A real-method based solution. Integrate
$\begin{align}
&\int_0^1\frac{\ln x \ln (1+x^2)}{1+x^2}dx \\
=& \int _0^1 \ln (1+x^2)\> d \left(\int_1^x \frac{\ln t}{1+t^2}dt\right)
\overset{IBP}=\int _0^1 \frac{-2x}{1+x^2}\left( \int_0^x \right. \overset{t=xs }{\left. \frac{\ln t}{1+t^2}dt+G\right)}dx \\
=& \int_0^1 \int_0^1 \frac{-2x^2\ln (xs) }{(1+x^2)(1+x^2s^2)}dsdx -G\int_0^1 \frac{2x}{1+x^2}dx
\\
= &\int_0^1 \int_0^1 \frac{-2x^2\ln x}{(1+x^2)(1+x^2s^2)}dsdx + \int_0^1 \int_0^1 \frac{-2x^2\ln s }{(1+x^2)(1+x^2s^2)}dsdx -G\ln2 \\
= &-2\int_0^1 \frac{x\ln x\> \tan^{-1}x}{1+x^2}dx -2\int_0^1 \frac{s\ln s\> \tan^{-1}s}{1-s^2}ds -G\ln2 \\
=& -4\int_0^1 \underset{=J}{\frac{x\ln x\> \tan^{-1}x}{1-x^4}}dx -G\ln2=-4J -G\ln2
\end{align}$
and, similarly, integrate
\begin{align}\int _0^1\frac{\ln x \ln (1-x^2)}{1+x^2}dx
&= \int _0^1 \ln (1-x^2)\> d \left(\int_1^x \frac{\ln t}{1+t^2}dt\right)
= 4J-2G\ln2 +\frac{\pi^3}{16}
\end{align}
Then, eliminate the common term $J$
$$\int_0^1\frac{\ln x\ln \left(1+x^2\right)}{1+x^2}dx
= -\int _0^1\frac{\ln x \ln (1-x^2)}{1+x^2}dx -3G\ln 2 +\frac{\pi^3}{16}\tag1 $$
Next, combine
\begin{align}
&\hspace{-3mm}\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx
\overset{\text{split}(0,\infty)}= 2\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx
-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx+\frac{\pi^3}{16}\\
&\hspace{-3mm}\int_0^1 \underset{x\to\frac{1-x}{1+x}}{\frac{\ln^2(1-x)}{1+x^2}dx}
= \hspace{-3mm}\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx
-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx-2G\ln2+\frac{\pi^3}{16}\\
\end{align}
to get
$$\hspace{-5mm}\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx
=\frac12\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx-\hspace{-3mm} \int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx -2G\ln2+\frac{\pi^3}{32}\tag2
$$
Also, per $ab = \frac12 (a^2 +b^2 -(a-b)^2)$
$$\hspace{-5mm}\int_0^1 \frac{\ln x\ln(1-x)}{1+x^2}dx
=\frac12\int_0^1\frac{\ln^2(1-x)}{1+x^2}dx-\frac12\int_0^1\frac{\ln^2\frac x{1-x}}{1+x^2}dx + \frac{\pi^3}{32} \tag3
$$
Then, (2) + (3)
$$
\int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx
=-\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx -2G\ln2+\frac{\pi^3}{16} +\frac12 K \tag4$$
where $ \int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx=2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)$ and
\begin{align}
K =& \int_0^1 \underset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx +\underset{t=1+x}{
\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}}dx -\int_0^1\underset{t=\frac x{1-x}}{\frac{\ln^2\frac x{1-x}}{1+x^2}}dx \\
=& \int_0^\infty {\frac{4t \ln^2 t}{4+t^4}}dt
\overset{t^2\to 2t}= \frac{\pi}{8}\ln^22+\frac{\pi^3}{32}
\end{align}
Plug into (4) and then into (1) to obtain
$$
\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx =2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) -G\ln2
-\frac{\pi}{16}\ln^22-\frac{\pi^3}{64}$$
