Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Constant.
I was able to express it in terms of Euler Sums but it does not seem to be of any use.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.

Presented below is a self-contained evaluation. With
$\int_0^1 \frac{\ln t}{1+t^2}dt =-G$
\begin{align*}
I & = \int_0^1 \frac{\ln x \ln (1-x^4 )}{1+x^2}dx \\
& = \int_0^1 \ln (1-x^4 ) d\left(\int_1^x \frac{\ln t}{1+t^2}dt \right) \overset{IBP}=\int_0^1 \frac{ 4x^3}{1-x^4} \underset{t=xs }{\left(\int_0^x \frac{\ln t}{1+t^2}dt +G \right) } dx \\
& =4\int_0^1 \left( \int_0^1 \frac{x^4 \ln x+x^4\ln s }{(1-x^4 )(1+x^2s^2)}ds +\frac{Gx^3}{1-x^4} \right) dx\\
& =4\int_0^1 \int_0^1 \frac{x^4\ln x}{(1-x^4)(1+x^2s^2)}dsdx -4 \int_0^1\int_0^1 \frac{\ln s}{1+x^2s^2}dx ds \\
& \>\>\>\>\>+ 4 \int_0^1 \left(\int_0^1 \frac{\ln s }{(1-x^4 )(1+x^2s^2)}ds +\frac{Gx ^3}{1-x^4} \right) dx\\
\end{align*}
Integrate the 2nd integral
\begin{align*}
& \int_0^1\int_0^1 \frac{\ln s}{1+x^2s^2}dx ds =\int_0^1 \frac{\ln s\tan^{-1}s}sds
\overset{IBP}=-\frac12 \int_0^1 \frac{\ln^2s}{1+s^2}ds=-\frac{\pi^3}{32}
\end{align*}
and apply the decomposition below in the 3rd integral
$$\frac{1 }{(1-x^4 )(1+x^2s^2)}
= \frac{-s^4}{(1-s^4)(1+x^2s^2)} +\frac1{2(1-s^2)(1+x^2)}+ \frac1{2(1+s^2)(1-x^2)}
$$
Then, the 1st integral cancels and
\begin{align*}
I =& -4\left(-\frac{\pi^3}{32}\right)
+2\int_0^1 \int_0^1 \frac{\ln s }{(1-s^2 )(1+x^2)}dx ds \\
&\>\>\> + 2\int_0^1 \left( \int_0^1 \frac{\ln s }{(1+s^2 )(1-x^2)}ds  +\frac{2Gx ^3}{1-x^4}\right) dx\\
= & \frac{\pi^3}8+ 2\int_0^1 \frac{\ln s ds }{1-s^2}\int_0^1\frac{dx }{1+x^2}
 -2G \int_0^1 \left( \frac{1}{1-x^2} -\frac{2x^3}{1-x^4}\right) dx\\
 = & \frac{\pi^3}8+ 2\left(-\frac{\pi^2}{8}\right) \frac\pi4
 -2G \int_0^1 \left( \frac{x}{1+x^2} +\frac{1}{1+x}\right) dx\\
 = & \frac{\pi^3}{16}
 -3G\ln2\\
\end{align*}
