Analytic guess for integral $\int_0^\infty \frac{\ln^2(\sqrt{x^2+1}+x) }{\sqrt{x^2 + 1}} \ln \frac{\cosh w+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } dx$ (I was directed here from mathoverflow.)
I am not able to do the following integral analytically, but after numerical evaluations was able to guess the result for all real $w$ values. Not only is the full $w$-dependence clear numerically, the actual coefficients are also correct to very high precision.
$$\int_0^\infty \frac{1}{\sqrt{x^2 + 1}} \log^2(\sqrt{x^2+1}+x) \log\left(\frac{\cosh(w)+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } \right) dx = \frac{1}{12} \left( 2\pi^2 w^2 + w^4 \right)$$
The fact that the result is a simple low order polynomial in $w$ is surprising enough to me.
I tried all kinds of tricks (new variables, partial integrations, etc.) to prove the above but was not able to. How would I go about proving it analytically?

Substitute $x=\sinh t$ to reduce the integral
\begin{align}
I(w)=&\int_0^\infty \frac{\ln^2(\sqrt{x^2+1}+x) }{\sqrt{x^2 + 1}} \ln \frac{\cosh w+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } dx\\
= &\int_0^\infty t^2 \ln \frac{\cosh w+\cosh t}{1+\cosh t}dt
\end{align}
and evaluate
\begin{align}
I’(w) =& \int_0^\infty \frac{t^2 \sinh w}{\cosh w+\cosh t}\overset{y=e^{-t}}{dt}\\
=& \ \sinh w\int_0^\infty \frac{\ln^2 y}{y^2 +2y \cosh w+1}dy
= \frac w3(\pi^2+w^2)
\end{align}
Then
$$I(w) =\int_0^w I’(a)da =\frac13 \int_0^w a(\pi^2+a^2)da
=\frac{1}{12} \left( 2\pi^2 w^2 + w^4 \right)$$
