(1-cosh(x))x^2e^(-x)dx
Notice that my idea was to choose the integration limits from $0$ to $\infty$ with the purpose to get $\zeta(3)$ as a summand in the output. But I don't know how to evaluate all terms, specially those involving the polylogarithm (I know that the other terms are tedious, and one could to calculate those with a CAS).

Question. Is it possible to get a closed-form of $$-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx.$$
  in terms of particular values of of well-known special functions? Many thanks.

If you prefer different calculations of previous explanation (get a closed-form and evaluate the integration limits as did Wolfram Alpha), feel free to tell us your approach.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&-\int_{0}^{\infty}\ln\pars{1 - \cosh\pars{x}}\,{x^{2} \over \expo{x}}\,\dd x =
-\int_{0}^{\infty}\bracks{\ln\pars{\cosh\pars{x} - 1} + \ic\pi}\,
x^{2}\expo{-x}\,\dd x
\\[5mm] = &\
-2\pi\ic - \int_{0}^{\infty}\ln\pars{{\expo{x} + \expo{-x} \over 2} - 1}
\,x^{2}\expo{-x}\,\dd x
\\[5mm] \stackrel{\substack{x\ =\ -\ln\pars{t}\\[0.25mm] t\ =\ \expo{-x}}\\ }{=}\,\,\,&
-2\pi\ic - \int_{1}^{0}\ln\pars{{1/t + t \over 2} - 1}\ln^{2}\pars{t}\,
t\,\,{\phantom{-}\dd t \over -t}
\\[5mm] = &\
-2\pi\ic - \int_{0}^{1}\ln\pars{\bracks{1 - t}^{2} \over 2t}\ln^{2}\pars{t}\,
\dd t
\\[5mm] = &\
-2\pi\ic - 2\
\underbrace{\int_{0}^{1}\ln^{2}\pars{t}\ln\pars{1 - t}\,\dd t}
_{\ds{-6 + {\pi^{2} \over 3} + 2\,\zeta\pars{3}}}\ +\
\ln\pars{2}\ \underbrace{\int_{0}^{1}\ln^{2}\pars{t}\,\dd t}_{\ds{2}}\ +\
\underbrace{\int_{0}^{1}\ln^{3}\pars{t}\,\dd t}_{\ds{-6}}\label{1}\tag{1}
\\[5mm] = &\
\bbx{6 + 2\ln\pars{2} - {2\pi^{2} \over 3} - 4\zeta\pars{3} - 2\pi\ic}
\approx -4.0015 + 6.2832\,\ic
\end{align}

Note that the integrals in \eqref{1} are evaluated as follows:

$$
\left\{\begin{array}{rcl}
\ds{\int_{0}^{1}\ln^{2}\pars{t}\ln\pars{1 - t}\,\dd t} & \ds{=} &
\ds{\left.\partiald[2]{}{\mu}\partiald{}{\nu}
{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over
\Gamma\pars{\mu + \nu + 2}}\right\vert_{\ \mu =\ \nu\ =\ 0}}
\\[5mm]
\ds{\int_{0}^{1}\ln^{k}\pars{t}\,\dd t} & \ds{=} &
\ds{\left.\partiald[k]{}{\mu}\int_{0}^{1}t^{\mu}\,\dd t\,
\right\vert_{\ \mu\ =\ 0} = \pars{-1}^{k}\, k!}
\end{array}\right.
$$
