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 $?$.

This is a partial solution.
Let us put, for $0\leq t\leq 1$,
$$F(t) = \int_0^1 \frac{\log x \log(1-tx^4)}{1+x^2} dx$$
Then
$$F'(t) = -\int_0^1 \frac{x^4\log x}{(1+x^2)(1-tx^4)} dx =  -\int_0^1 \frac{x^4\log x}{1+x^2} \sum_{n=0}^\infty t^nx^{4n} dx$$
$$=-\sum_{n=0}^\infty t^{n} C_{4(n+1)}$$
where $$C_m = \int_0^1 \frac{x^{m}\log x}{1+x^2} dx.$$
One has $C_0 = -G$. Multiplying both sides of the identity
$$x^m = \frac{x^m}{1+x^2} + \frac{x^{m+2}}{1+x^2}$$
by $\log x$ and integrating from $0$ to $1$, one finds the recurrence formula
$$C_m + C_{m+2} = \frac{-1}{(1+m)^2}$$
and therefore
$$C_{m+4} - C_m = \frac{-1}{(3+m)^2} + \frac{1}{(1+m)^2}.$$
Therefore, 
$$C_0 = -G$$
$$C_4 = -G +1 - \frac{1}{3^2}$$
$$C_8 = -G + 1 - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2}.$$
and so on. (Remark that $C_{4n} \to 0$ by definition of $G$.) Now, remark that $F(0) = 0$, so your integral is
$$F(1) = \int_0^1 F'(t) dt = -\sum_{n=0}^\infty \frac{C_{4(n+1)}}{n+1} = -\sum_{n=1}^\infty \frac{C_{4n}}{n}.$$
Now, it should be a matter of partial summation to transform the sum $-\sum_{n=1}^\infty \frac{C_{4n}}{n}$ into $\pi^3/16 -3G\log 2$ (in a manner similar to this), but I don't see it right away. I'll think about it a bit more later.
