<s>[INST] 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.
 [/INST] 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}
</s>