Let $a=\ln x, b=\ln(1-x), c=\ln(1+x), d=\ln(1+x^2)$. I use the following notations:
$$I_{aa} = \int_0^1 \frac{\ln^2 x}{1+x^2}dx \qquad I_{ab} = \int_0^1 \frac{\ln x \ln(1-x)}{1+x^2}dx \qquad \cdots \qquad I_{cd} = \int_0^1 \frac{\ln (1+x) \ln(1+x^2)}{1+x^2}dx$$
Hence we get $10$ integrals. My goal is to find $9$ linearly independent relations between them, so your desired value $2I_{ad}+I_{cd}$ falls out easily.
Let $x=(1-u)/(1+u)$, then $dx/(1+x^2) = du/(1+u^2)$, and we have the following transformation rules:
$$\begin{aligned}a &\mapsto b-c \\
b &\mapsto \ln 2 + a - c \\
c &\mapsto \ln 2 - c \\
d &\mapsto \ln 2 + d - 2c \end{aligned}$$
For example, we apply this on $I_{aa}$,we have
$$\tag{1}I_{aa} = I_{bb} - 2I_{bc} + I_{cc}$$
We can apply this transformation on each of the ten integrals, but we only yield four linearly independent relations:
$$\tag{2} I_{bb}=I_{aa}-2 I_{ac}-2 G \ln 2+I_{cc}$$
$$\tag{3} I_{dd}=2 \ln (2) \left(\frac{1}{2} \pi \ln (2)-G\right)+4 I_{cc}-4 I_{cd}+I_{dd}-\frac{1}{4} \pi \ln ^2(2)$$
$$\tag{4} I_{bd}=-2 I_{ac}+I_{ad}+\ln (2) \left(\frac{1}{2} \pi \ln (2)-G\right)-G \ln (2)+2 I_{cc}-I_{cd}-\frac{1}{8} \pi \ln ^2(2)$$
Of course, we have explicit evaluation of $I_{aa}$, which can be our fifth linearly independent relation: $$\tag{5} I_{aa} = \frac{\pi^3}{16}$$
To find more relations, we must rely on other methods. Here I use contour integration. Let $\log_1$ denote logarithm with branch cut at negative $x$-axis, while $\log_2$ denote logarithm with cut at positive $x$-axis. Integrate the function
$$\frac{(\log_1 z)^a(\log_2 (z-1))^b}{1+z^2}$$
around a contour with two keyhole, wrapping around the two cuts: $(1,\infty)$ and $(-\infty,0)$. Then we obtain
$$\int_1^\infty \cdots + \int_{-\infty}^0 \cdots = 2\pi i \text{(Sum of residues)}$$
The first integral's range can be brought back to $(0,1)$ via $x\mapsto 1/x$. The second integral, we first bring it back to $(0,\infty)$, then split intervals, finally apply $x\mapsto 1/x$ for the one with range $(1,\infty)$. After all these,
We have $$\int_0^1 \frac{f_{a,b}(x)}{1+x^2} dx = 2\pi i \text{(Sum of residues)}$$
where $$f_{a,b}(x) = (-\ln (x))^a \left[(\ln (1-x)-\ln (x))^b-(\ln (1-x)-\ln (x)+2 \pi i)^b\right]-\left[(-\ln (x)-\pi i)^a-(-\ln (x)+\pi i)^a\right] (\ln (x+1)-\ln (x)+\pi i)^b-\left[(\ln (x)-\pi i)^a-(\ln (x)+\pi i)^a\right] (\ln (x+1)+\pi i)^b$$
Now apply this to $a=1,b=2$:
$$\int_0^1 \frac{f_{1,2}(x)}{1+x^2}dx = -\frac{17 i \pi ^4}{16}+\frac{1}{4} i \pi ^2 \ln^2(2)-\pi ^3 \ln(2)$$
Hence comparing imaginary part:$$\tag{6}-2 \pi I_{aa}+4 \pi I_{ab}-4 \pi I_{ac}+4 \pi I_{cc}-\pi ^4=\frac{1}{4} \pi ^2 \ln ^2(2)-\frac{17 \pi ^4}{16}$$
This this our sixth linearly independent relation.
Apply above method again to $a=0,b=3$:
$$\tag{7}-6 \pi I_{bb}-6 \pi I_{aa}+12 \pi I_{ab}+2\pi^4 =-\frac{3}{4} \pi ^2 \ln (2)$$
The final two relations come from gamma/zeta function.
Note that $$\int_1^\infty \frac{\ln^2(1+x^2)}{1+x^2}dx = I_{dd}-4I_{ad}+4I_{aa}$$
Hence $$\tag{8}2I_{dd}-4I_{ad}+4I_{aa} = \int_0^\infty \frac{\ln^2(1+x^2)}{1+x^2}dx = 4\int_0^{\pi/2} \ln^2(\cos x)dx = \frac{1}{6} \left(\pi ^3+12 \pi \ln ^2 2\right)$$
The last relation is more nontrivial:
$$I_{ad}+I_{ab}+I_{ac} = \int_0^1 \frac{\ln x \ln \left(1-x^4 \right)}{1+x^2}dx = \frac{\pi^3}{16}-3G\ln 2 \tag{9}$$
which uses, in a critical way, values of digamma function.
Now solve those $9$ equations, we have one free variable (this involves a new constant, see below), and that free variable cancels for $2I_{ad}+I_{cd}$, proving your claim.
The new constant comes from $$\tag{10} I_{bb} = \int_0^1 \frac{\ln^2 x}{x^2-2x+2}dx = 2 \Im\left[\text{Li}_3\left(\frac{1+i}{2}\right)\right]$$
This follows directly from the indefinite integration:
$$\int \frac{\ln^2 x}{x-a} = -2 \text{Li}_3\left(\frac{x}{a}\right)+2 \ln (x) \text{Li}_2\left(\frac{x}{a}\right)+\ln^2(x) \ln\left(1-\frac{x}{a}\right)$$
To consummate this approach, we obtain simultaneous evaluation of all $10$ integrals, all are nontrivial (except $I_{aa}, I_{bb}$) when considered individually.
$$\begin{aligned}
\int_0^1 \frac{\ln^2(1+x)}{1+x^2} dx &= -2 G \ln (2)-4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{7 \pi ^3}{64}+\frac{3}{16} \pi \ln ^2(2) \\
\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2} dx &= -2 G \ln (2)+4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{7 \pi ^3}{96}+\frac{7}{8} \pi \ln ^2(2) \\
\int_0^1 \frac{\ln x \ln(1-x)}{1+x^2} dx &= \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{\pi ^3}{128}-\frac{1}{32} \pi \ln ^2(2) \\
\int_0^1 \frac{\ln x \ln(1+x)}{1+x^2} dx &= -2 G \ln (2)-3 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{11 \pi ^3}{128}+\frac{3}{32} \pi \ln ^2(2) \\
\int_0^1 \frac{\ln x \ln(1+x^2)}{1+x^2} dx &= -G \ln (2)+2 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{\pi ^3}{64}-\frac{1}{16} \pi \ln ^2(2) \\
\int_0^1 \frac{\ln (1-x) \ln(1+x)}{1+x^2} dx &= -G \ln (2)-\Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{3 \pi ^3}{128}+\frac{3}{32} \pi \ln ^2(2) \\
\int_0^1 \frac{\ln (1-x) \ln(1+x^2)}{1+x^2} dx &= -\frac{1}{2} G \ln (2)+4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{5 \pi ^3}{64}+\frac{1}{8} \pi \ln ^2(2) \\
\int_0^1 \frac{\ln (1+x) \ln(1+x^2)}{1+x^2} dx &= -\frac{5}{2} G \ln (2)-4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{7 \pi ^3}{64}+\frac{3}{8} \pi \ln ^2(2)
\end{aligned}$$
The Mathematica input is:
{aa -> \[Pi]^3/16, bb -> 2 Im[PolyLog[3, 1/2 + I/2]], cc -> (7 \[Pi]^3)/64 - 4 Im[PolyLog[3, 1/2 + I/2]] - 2 Catalan Log[2] + 3/16 \[Pi] Log[2]^2, dd -> -((7 \[Pi]^3)/96) + 4 Im[PolyLog[3, 1/2 + I/2]] - 2 Catalan Log[2] - 1/8 \[Pi] Log[2]^2 + 1/4 \[Pi] Log[4]^2, ab -> -(\[Pi]^3/128) + Im[PolyLog[3, 1/2 + I/2]] - 1/32 \[Pi] Log[2]^2, ac -> (11 \[Pi]^3)/128 - 3 Im[PolyLog[3, 1/2 + I/2]] - 2 Catalan Log[2] + 3/32 \[Pi] Log[2]^2, ad -> -(\[Pi]^3/64) + 2 Im[PolyLog[3, 1/2 + I/2]] - Catalan Log[2] - 1/16 \[Pi] Log[2]^2, bc -> (3 \[Pi]^3)/128 - Im[PolyLog[3, 1/2 + I/2]] - Catalan Log[2] + 3/32 \[Pi] Log[2]^2, bd -> -((5 \[Pi]^3)/64) + 4 Im[PolyLog[3, 1/2 + I/2]] - 1/2 Catalan Log[2] + 1/8 \[Pi] Log[2]^2, cd -> (7 \[Pi]^3)/64 - 4 Im[PolyLog[3, 1/2 + I/2]] - 5/2 Catalan Log[2] + 3/8 \[Pi] Log[2]^2}
Best Answer
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 $y=\dfrac{1}{x}$ )