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.

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. 
