Expanding the inverse tangent in logarithms, writing $\frac{x}{1+x^2}=\Re\frac1{x-i}$, and expanding $\log(1-x^2)=\log(1-x)+\log(1+x)$, each of the resulting four indefinite integrals has a closed form. Each term is amenable to automatic integration, (an example), which means that after taking limits, slogging through simplifications and special values, such as those found here, the closed form can be computed.
For example, for the term above,
$$ \int_0^1\frac{\log(1-ix)\log(1-x)}{x-i}\,dx =
-\frac{K\pi }{4}-\frac{17 i \pi ^3}{384}-\frac{1}{2} i K \log2+\frac{13}{192} \pi ^2 \log2+\frac{3}{32} i \pi (\log2)^2-\frac{(\log2)^3}{48}+3 \,\text{Li}_3({\textstyle\frac{1+i}{2}})-\frac{45 \zeta(3)}{32}.
$$
Now, the integrand of the integral in question is the real part of the sum
$$ \frac i2 \frac{\log(1-ix)\log(1-x)}{x-i} - \frac i2\frac{\log(1+i x)\log(1-x)}{x-i}+\frac i2\frac{\log(1-ix)\log(1+x)}{x-i}-\frac i2\frac{\log(1+ix)\log(1+x)}{x-i},
$$
where each term has a closed form for its integral, as above, in terms of $\pi$, $K$, $\log 2$ and $\text{Li}_3$.
After sufficient simplification, the integral of that sum is
$$\begin{aligned} &\int_0^1 \frac{\arctan x\log(1-x^2)}{x-i}\,dx = \\
&-\frac{1}{4} i K\pi -\frac{\pi ^3}{48}+\frac{1}{32} i \pi ^2 \log2-\frac{1}{8} \pi (\log2)^2+K \log2+\frac{7}{32} i \zeta(3), \end{aligned}$$
of which the real part gives the answer
$$ -\frac{\pi ^3}{48}-\frac{1}{8} \pi (\log2)^2+ K \log2$$
Using the Fourier series of $\ln(\tan{x})$,
\begin{align}
&\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\
=&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\
=&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\
=&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\
&\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\\
=&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{G}}\color{#E2062C}{-\frac{1}{2}G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}G}}\\
=&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)G=\large{-\frac{2}{3}G}
\end{align}
Things could be made clearer if we explicitly write out the terms of the sums. For the red sums,
\begin{align}
&-\frac{1}{2}\left(\frac{1}{1^2}+\frac{1}{5^2}-\frac{1}{7^2}-\frac{1}{11^2}+\cdots\right)\\
=&-\frac{1}{2}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2}-\frac{1}{11^2}+\cdots\right)-\frac{1}{2}\left(\frac{1}{3^2}-\frac{1}{9^2}+\frac{1}{15^2}-\cdots\right)\\
=&-\frac{1}{2}G-\frac{1}{2}\cdot\frac{1}{9}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\cdots\right)=-\frac{5}{9}G
\end{align}
Best Answer
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: