Now let us continue with an extension of the problem. Well, let us consider the double parametric family of integrals, that is:
$$\mathcal{J}(m, k)= \int_0^1 \frac{\log^m (1+x) \log x \log^k (1-x)}{x} \, {\rm d}x$$
Well , here is a list of results that I leave them as an exercise.
$$\begin{array}{||c|c|c|c|c||}
\hline \text{Integral}&\mathcal{J}(m, k)& m& k &\text{Value} \\
\hline
\int_{0}^{1} \frac{\log x \log(1-x)}{x} \, {\rm d}x&\mathcal{J}(0, 1)&0 &1& \zeta(3)\\\\
\int_0^1 \frac{\log x \log^2 (1-x)}{x} \, {\rm d}x&\mathcal{J}(0, 2)&0 &2 & - \frac{\pi^4}{180} \\\\
\int_0^1 \frac{\log (1+x) \log x \log(1-x)}{x}\, {\rm d}x&\mathcal{J}(1, 1) &1 &1 & -\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} + 2 \text{Li}_4 \left(\frac{1}{2} \right) \sim 0.290721 \\\\
\int_{0}^{1} \frac{\log x \log(1-x) \log^2 (1+x)}{x} \, {\rm d}x&\mathcal{J}(2,1)&2 &1& \frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5) \\\\
\hline
\end{array}$$
It is quite interesting for someone to prove that
$$\int_{0}^{1} \frac{\log x \log(1-x) \log^2 (1+x)}{x} \, {\rm d}x= \frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)$$
P.S 1: We have seen $\displaystyle \int_0^1 \frac{\log x \log (1-x)}{x} \, {\rm d}x$
here.
P.S 2: A similar problem by Ovidiu Furdui is the following:
$$\int_{0}^{1}\ln^k (1-x)\ln x\, {\rm d}x =(-1)^{k+1}k! \left ( k+1 - \zeta(2)-\zeta(3)- \cdots -\zeta(k+1) \right )$$
where $\mathbb{Z} \ni k \geq 1$ and $\zeta$ is the Riemann zeta function.