Integral $\int_0^1 dx \frac{\ln x \ln^2(1-x)\ln(1+x)}{x}$
We will use similar approach as sos440's answer in I&S. Using the simple algebraic identity $$ ab^2=\frac{(a+b)^3+(a-b)^3-2a^3}{6}, $$ it follows that \begin{align} \int_0^1 \frac{\ln x\ln(1+x)\ln^2(1-x)}{x}\ dx &=\frac16I_1+\frac16I_2-\frac13I_3\ ,\tag1 \end{align} where \begin{align} I_1&=\int_0^1\frac{\ln x\ln^3(1-x^2)}{x}\ dx\\[12pt] I_2&=\int_0^1\frac{\ln x}{x}\ln^3\left(\frac{1+x}{1-x}\right)\ dx\\[12pt] I_3&=\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx \end{align}
Evaluation of $I_1$
Setting $t=x^2$ followed by $t\mapsto1-t$, we have \begin{align} I_1&=\frac14\int_0^1\frac{\ln t\ln^3(1-t)}{t}\ dt\\ &=\frac14\int_0^1\frac{\ln (1-t)\ln^3t}{1-t}\ dt\\ \end{align} To evaluate the integral above, we can use multiple derivative of beta function
$$ I_1=\frac14\lim_{x\to1}\lim_{y\to0^+}\frac{\partial^4\text{B}(x,y)}{\partial x^3\partial y}=3\zeta(5)-\frac32\zeta(2)\zeta(3).\tag2 $$
Alternatively, we can use generating function for the harmonic numbers for $|z|<1$ $$ \sum_{n=1}^\infty H_n z^n=-\frac{\ln(1-z)}{1-z}, $$ identity of the harmonic numbers $$ H_{n+1}-H_n=\frac1{n+1}, $$ and $$ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots\tag3 $$ We may refer to the following answer to see the complete approach for evaluating $I_1$.
Evaluation of $I_2$
$I_2$ has been evaluated by sos440 here and it is equal to
$$ I_2=\frac{21}{4}\zeta(2)\zeta(3)-\frac{93}{8}\zeta(5).\tag4 $$
Alternatively, we can use the following technique. Setting $t=\dfrac{1-x}{1+x}\ \color{red}{\Rightarrow}\ x=\dfrac{1-t}{1+t}$ and $dx=-\dfrac{2}{(1+t)^2}\ dt$, then \begin{align} I_2&=-\int_0^1\frac{\ln x}{x}\ln^3\left(\frac{1-x}{1+x}\right)\ dx\\ &=2\int_0^1\frac{\ln^3 t\ln(1+t)}{(1-t)(1+t)}\ dt-2\int_0^1\frac{\ln^3 t\ln(1-t)}{(1-t)(1+t)}\ dt.\tag5 \end{align} Using the fact that $$ \frac{2}{(1-t)(1+t)}=\frac1{1-t}+\frac1{1+t} $$ and $(5)$ can be evaluated by performing some tedious calculations involving series expansion (double summation or generating function for the harmonic numbers) of the form $\dfrac{\ln(1\pm t)}{1\pm t}$ and equation $(3)$.
Another alternative way to evaluate $I_2$ without using complex analysis and dividing integral into four separated integrals is substituting $t=\dfrac{1-x}{1+x}$ and $I_2$ turns out to be $$ I_2=-2\int_0^1\frac{\ln^3t}{1-t^2}\ln\left(\frac{1-t}{1+t}\right)\ dt,\tag6 $$ where $(6)$ has been evaluated by Omran Kouba (see evaluation of $K$).
Evaluation of $I_3$
$I_3$ has been evaluated here and it is equal to
\begin{align} I_3=&\ \frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac25\ln^52+\frac{\pi^2}3\ln^32-\frac{21}4\zeta(3)\ln^22\\&-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right).\tag7 \end{align}
Thus, putting altogether we obtain
$$ I=\color{blue}{\small{\frac{2}{15}\ln^5 2-\frac{2}{3}\zeta(2)\ln^3 2 +\frac{7}{4}\zeta(3)\ln^2 2−\frac{3}{8}\zeta(2)\zeta(3) -\frac{7}{2}\zeta(5)+4\operatorname{Li}_4\left(\frac12\right)\ln2+4\operatorname{Li}_5\left(\frac{1}{2}\right)}}. $$
$$ \large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}} $$