About the integral $\int\arctan\left(\frac{1}{\sinh^2 x}\right)dx$, some idea or feedback While I was playing with Wolfram Alpha calculator I wondered if it is known a closed-form for $$\int_0^\infty\arctan\left(\frac{1}{\sinh^2 x}\right)dx.\tag{1}$$
Wolfram Alpha provide me the coressponding indefinite integral using this code 
int  arctan(1/sinh^2(x))dx
but it seems like as science fiction that I can to understand what did this CAS since the integral is very difficult.

Question. Can you provide me an idea to get such indefinite integral $$\int\arctan\left(\frac{1}{\sinh^2 x}\right)dx?$$
  Of course, if it is a known integral and closed-form $(1)$ answer as a reference request, refering the literature and I try to find and read such exercise from the literature. Thanks you in advance.

 A: By substituting $x=\text{arcsinh}(u)$ we have
$$ \int_{0}^{+\infty}\arctan\left(\frac{1}{\sinh t}\right)\,dt = \int_{0}^{+\infty}\frac{\arctan\frac{1}{u}}{\sqrt{1+u^2}}\,du \stackrel{u\mapsto 1/v}{=}\int_{0}^{+\infty}\frac{\arctan v}{v\sqrt{1+v^2}}\,dv$$
and by substituting $v=\tan\theta$ we are left with
$$ \int_{0}^{\pi/2}\frac{\theta\,d\theta}{\sin\theta} $$
which is a notorious integral, equal to twice the Catalan constant $G$:
$$ \int_{0}^{+\infty}\arctan\left(\frac{1}{\sinh t}\right)\,dt = 2\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}\approx 1.831931188354438. $$
The indefinite integral is similarly related to the dilogarithm function $\text{Li}_2$.
If we have an extra square there is little to change:
$$ \int_{0}^{+\infty}\arctan\left(\frac{1}{\sinh^2 t}\right)\,dt = \int_{0}^{+\infty}\frac{\arctan u}{2u\sqrt{1+u}}\,du=\int_{0}^{1}\int_{0}^{+\infty}\frac{1}{2(1+a^2 u^2)\sqrt{1+u}}\,du\,da$$
we "simply" have a linear combination of squared logarithms evaluated at ugly points.
This is not the only case for the geometry of the dilogarithm to lead to unexpected simple closed forms.
See here, for instance.
