Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$ 
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?

I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
Attempt
$$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx \\ =\int_0^\infty\frac{\operatorname{arcsinh}(2x)}{1+x^2} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{4+\sinh^2x} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{3+\cosh^2x} \, dx\\
=2\int_0^\infty\sum_{n=0}^\infty x(-3)^n\cosh^{-2n-1}(x) \, dx$$
I failed to integrate $x\cosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
 A: Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign. 
Let
$$I(a) = \int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (a \tan x) \, dx, \qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.
For $a = 1$ we have
\begin{align}
I(1) &= \int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (\tan x) \, dx\\
&= \int_0^{\frac{\pi}{2}} \ln \left (\frac{1 + \sin x}{\cos x} \right ) \, dx\\
&= \int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx - \int_0^{\frac{\pi}{2}} \ln (\cos x) \, dx.
\end{align}
Now the first of these integrals can be found, for example, by rewriting it as
\begin{align}
\int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx &= \int_0^{\frac{\pi}{2}} \ln (1 + \cos x) \, dx\\
&= \int_0^{\frac{\pi}{2}} \ln \left (\frac{1}{2} \cos^2 \frac{x}{2} \right ) \, dx\\
&= \frac{\pi}{2} \ln 2 + 4 \int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx
\end{align} 
and then employing the Fourier series representation for $\ln (\cos x)$ found here. The final result is
$$\int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx = 2 \mathbf{G} - \frac{\pi}{2} \ln 2.$$
Here $\mathbf{G}$ is Catalan's constant.
The second of the integrals is very well known. Here
$$\int_0^{\frac{\pi}{2}} \ln (\cos x) \, dx = - \frac{\pi}{2} \ln 2.$$
Thus $I(1) = 2 \mathbf{G}$.
Moving to the main event, by differentiating under the integral sign with respect to $a$ we have
\begin{align}
I'(a) &= \int_0^{\frac{\pi}{2}} \frac{\tan x}{\sqrt{a^2 \tan^2 x + 1}} \, dx\\
&= \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{a^2 - (a^2 - 1) \cos^2 x}} \, dx.
\end{align}
Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = \cos x$ one has
$$I'(a) = \int_0^1 \frac{du}{\sqrt{a^2 - (a^2 - 1) u^2}} = \frac{1}{\sqrt{a^2 - 1}} \sin^{-1} \left (\frac{\sqrt{a^2 - 1}}{a} \right ).$$
As we require $I(2)$ observe that 
$$I(2) - I(1) = \int_1^2 I'(a) \, da.$$
Thus
$$I(2) = 2 \mathbf{G} + \int_1^2 \frac{1}{\sqrt{a^2 - 1}} \sin^{-1} \left (\frac{\sqrt{a^2 - 1}}{a} \right ) \, da.$$ 
Integrating by parts leads to
$$I(2) = 2 \mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - \int_1^2 \frac{\cosh^{-1} a}{a \sqrt{a^2 - 1}} \, da.$$
Now let $a = \cosh t$. This gives
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - \int_0^{\ln(2 + \sqrt{3})} \frac{t}{\cosh t} \, dt.$$
Then let $t = \ln y$. This gives
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\int_1^{2 + \sqrt{3}} \frac{\ln y}{1 + y^2} \, dy.$$
Now let $y = \tan \theta$. Then we have
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\int_{\frac{\pi}{4}}^{\frac{5\pi}{12}} \ln (\tan \theta) \, d\theta.$$
Finally, enforcing a substitution of $\theta \mapsto \dfrac{\pi}{2} - \theta$ leads to
\begin{align}
I(2) &= 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) + 2 \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \ln (\tan \theta) \, d\theta\\
&= 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) + 2\int_0^{\frac{\pi}{4}} \ln (\tan \theta) \, d\theta - 2 \int_0^{\frac{\pi}{12}} \ln (\tan \theta) \, d\theta.
\end{align}
For the first of the integrals we have
$$\int_0^{\frac{\pi}{4}} \ln (\tan x) \, dx = -\mathbf{G}.$$
While for the second of the integrals we have
$$\int_0^{\frac{\pi}{12}} \ln (\tan x) \, dx = -\frac{2}{3} \mathbf{G}.$$
So finally
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\mathbf{G} + \frac{4}{3} \mathbf{G},$$
or
$$\int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (2 \tan x) \, dx = \frac{\pi}{3} \ln (2 + \sqrt{3}) + \frac{4}{3} \mathbf{G},$$
as announced. 
A: Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
\begin{equation}
 I = \int_0^{\infty}\frac{\operatorname{arcsinh(2x)}}{1 + x^2}\:dx
\end{equation}
Let 
\begin{equation}
 J(t) = \int_0^{\infty}\frac{\operatorname{arcsinh(tx)}}{1 + x^2}\:dx
\end{equation}
We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':
\begin{equation}
 J'(t) = \int_0^{\infty}\frac{x}{\sqrt{1 + t^2x^2}}\frac{1}{1 + x^2}\:dx = \left[\frac{1}{\sqrt{t^2 - 1}} \cdot \arctan\left(\sqrt{\frac{1 + t^2x^2}{t^2 - 1}} \right) \right]_0^{\infty} = \frac{1}{\sqrt{t^2 - 1}}\left[\frac{\pi}{2} - \arctan\left(\frac{1}{\sqrt{t^2 -1}} \right) \right]
\end{equation}
Thus, 
\begin{align}
J(t) &= \int \frac{1}{\sqrt{t^2 - 1}}\left[\frac{\pi}{2} - \arctan\left(\frac{x}{\sqrt{t^2 -1}} \right) \right]\:dt \\
&= \int \frac{\pi}{2}\cdot \frac{1}{\sqrt{t^2 - 1}}\:dt - \int  \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt = I_1 - I_2
\end{align}
For $I_1$:
\begin{equation}
 I_1 = \int \frac{\pi}{2}\cdot \frac{1}{\sqrt{t^2 - 1}}\:dt = \frac{\pi}{2}\ln\left| \sqrt{t^2 - 1} + t\right| + C_1
\end{equation}
Where $C_1$ is the constant of integration. Note that $I_1(2) = \ln\left|2 + \sqrt{3} \right|$
For $I_2$:
\begin{equation}
I_2 = \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt
\end{equation}
Unfortunately this is not so easy to evaluate. I will first make the substitution $u = \frac{1}{\sqrt{t^2 - 1}}$:
\begin{align}
I_2 &= \int \frac{1}{\sqrt{t^2 - 1}}\arctan\left(\frac{1}{\sqrt{t^2 -1}} \right)\:dt = \int u \cdot \arctan(u) \cdot \frac{-u^4}{\sqrt{1 +u^2}}\:du\\
& = - \int \frac{-u^5}{\sqrt{1 +u^2}} \cdot \arctan(u)  \:du
\end{align}
A: On the path of Kemono Chen...
\begin{align}J&=\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx\end{align}
Perform the change of variable $y=\operatorname{arcsinh}(2\tan x)$,
\begin{align}J&=\int_0^{+\infty}\frac{2x\cosh x}{4+\sinh^2 x}\,dx\\
&=\int_0^{+\infty}\frac{4x\left(\text{e}^{x}+\text{e}^{-x}\right)}{14+\text{e}^{2x}+\text{e}^{-2x}}\,dx\\
&=\int_0^{+\infty}\frac{4x\text{e}^{-x}\left(\text{e}^{2x}+1\right)}{14+\text{e}^{2x}+\text{e}^{-2x}}\,dx\\
\end{align}
Perform the change of variable $y=\text{e}^{-x}$,
\begin{align}J&=-\int_0^1 \frac{4\ln x\left(1+\frac{1}{x^2}\right)}{14+x^2+\frac{1}{x^2}}\\
&=-\int_0^1 \frac{4\ln x\left(1+x^2\right)}{x^4+14x^2+1}\\
&=\left[-\arctan\left(\frac{4x}{1-x^2}\right)\ln x\right]_0^1+\int_0^1 \frac{\arctan\left(\frac{4x}{1-x^2}\right)}{x}\,dx\\
&=\int_0^1 \frac{\arctan\left(\frac{4x}{1-x^2}\right)}{x}\,dx\\
&=\int_0^1 \frac{\arctan\left(\left(2+\sqrt{3}\right)x\right)}{x}\,dx+\int_0^1 \frac{\arctan\left(\left(2-\sqrt{3}\right)x\right)}{x}\,dx\\
\end{align}
In the first integral perform the change of variable $y=\left(2+\sqrt{3}\right)x$,
In the second integral perform the change of variable $y=\left(2-\sqrt{3}\right)x$,
\begin{align}J&=\int_0^{2+\sqrt{3}}\frac{\arctan x}{x}\,dx+\int_0^{2-\sqrt{3}}\frac{\arctan x}{x}\,dx\\
&=\Big[\arctan x\ln x\Big]_0^{2+\sqrt{3}}-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx+\Big[\arctan x\ln x\Big]_0^{2-\sqrt{3}}-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{5\pi}{12}\ln\left(2+\sqrt{3}\right)-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx+\frac{\pi}{12}\ln\left(2-\sqrt{3}\right)-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-\int_0^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx
\end{align}
In the first integral perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\int_{2-\sqrt{3}}^{+\infty}\frac{\ln x}{1+x^2}\,dx-\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\int_0^{+\infty}\frac{\ln x}{1+x^2}\,dx-2\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\int_0^{2-\sqrt{3}}\frac{\ln x}{1+x^2}\,dx\\
\end{align}
Perform the change of variable $y=\tan x$,
\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\int_0^{\frac{\pi}{12}}\ln\left(\tan x\right)\,dx\\
\end{align}
It is well known that,
\begin{align}
\int_0^{\frac{\pi}{12}}\ln\left(\tan x\right)\,dx=-\frac{2}{3}\text{G}
\end{align}
(see: Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ )
Thus,
\begin{align}J&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)-2\times -\frac{2}{3}\text{G}\\
&=\boxed{\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)+\frac{4}{3}\text{G}}
\end{align}
NB:
Observe that,
\begin{align}2-\sqrt{3}&=\frac{1}{2+\sqrt{3}}\\
\ln\left(2-\sqrt{3}\right)&=-\ln\left(2+\sqrt{3}\right)\\
\int_0^\infty \frac{\ln x}{1+x^2}\,dx&=0
\end{align}
(perform the change of variable $y=\dfrac{1}{x}$ )
A: I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = \int_{0}^{\infty} \frac{\operatorname{arsinh}(tx)}{1+x^2} \, \mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=\sqrt{u^{-2}-1}$, we obtain
\begin{align*}
J'(t)
= \int_{0}^{\infty} \frac{x}{(1+x^2)\sqrt{1+t^2x^2}} \, \mathrm{d}x
= \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u.
\end{align*}
So it follows that
\begin{align*}
J(2)
&= \int_{0}^{2} \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t \\
&= \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t
+ \int_{1}^{2} \int_{0}^{1} \frac{1}{\sqrt{1 + (t^2-1)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t.
\end{align*}
The inner integral is easily computed, yielding
\begin{align*}
J(2)
&= \int_{0}^{1} \frac{\operatorname{artanh}\left( \sqrt{1 - t^2} \right)}{\sqrt{1 - t^2}} \, \mathrm{d}t
+ \int_{1}^{2} \frac{\arctan\left(\sqrt{t^2-1}\right)}{\sqrt{t^2 - 1}} \, \mathrm{d}t.
\end{align*}
Now we substitute $t = \operatorname{sech} \varphi$ for the first integral and $t = \sec \theta$ for the second integral. This yields
\begin{align*}
J(2)
&= \int_{0}^{\infty} \frac{\varphi}{\cosh\varphi} \, \mathrm{d}\varphi
+ \int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta.
\end{align*}
These integrals can be computed as follows:


*

*Using $ \operatorname{sech}\varphi = \frac{2e^{-\varphi}}{1 + e^{-2\varphi}} = 2 \sum_{n=0}^{\infty} (-1)^n e^{-(2n+1)\varphi} $, we obtain
$$ \int_{0}^{\infty} \frac{\varphi}{\cosh\varphi} \, \mathrm{d}\varphi
= 2 \sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} \varphi e^{-(2n+1)\varphi} \, \mathrm{d}\varphi
= 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}
= 2G. $$

*Taking integration by parts,
\begin{align*}
\int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta
&= \left[ - \theta \log \left( \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right) \right]_{0}^{\frac{\pi}{3}} + \int_{0}^{\frac{\pi}{3}} \log \left( \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right)   \, \mathrm{d}\theta \\
&= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) + 2 \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \log \left( \tan \theta \right)   \, \mathrm{d}\theta.
\end{align*}
This can be computed by using the Fourier series $\log \left( \tan \theta \right) = - 2 \sum_{n=0}^{\infty} \frac{\cos(4n+2)\theta}{2n+1} $ to yield
\begin{align*}
\int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta
&= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) - 2 \sum_{n=0}^{\infty} \frac{\sin\left( \frac{\pi}{2} (2n+1) \right) - \sin\left( \frac{\pi}{6} (2n+1) \right)}{(2n+1)^2} \\
&= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) - \frac{2}{3}G.
\end{align*}
Combining two result, we obtain the desired answer.
A: This is not an answer.
Your premonition seems to be good. Using another CAS,
$$2\int_0^\infty\frac{x\cosh( x)}{3+\cosh^2(x)}\,dx=-\frac{\pi}{4}   \log \left(7-4 \sqrt{3}\right)-i \left(\text{Li}_2\left(-i \left(-2+\sqrt{3}\right)\right)-\text{Li}_2\left(i
   \left(-2+\sqrt{3}\right)\right)\right)$$ Now, ???
A: Let $J(a)=\int_0^{\frac\pi2}\sinh^{-1}(\sec a\tan x)dx$, along with
\begin{align}
J’(a)=\int_0^{\frac\pi2}\frac{\tan a \sin x}{\sqrt{1-(\sin a\cos x)^2}}dx= a\sec a
\end{align}
Then
\begin{align}
\int_0^{\frac\pi2}\sinh^{-1}(2\tan x)dx
&= J(\frac\pi3)=J(0)+\int_0^{\frac\pi3} J’(a)da \\&=\int_0^{\frac\pi2}\sinh^{-1}(\tan x)dx+{\int_0^{\frac\pi3} a\sec ada }\\
 &=\int_0^{\frac\pi2}\ln (\tan x+\sec x) dx
+ \int_0^{\frac\pi3} a\>d[\ln (\tan a+\sec a)]\\
&= a\ln(\tan a+\sec a)\bigg|_0^{\frac\pi3} + \int_{ \frac\pi3} ^{\frac\pi2} 
{\ln(\tan a+\sec a) da}\\&= \frac\pi3 \ln(2+\sqrt3)-2 \int^{ \frac\pi{12}}_{0} 
\ln\tan\theta \>d\theta\>\>\>\>\>\>\>(a=\frac\pi2-2\theta)\\
 &= \frac\pi3 \ln(2+\sqrt3)+\frac43G
\end{align}
