Start with integration by parts (IBP) by setting $u=\ln^3(1+x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields
\begin{align}
I&=-\frac32\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}\ dx\\
&=-\frac32\int_1^2\frac{\ln^2x\ln^2 (x-1)}{x}\ dx\quad\Rightarrow\quad\color{red}{x\mapsto1+x}\\
&=-\frac32\int_{\large\frac12}^1\left[\frac{\ln^2x\ln^2 (1-x)}{x}-\frac{2\ln^3x\ln(1-x)}{x}+\frac{\ln^4x}{x}\right]\ dx\quad\Rightarrow\quad\color{red}{x\mapsto\frac1x}\\
&=-\frac32\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\left.\frac3{10}\ln^5x\right|_{\large\frac12}^1\\
&=-\frac32\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\frac3{10}\ln^52.
\end{align}
Applying IBP again to evaluate the red integral by setting $u=\ln^2(1-x)$ and $dv=\dfrac{\ln^2 x}{x}\ dx$ yields
\begin{align}
\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}&=\frac13\ln^52+\frac23\color{blue}{\int_{\large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx}.
\end{align}
For the simplicity, let
$$
\color{blue}{\mathbf{H}_{m}^{(k)}(x)}=\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^m}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=\sum_{n=1}^\infty H_{n}x^n,
$$
Introduce a generating function for the generalized harmonic numbers for $|x|<1$
$$
\color{blue}{\mathbf{H}^{(k)}(x)}=\sum_{n=1}^\infty H_{n}^{(k)}x^n=\frac{\operatorname{Li}_k(x)}{1-x}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=-\frac{\ln(1-x)}{1-x}
$$
and the following identity
$$
H_{n+1}^{(k)}-H_{n}^{(k)}=\frac1{(n+1)^k}\qquad\Rightarrow\qquad H_{n+1}-H_{n}=\frac1{n+1}
$$
Let us integrating the indefinite form of the blue integral.
\begin{align}
\color{blue}{\int\frac{\ln^3x\ln (1-x)}{1-x}\ dx}=&-\int\sum_{n=1}^\infty H_nx^n\ln^3x\ dx\\
=&-\sum_{n=1}^\infty H_n\int x^n\ln^3x\ dx\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\int x^n\ dx\right]\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\frac{x^{n+1}}{n+1}\right]\\
=&-\sum_{n=1}^\infty H_n\left[\frac{x^{n+1}\ln^3x}{n+1}-\frac{3x^{n+1}\ln^2x}{(n+1)^2}+\frac{6x^{n+1}\ln x}{(n+1)^3}-\frac{6x^{n+1}}{(n+1)^4}\right]\\
=&-\ln^3x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{n+1}+\ln^3x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^2}+3\ln^2x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^2}\\&-3\ln^2x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^3}-6\ln x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^3}+6\ln x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^4}\\&+6\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^4}-6\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^5}\\
=&\ -\sum_{n=1}^\infty\left[\frac{H_nx^{n}\ln^3x}{n}-\frac{x^{n}\ln^3x}{n^2}-\frac{3H_nx^{n}\ln^2x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}\right.\\& \left.\ +\frac{6H_nx^{n}\ln x}{n^3}-\frac{6x^{n}\ln x}{n^4}-\frac{6H_nx^{n}}{n^4}+\frac{6x^{n}}{n^5}\right]\\
=&\ -\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+\operatorname{Li}_2(x)\ln^3x+3\color{blue}{\mathbf{H}_{2}(x)}\ln^2x-3\operatorname{Li}_3(x)\ln^2x\\&\ -6\color{blue}{\mathbf{H}_{3}(x)}\ln x+6\operatorname{Li}_4(x)\ln x+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x).
\end{align}
Therefore
\begin{align}
\color{blue}{\int_{\Large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx}
=&\ 6\color{blue}{\mathbf{H}_{4}(1)}-6\operatorname{Li}_5(1)-\left[\color{blue}{\mathbf{H}_{1}\left(\frac12\right)}\ln^32-\operatorname{Li}_2\left(\frac12\right)\ln^32\right.\\&\left.\ +3\color{blue}{\mathbf{H}_{2}\left(\frac12\right)}\ln^22-3\operatorname{Li}_3\left(\frac12\right)\ln^22+6\color{blue}{\mathbf{H}_{3}\left(\frac12\right)}\ln 2\right.\\&\ -6\operatorname{Li}_4(x)\ln 2+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x)\bigg]\\
=&\ 12\zeta(5)-\pi^2\zeta(3)+\frac{3}8\zeta(3)\ln^22-\frac{\pi^4}{120}\ln2-\frac{1}
{4}\ln^52\\&\ -6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}+6\operatorname{Li}_4\left(\frac12\right)\ln 2+6\operatorname{Li}_5\left(\frac12\right).
\end{align}
Using the similar approach as calculating the blue integral, then
\begin{align}
\int\frac{\ln^3x\ln (1-x)}{x}\ dx&=-\int\sum_{n=1}^\infty \frac{x^{n-1}}{n}\ln^3x\ dx\\
&=-\sum_{n=1}^\infty \frac{1}{n}\int x^{n-1}\ln^3x\ dx\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\int x^{n-1}\ dx\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\frac{x^{n}}{n}\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\left[\frac{x^{n}\ln^3x}{n}-\frac{3x^{n}\ln^2x}{n^2}+\frac{6x^{n}\ln x}{n^3}-\frac{6x^{n}}{n^4}\right]\\
&=\sum_{n=1}^\infty \left[-\frac{x^{n}\ln^3x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}-\frac{6x^{n}\ln x}{n^4}+\frac{6x^{n}}{n^5}\right]\\
&=6\operatorname{Li}_5(x)-6\operatorname{Li}_4(x)\ln x+3\operatorname{Li}_3(x)\ln^2x-\operatorname{Li}_2(x)\ln^3x.
\end{align}
Hence
$$
\int_{\large\frac{1}{2}}^1\frac{\ln^3x\ln (1-x)}{x}\ dx=\frac{\pi^2}{6}\ln^32-\frac{21}{8}\zeta(3)\ln^22-6\operatorname{Li}_4\left(\frac{1}{2}\right)\ln2-6\operatorname{Li}_5\left(\frac{1}{2}\right)+6\zeta(5).
$$
Combining altogether, we have
\begin{align}
I=&\ \frac{\pi^4}{120}\ln2-\frac{33}4\zeta(3)\ln^22+\frac{\pi^2}2\ln^32-\frac{11}{20}\ln^52+6\zeta(5)+\pi^2\zeta(3)\\
&\ +6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}-18\operatorname{Li}_4\left(\frac12\right)\ln2-24\operatorname{Li}_5\left(\frac12\right).
\end{align}
Continuing my answer in: A sum containing harmonic numbers $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, we have
\begin{align}
\color{blue}{\mathbf{H}_{3}\left(x\right)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag1
\end{align}
Dividing $(1)$ by $x$ and then integrating yields
$$\small\begin{align}
\color{blue}{\mathbf{H}_{4}\left(x\right)}=&\frac14\zeta(3)\ln^2 x-\frac18\int\frac{\ln^2x\ln^2(1-x)}x\ dx+\frac12\int\frac{\ln x}x\bigg[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\bigg]\ dx\\&+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac12\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx+\frac{\pi^4}{60}\ln x\\
=&\frac14\zeta(3)\ln^2 x+\frac{\pi^4}{60}\ln x+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac18\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}\\&+\frac12\left[\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}-\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}-\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}\right].\tag2
\end{align}$$
Evaluating the red integral using the same technique as the previous one yields
\begin{align}
\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}&=\frac13\ln^3x\ln^2(1-x)-\frac23\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}
Evaluating the purple integral yields
\begin{align}
\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\frac{\partial}{\partial n}\left[\int x^{n-1}\ dx\right]\\
&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\left[\frac{x^n\ln x}{n}-\frac{x^n}{n^2}\right]\\
&=\color{blue}{\mathbf{H}_{3}(x)}\ln x-\color{blue}{\mathbf{H}_{4}(x)}.
\end{align}
Evaluating the green integral using IBP by setting $u=\ln x$ and $dv=\dfrac{\operatorname{Li}_3(x)}{x}\ dx$ yields
\begin{align}
\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}&=\operatorname{Li}_4(x)\ln x-\int\frac{\operatorname{Li}_4(x)}x\ dx\\
&=\operatorname{Li}_4(x)\ln x-\operatorname{Li}_5(x).
\end{align}
Evaluating the orange integral using IBP by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields
\begin{align}
\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}&=\frac12\operatorname{Li}_3(1-x)\ln^2 x+\frac12\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}.
\end{align}
Applying IBP again to evaluate the maroon integral by setting $u=\operatorname{Li}_2(1-x)$ and
$$
dv=\dfrac{\ln^2 x}{1-x}\ dx\quad\Rightarrow\quad
v=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x,
$$
we have
$$\small{\begin{align}
\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}=&\left[2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)\\
&-2\int\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx+2\int\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ dx+\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}}$$
We use the generating function for the generalized harmonic numbers evaluate the above integrals involving polylogarithm.
\begin{align}
\int\frac{\operatorname{Li}_k(x)\ln x}{1-x}\ dx&=\sum_{n=1}^\infty H_{n}^{(k)}\int x^n\ln x\ dx\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\frac{\partial}{\partial n}\left[\int x^n\ dx\right]\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\left[\frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^2}\right]\\
&=\sum_{n=1}^\infty\left[\frac{H_{n+1}^{(k)}x^{n+1}\ln x}{n+1}-\frac{x^{n+1}\ln x}{(n+1)^{k+1}}-\frac{H_{n+1}^{(k)}x^{n+1}}{(n+1)^2}+\frac{x^{n+1}}{(n+1)^{k+2}}\right]\\
&=\sum_{n=1}^\infty\left[\frac{H_{n}^{(k)}x^{n}\ln x}{n}-\frac{x^{n}\ln x}{n^{k+1}}-\frac{H_{n}^{(k)}x^{n}}{n^2}+\frac{x^{n}}{n^{k+2}}\right]\\
&=\color{blue}{\mathbf{H}_{1}^{(k)}(x)}\ln x-\operatorname{Li}_{k+1}(x)\ln x-\color{blue}{\mathbf{H}_{2}^{(k)}(x)}+\operatorname{Li}_{k+2}(x).
\end{align}
Dividing generating function of $\color{blue}{\mathbf{H}^{(k)}(x)}$ by $x$ and then integrating yields
\begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n}&=\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{1}^{(k)}(x)}&=\int\frac{\operatorname{Li}_k(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx\\
&=\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx.
\end{align}
Repeating the process above yields
\begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^2}
&=\int\frac{\operatorname{Li}_{k+1}(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{2}^{(k)}(x)}&=\operatorname{Li}_{k+2}(x)+\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx,
\end{align}
where it is easy to show by using IBP that
\begin{align}
\int\frac{\operatorname{Li}_2(x)}{1-x}\ dx&=-\int\frac{\operatorname{Li}_2(1-x)}{x}\ dx\\
&=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln(x)-\operatorname{Li}_2(1-x)\ln x-\ln (1-x)\ln^2x
\end{align}
and
$$
\int\frac{\operatorname{Li}_3(x)}{1-x}\ dx=-\int\frac{\operatorname{Li}_3(1-x)}{x}\ dx=-\frac12\operatorname{Li}_2^2(1-x)-\operatorname{Li}_3(1-x)\ln x.
$$
Now, all unknown terms have been obtained. Putting altogether to $(2)$, we have
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+C.\tag3
\end{align}}$$
The next step is finding the constant of integration. Setting $x=1$ to $(3)$ yields
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(1)}
&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac65\operatorname{Li}_5(1)-\frac15\operatorname{Li}_4(1)-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(1)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(1)}+C\\
3\zeta(5)+\zeta(2)\zeta(3)&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac{19}{30}\operatorname{Li}_5(1)+\frac{3}{5}\operatorname{Li}_3(1)+C\\
C&=\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5).
\end{align}}$$
Thus
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\tag4
\end{align}}$$
and setting $x=\frac12$ to $(4)$ yields
\begin{align}
\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}=&\ \frac{\ln^52}{40}-\frac{\pi^2}{36}\ln^32+\frac{\zeta(3)}{2}\ln^22-\frac{\pi^2}{12}\zeta(3)\\&+\frac{\zeta(5)}{32}-\frac{\pi^4}{720}\ln2+\operatorname{Li}_4\left(\frac12\right)\ln2+2\operatorname{Li}_5\left(\frac12\right).\tag5
\end{align}
Finally, we obtain
\begin{align}
\int_0^1\frac{\ln^3(1+x)\ln x}x\ dx=&\ \color{blue}{\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}\\&\color{blue}{-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right)},
\end{align}
which again matches @Cleo's answer.
References :
$[1]\ $ Harmonic number
$[2]\ $ Polylogarithm
The value of $I$ is a $\mathbb{Q}$-linear combination of values of the multiple polylogarithm at rational arguments. I'll explain how to compute this.
Expanding each logarithm in the integrand as an integral, multiplying out, and dividing into regions, and making the substitution $x\leftrightarrow 1-x$, we get that $I$ is a
$\mathbb{Q}$-linear combination of iterated integrals of the form
$$
\int_{1\geq t_1\geq t_2\geq t_3\geq t_4\geq 0} \frac{dt_4}{f_4(t_4)}\frac{dt_3}{f_3(t_3)}\frac{dt_2}{f_2(t_2)}\frac{dt_1}{f_1(t_1)},
$$
where each $f_i(t)$ is either $t$ or $1-wt$ for some $w\in\{1/2,2/3\}$.
Claim: each iterated integral of this form is a value of the multiple polylogarithm, defined by
$$
Li_{s_1,\ldots,s_k}(z_1,\ldots,z_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{z_1^{n_1}\ldots z_k^{n_k}}{n_1^{s_1}\ldots n_k^{s_k}}.
$$
For $k=1$ this is the ordinary polylogarithm, and $Li_{s_1,\ldots,s_k}(1,\ldots,1)=\zeta(s_1,\ldots,s_k)$ is the multiple zeta value.
The claim isn't too hard to see by induction on the number of terms in the iterated integral: we have
$$
\int_0^{z_1} \frac{Li_{s_1,\ldots,s_k}(t,z_2,\ldots,z_k)}{t}\,dt=Li_{s_1+1,\ldots,s_k}(z_1,z_2,\ldots,z_k),
$$
$$
\int_0^{z_1} \frac{Li_{s_1,\ldots,s_k}(t,z_2,\ldots,z_k)}{1-wt}\,dt=\frac{1}{w}Li_{1,s_1,\ldots,s_k}(wz_1,1/w,z_2,z_3,\ldots,z_k).
$$
(I hope I wrote this all out correctly)
Values of multiple polylogarithms satisfy many relations, so it's possible the expression one gets can be simplified.
Iterated integrals like this show up when computing the action of parallel transport on algebraic vector bundles with nilpotent connection on open subsets of $\mathbb{P}^1$. It's not so hard to write down such a thing on $\mathbb{P}^1\backslash\{1,-1,-1/2,\infty\}$ giving $I$ as a matrix coefficient for transport along $[0,1]$.
There's a thing called the unipotent fundamental group of a variety, which has the structure of a motive. Without getting into what exactly this is, I'll just say that the observation about parallel transport essentially amounts to $I$ being a period of $\pi_{1,\cdot}(\mathbb{P}^1\backslash\{1,-1,-1/2,\infty\})$. One doesn't get a good model of $X:=\mathbb{P}^1\backslash\{1,-1,-1/2,\infty\}$ over $\mathbb{Z}$ because the removed points collide mod 2 and mod 3, but there is a good model over $\mathbb{Z}[1/6]$. It is known that the fundamental group of a rational curve has the structure of a mixed Tate motive, so $I$ is the period of a mixed Tate motive over $\mathbb{Z}[1/6]$. I don't really understand the construction of mixed Tate motives, so I'm just viewing it as a black box. Probably someone who understood them better than I could see directly that $I$ is the period of a mixed Tate motive without thinking about $\pi_1$.
For comparison: if the only denominators appearing in the iterated integral were $t$ and $1-t$, then the value of the integral is a multiple zeta value. These numbers are periods of the fundamental group of $\mathbb{P}^1\backslash\{0,1,\infty\}$, which is a mixed Tate motive over $\mathbb{Z}$. It's a theorem that the space of all periods of mixed Tate motives over $\mathbb{Z}$ is the $\mathbb{Q}[(2\pi i)^{-1}]$ span of the multiple zeta values. I think in the case of $I$ the mixed Tate motive we need is only defined over $\mathbb{Z}[1/6]$, so that $I$ can't necessarily be written in terms of multiple zeta values.
There's a conjecture that all periods of mixed Tate motives over any ring $\mathbb{Z}[1/N]$ are linear combinations of values of multiple polylogarithms.
Best Answer
This answer is split into 3 main steps.
Step 1: Expressing the integral as a sum
\begin{align} &\ \ \ \ \ \int^1_0\ln(1+x)\ln(1-x)\ln^2{x} \ {\rm d}x\\ &=\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{k}\int^1_0x^{j+k}\ln^2{x} \ {\rm d}x\\ &=2\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{k(k+j+1)^3}\\ &=\small{2\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{(j+1)^3k}-\frac{1}{(j+1)^3(k+j+1)}-\frac{1}{(j+1)^2(k+j+1)^2}-\frac{1}{(j+1)(k+j+1)^3}}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}-2\sum^\infty_{j=1}\frac{(-1)^j\left[\zeta(2)-H_{j+1}^{(2)}\right]}{j(j+1)^2}-2\sum^\infty_{j=1}\frac{(-1)^j\left[\zeta(3)-H_{j+1}^{(3)}\right]}{j(j+1)} \end{align}
Step 2a: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n}$ \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n} &=\frac{1}{2}\ln^2{2}-\frac{\pi^2}{12} \end{align} See here for the details.
Step 2b: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2}$ \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2} &=-\frac{5}{8}\zeta(3) \end{align} See here for the details.
Step 2c: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}$ \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3} &=\int^{-1}_0\frac{1}{y}\left[\int^y_0\frac{1}{x}\left[\int^x_0\frac{\ln(1-t)}{t(t-1)}{\rm d}t\right]{\rm d}x\right]{\rm d}y\\ &=2{\rm Li}_4\left(\frac{1}{2}\right)-\frac{11\pi^4}{360}+\frac{1}{12}\ln^4{2}+\frac{7}{4}\zeta(3)\ln{2}-\frac{\pi^2}{12}\ln^2{2} \end{align} Tunk-Fey did a calculation of this type here.
Step 2d: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n}$ \begin{align} \sum^\infty_{n=1}\frac{H_n^{(2)}}{n}x^n &=\int^x_0\frac{{\rm Li}_2(t)}{t(1-t)}{\rm d}t\\ &={\rm Li}_3(x)+\int^x_0\frac{{\rm Li}_2(t)}{1-t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\int^x_0\frac{\ln^2(1-t)}{t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)+\int^{1-x}_1\frac{\ln^2{t}}{1-t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}+\int^{1-x}_1\frac{2\ln(1-t)\ln{t}}{t}{\rm d}t\\ &\small{={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln(1-x)+\int^{1-x}_1\frac{2{\rm Li}_2(t)}{t}{\rm d}t}\\ &\small{={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln(1-x)+2{\rm Li}_3(1-x)-2\zeta(3)} \end{align} Therefore \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n} &={\rm Li}_3(-1)-{\rm Li}_2(-1)\ln{2}-\ln^2{2}\ln(-1)-2{\rm Li}_2(2)\ln{2}+2{\rm Li}_3(2)-2\zeta(3)\\ &=-\zeta(3)+\frac{\pi^2}{12}\ln{2} \end{align} You can use polylogarithm identities to simplify the last equation. I took the easy way out and used Wolfram Alpha. Note that contour integration is a slightly more efficient method to solve this sum, however this method is required if I want to solve $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$ as well.
Step 2e: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(3)}}{n}$
\begin{align} \sum^\infty_{n=1}\frac{H_n^{(3)}}{n}x^n &=\int^x_0\frac{{\rm Li}_3(t)}{t(1-t)}{\rm d}t\\ &={\rm Li}_4(x)+\int^x_0\frac{{\rm Li}_3(t)}{1-t}{\rm d}t\\ &={\rm Li}_4(x)-{\rm Li}_3(x)\ln(1-x)-\int^x_0\frac{-\ln(1-t){\rm Li}_2(t)}{t}{\rm d}t\\ &={\rm Li}_4(x)-{\rm Li}_3(x)\ln(1-x)-\frac{1}{2}{\rm Li}^2_2(x) \end{align} Therefore \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n^{(3)}}{n} &={\rm Li}_4(-1)-{\rm Li}_3(-1)\ln{2}-\frac{1}{2}{\rm Li}^2_2(-1)\\ &=-\frac{19\pi^4}{1440}+\frac{3}{4}\zeta(3)\ln{2} \end{align}
Step 2f: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$
This part is rather similar to Tunk-Fey's answer, so he certainly deserves credit. \begin{align} &\ \ \ \ \ \sum^\infty_{n=1}\frac{H_n^{(2)}}{n^2}x^n\\ &=\small{{\rm Li}_4(x)-2\zeta(3)\ln{x}+\frac{1}{2}{\rm Li}_2^2(x)+\color{blue}{\int\frac{-\ln^2(1-x)\ln{x}}{x}{\rm d}x}+\color{\orange}{2\int\frac{{\rm Li}_3(1-x)-{\rm Li}_2(1-x)\ln(1-x)}{x}{\rm d}x}} \end{align} The blue integral is \begin{align} &\ \ \ \ \ \color{blue}{\int\frac{-\ln^2(1-x)\ln{x}}{x}{\rm d}x}\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)-\int\frac{\ln^2{x}\ln(1-x)}{1-x}{\rm d}x\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\sum^\infty_{n=1}H_n\int x^n\ln^2{x} \ {\rm d}x\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\sum^\infty_{n=1}H_n\partial^2_n\frac{x^{n+1}}{n+1}\\ &=\color\grey{-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\ln^2{x}\sum^\infty_{n=1}\frac{H_nx^{n+1}}{n+1}}-2\ln{x}\sum^\infty_{n=1}\frac{H_nx^{n+1}}{(n+1)^2}+2\sum^\infty_{n=1}\frac{H_{n}x^{n+1}}{(n+1)^3}\\ &=\color{blue}{2\ln{x}{\rm Li}_3(x)-2{\rm Li}_4(x)-2\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n+2\sum^\infty_{n=1}\frac{H_n}{n^3}x^n} \end{align} The orange integral is \begin{align} &\ \ \ \ \ \ \color{orange}{2\int\frac{{\rm Li}_3(1-x)-{\rm Li}_2(1-x)\ln(1-x)}{x}{\rm d}x}\\ &=2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)+2\int\frac{\ln(1-x)\ln^2{x}}{1-x}{\rm d}x\\ &=\color{orange}{2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)-\ln^2{x}\ln^2(1-x)-4\ln{x}{\rm Li}_3(x)+4{\rm Li}_4(x)+4\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n-4\sum^\infty_{n=1}\frac{H_n}{n^3}x^n} \end{align} So \begin{align} & \ \ \ \ \ \sum^\infty_{n=1}\frac{H_n^{(2)}}{n^2}x^n\\ &=3{\rm Li}_4(x)+2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_3(x)\ln{x}-2\zeta(3)\ln{x}+\frac{1}{2}{\rm Li}_2^2(x)-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)-\ln^2{x}\ln^2(1-x)+2\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n-2\sum^\infty_{n=1}\frac{H_n}{n^3}x^n+C \end{align} Therefore \begin{align} & \ \ \ \ \ \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}\\ &=3{\rm Li}_4(-1)+\color\grey{2{\rm Li}_3(2)\ln(-1)-2{\rm Li}_3(-1)\ln(-1)-2\zeta(3)\ln(-1)}\\ &+\frac{1}{2}{\rm Li}_2^2(-1)\color\grey{-2{\rm Li}_2(2)\ln(-1)\ln(2)-\ln^2(-1)\ln^2{2}+2\ln(-1)\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2}}-2\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}\\ &=\frac{17\pi^4}{480}-4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{1}{6}\ln^4{2}-\frac{7}{2}\zeta(3)\ln{2}+\frac{\pi^2}{6}\ln^2{2} \end{align} The grey terms miraculously cancel.
Step 3a: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}$ \begin{align} & \ \ \ \ \ 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j}-\frac{(-1)^jH_{j+1}}{(j+1)^3}-\frac{(-1)^jH_{j+1}}{(j+1)^2}-\frac{(-1)^jH_{j+1}}{j+1}\\ &=\small{2\sum^\infty_{j=1}\frac{(-1)^jH_{j}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^3}+2+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^2}+2+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^3}+2}\\ &=4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{11\pi^4}{180}+\frac{1}{6}\ln^4{2}+\frac{7}{2}\zeta(3)\ln{2}-\frac{5}{4}\zeta(3)-\frac{\pi^2}{6}\ln^2{2}-\frac{\pi^2}{3}+2\ln^2{2}-4\ln{2}+8 \end{align}
Step 3b: Evaluating $\displaystyle -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}$ \begin{align} -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2} &=-\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j}-\frac{(-1)^j}{(j+1)^2}-\frac{(-1)^j}{j+1}\\ &=\frac{\pi^2}{3}\ln{2}+\frac{\pi^4}{36}-\frac{\pi^2}{3}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{3}\\ &=\frac{\pi^4}{36}+\frac{2\pi^2}{3}\ln{2}-\frac{2\pi^2}{3} \end{align}
Step 3c: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}$ \begin{align} & \ \ \ \ \ 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j}-\frac{(-1)^jH_{j+1}^{(2)}}{(j+1)^2}-\frac{(-1)^jH_{j+1}^{(2)}}{j+1}\\ &=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(2)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}+2\sum^\infty_{j=1}\frac{(-1)^jH_j^{(2)}}{j^2}+2+2\\ &=-8{\rm Li}_4\left(\frac{1}{2}\right)+\frac{17\pi^4}{240}-\frac{1}{3}\ln^4{2}-7\zeta(3)\ln{2}-4\zeta(3)+\frac{\pi^2}{3}\ln^2{2}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{6}-4\ln{2}+8\\ \end{align}
Step 3d: Evaluating $\displaystyle -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}$ \begin{align} -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)} &=-2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j}+2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j+1}\\ &=4\zeta(3)\ln{2}-2\zeta(3)\\ \end{align}
Step 3e: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)}$ \begin{align} 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)} &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j}-2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j+1}\\ &=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(3)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^3}+2\\ &=-\frac{19\pi^4}{360}+3\zeta(3)\ln{2}-\frac{3}{2}\zeta(3)-\frac{\pi^2}{6}-4\ln{2}+8 \end{align}
Step 4: Obtaining the final result
Summing the results from steps 3a, 3b, 3c, 3d and 3e gives $$\int^1_0\ln(1+x)\ln(1-x)\ln^2{x} \ {\rm d}x=24-\frac{4\pi^2}3-\frac{11\pi^4}{720}-12\ln2\\+2\ln^22-\frac16\ln^42+\pi ^2\ln2+\frac{\pi^2}6\ln^22-4\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{35}4\zeta(3)+\frac72\zeta(3)\ln2.$$ hence completing the proof.