$$I=\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx$$
Let,
\begin{align*}
\displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\
\displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\
\displaystyle C&=\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx\\
\end{align*}
From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,
it follows that,
$$\displaystyle \int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx=A-\dfrac{1}{2}B-C-2G\ln 2+\beta(3)$$
and,
$$(1)\boxed{A=\dfrac{1}{64}\pi^3-B-G\ln 2}$$
with,
$$\displaystyle \beta(3)=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{(2n+1)^3}$$
Thus,
$$(2)\boxed{\displaystyle \int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)}$$
$G$, being the Catalan constant,
From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,
it follows that,
$$\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=A+\dfrac{1}{2}B+\beta(3)+\int_0^1\dfrac{\ln x\arctan x }{1-x}dx$$
and,
$$\displaystyle \int_0^1\dfrac{\ln x\arctan x}{1-x}dx=B-C+2G\ln 2-\dfrac{1}{16}\pi^3$$
Thus, using (1),
$$(3)\boxed{\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)}$$
Therefore,
\begin{align}
\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx&=\int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1-x)}{1+x^2}dx\\
&=B+\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)+\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)\\
&=2\beta(3)-\dfrac{\pi^3}{32}-2G\ln 2-2C
\end{align}
$C$ have been already evaluated (see Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ )
\begin{equation}
\boxed{\displaystyle C=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}}
\end{equation}
and, knowing that,
\begin{equation}
\beta(3)=\dfrac{\pi^3}{32}
\end{equation}
it follows that,
$$\boxed{I=\dfrac{\pi^3}{16}-3G\ln 2}$$
