I will follow @user15302's idea. In this answer, I showed that
$$ \int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv, $$
where $b = \frac{a-1}{a+1}$. Now let $I$ denote the Vladimir's integral and set $s = \frac{1}{4}$ and $a = 7$. Then we have $b = \frac{3}{4}$ and
$$ I = 2^{-3/4} \int_{0}^{1} \frac{1}{v^{3/4}\sqrt{(1-v)(1-\frac{3}{4}v)}} \, dv. $$
The reason why the case $b = \frac{3}{4}$ is special is that, if we plug $v = \operatorname{sech}^2 t$ then we can utilize the triple angle formula to get the following surprisingly neat integral
$$ I = 2^{5/4} \int_{0}^{\infty} \frac{\cosh t}{\sqrt{\cosh 3t}} \, dt. $$
Now using the substitution $x = e^{-6t}$, we easily find that
$$ I = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{1} \frac{x^{-11/12} + u^{-7/12}}{\sqrt{1+x}} \, dx = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{\infty} \frac{dx}{x^{11/12}\sqrt{1+x}}. $$
The last integral can be easily calculated by the following formula
$$ \int_{0}^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}} \, dx = \beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$
Therefore we obtain the following closed form
$$ I = \frac{\Gamma(\frac{1}{12})\Gamma(\frac{5}{12})}{3 \sqrt[4]{2}\sqrt{\pi}}. $$
In order to verify that this is exactly the same as Vladimir's result, We utilize the Legendre multiplication formula and the reflection formula to find that
$$ \Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12})
= \frac{\Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12})\Gamma(\tfrac{9}{12})}{\Gamma(\tfrac{3}{4})}
= \frac{2 \pi \cdot 3^{1/4} \Gamma(\frac{1}{4})}{\Gamma(\tfrac{3}{4})}
= 2^{1/2} 3^{1/4} \Gamma(\tfrac{1}{4})^2. $$
This completes the proof.
Since
$$
\frac{\sin(x)}{\cosh(x)+\cos(x)}=2\sum_{n\ge 1}(-1)^{n-1}e^{-nx}\sin(nx
)$$
(shown here), we can combine the above with $\sin^2(x)\sin(nx) = \frac14 \left[2 \sin(n x) + \sin((2 - n) x) - \sin((2 + n) x)\right]$ and $\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}\mathrm{d}x = \arctan\left(\frac{b}{a}\right)$ to get
\begin{align*}
I &= \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1}\int_{0}^{\infty}\frac{e^{-nx}}{x}\left[2 \sin(n x) + \sin((2 - n) x) - \sin((2 + n) x)\right] \, \mathrm{d}x\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1} \left[\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right)\right]\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1} \left[\arctan(n+1) - \arctan(n-1)\right]\\
& = \frac{1}{2}\left[-\arctan(0) + \arctan(1) \right]\\
& = \frac{\pi}{8}
\end{align*}
as desired.
Note that to re-write the arctangent expressions from equations $2\to 3$ we combine $\arctan(\alpha) - \arctan(\beta) = \arctan\left(\frac{\alpha - \beta}{1 + \alpha\beta} \right)$ and $\arctan(x) + \arctan\left( \frac1x\right)= \frac\pi2,\ x>0$. Thus
\begin{align*}
\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right) &=\frac\pi2 + \arctan\left(\frac{-n^2}{2}\right) \\
&\overset{\color{blue}{n^2/2 >0}}{=} \arctan\left(\frac{2}{n^2}\right) \\
&=\arctan(n+1) - \arctan(n-1)
\end{align*}
also exploiting the fact that $\arctan(x)$ is odd.
Best Answer
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 $$I(w) =\int_0^w I’(a)da =\frac13 \int_0^w a(\pi^2+a^2)da =\frac{1}{12} \left( 2\pi^2 w^2 + w^4 \right)$$