TITLE: A Challenging Logarithmic Integral $\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx$
QUESTION [58 upvotes]: How can we prove that:

$$\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx=\frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)$$ where $\zeta(z)$ is the Riemann Zeta Function.

The best I could do was to express it in terms of Euler Sums. Let $I$ denote the integral.
$$I=-\frac{\pi^2}{24}\zeta(3)+2\sum_{r=2}^\infty \frac{(-1)^r (H_r)^2}{r^3}-2\sum_{r=2}^\infty \frac{(-1)^r H_r}{r^4}+2 \sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}H_r}{r^2}-2\sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}}{r^3}$$
where $\displaystyle H_r^{(n)}=\sum_{n=1}^r \frac{1}{k^n}$. I am unable to simplify these sums further. Does anyone have any idea on how to solve this integral?
Please see here for more details.

Some time ago I was able to solve the simpler integral:

\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \end{align*} where $\text{Li}_n(z)$ is the Polylogarithm.

REPLY [3 votes]: On the path of Omran Kouba,
\begin{align}
K_1&=\int_0^1 \frac{\ln^3(x)\ln^2(1-x)}{1-x}dx\\
K_2&=\int_0^1 \frac{\ln^3(x)\ln^2(1+x)}{1+x}dx\\
U&=\int_0^1 \frac{\ln^3(x)\ln^2(1+x)}{1-x}dx\\
V&=\int_0^1 \frac{\ln^3(x)\ln^2(1-x)}{1+x}dx\\
Z&=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3 x}{1-x^2}dx\\
&=\frac{1}{2}\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3 x}{1-x}dx+\frac{1}{2}\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3 x}{1+x}dx\\
&=\frac{1}{2}K_1-\frac{1}{2}U+\frac{1}{2}V-\frac{1}{2}K_2\\
J&=\int_0^1 \frac{\ln(x)\ln(1-x)\ln^2(1+x)}{x}dx&=\\
J&=-\frac{7}{24}K_1+\frac{1}{3}Z\\
&=-\frac{7}{24}K_1+\frac{1}{3}\left(\frac{1}{2}K_1-\frac{1}{2}U+\frac{1}{2}V-\frac{1}{2}K_2\right)\\
&=\frac{1}{6}V-\frac{1}{6}U-\frac{1}{8}K_1-\frac{1}{6}K_2
\end{align}
From Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques
One obtains,
\begin{align*}K_1&=12\zeta(5)-6\zeta(2)\zeta(3)\\
K_2&=\frac{1}{2}K_1-\frac{9}{16}\zeta(5)\\
U&=-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\\
V&=\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)\\
\end{align*}
Therefore,
\begin{align*}J&=\frac{1}{6}\left(\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)\right)-\frac{1}{6}\left(-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\right)-\\
&\frac{1}{8}\Big(12\zeta(5)-6\zeta(2)\zeta(3)\Big)-\frac{1}{6}\left(\frac{87}{16}\zeta(5)-3\zeta(2)\zeta(3)\right)\\
&=\frac{7}{8}\zeta(2)\zeta(3)-\frac{25}{16}\zeta(5)\\
&=\boxed{\frac{7}{48}\pi^2\zeta(3)-\frac{25}{16}\zeta(5)}
\end{align*}