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\newcommand{\iff}{\Longleftrightarrow}
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$\ds{I \equiv \int_{0}^{\infty}x^{2}\expo{-x^{2}}{\rm erf}\pars{x}\ln\pars{x}
\,\dd x}$. Let's
$\ds{{\cal I}\pars{\mu}
\equiv \int_{0}^{\infty}x^{\mu}\expo{-x^{2}}{\rm erf}\pars{x}\,\dd x}$ such that $\ds{I = \lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}}$
Since
$\ds{{\rm erf}\pars{x}
\stackrel{{\rm def.}}{=}{2 \over \root{\pi}}\int_{0}^{x}\expo{-y^{2}}\,\dd y}$:
\begin{align}
{\cal I}\pars{\mu}&=\int_{0}^{\infty}x^{\mu}\expo{-x^{2}}
{2 \over \root{\pi}}\int_{0}^{\infty}\Theta\pars{x - y}\expo{-y^{2}}\,\dd y\,\dd x
\\[3mm]&=
{2 \over \root{\pi}}\int_{0}^{\pi/2}\dd\theta\,\cos^{\mu}\pars{\theta}
\Theta\pars{\cos\pars{\theta} - \sin\pars{\theta}}
\overbrace{\int_{0}^{\infty}\dd r\,r^{\mu + 1}\expo{-r^{2}}}
^{\Gamma\pars{1 + \mu/2}/2}
\\[3mm]&={1 \over \root{\pi}}\,\Gamma\pars{1 + {\mu \over 2}}
\int_{0}^{\pi/4}\dd\theta\,\cos^{\mu}\pars{\theta}
\end{align}
\begin{align}
I&=\lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}=
{1 \over 2\root{\pi}}\,\overbrace{\Psi\pars{2}}^{\ds{1 - \gamma}}\
\overbrace{\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}}^{\ds{\pars{\pi + 2}/8}}
\\[3mm]&+
{1 \over \root{\pi}}\,\overbrace{\Gamma\pars{2}}^{\ds{1}}
\overbrace{%
\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}\ln\pars{\cos\pars{\theta}}}
^{\ds{\braces{4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}}/16}}
\end{align}
$\Gamma\pars{z}$ and $\Psi\pars{z}$ are the $\it Gamma$ and $\it Digamma$ functions, respectively. $\gamma$ and $G$ are the $\it Euler-Mascheroni$ and Catalan constants, respectively.
$$
\begin{array}{|l|}\hline \mbox{}\\
\quad{\displaystyle\int_{0}^{\infty}x^{2}\expo{-x^{2}}
\,\mathrm{erf}\pars{x}\ln\pars{x}\,\dd x}\quad
\\[2mm] =
\quad{{\displaystyle\quad\pars{\pi + 2}\pars{1 - \gamma} + 4G + \pi -
2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}\quad} \over
{\displaystyle 16\root{\pi}}}\quad
\\ \mbox{}\\ \hline
\end{array}
\approx 0.0436462
$$
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
Best Answer
By the substitution $x\mapsto x^{-1}$ it is not hard to see that $$I(\nu^{-1},1)=\int^\infty_0\frac{\arctan^2{x}\arctan(\nu x)}{x^2}{\rm d}x$$
First, start off by re-expressing the integral \begin{align} \int^\pi_0x^2\cos(nx)\cot\left(\frac{x}{2}\right)\ {\rm d}x &=-{\rm Re}\int_C\frac{z+1}{z-1}z^{n-1}\ln^2{z}\ {\rm d}z\\ &=(-1)^{n}\int^1_0\frac{1-x}{1+x}x^{n-1}(\ln^2{x}-\pi^2)\ {\rm d}x-\int^1_0\frac{1+x}{1-x}x^{n-1}\ln^2{x}\ {\rm d}x\\ \end{align} where $C$ is the arc joining $z=1$ to $z=-1$. (The first equality follows from $z=e^{ix}$, whereas the second follows from the fact that the integral along $[-1,\epsilon]\cup\epsilon\exp(i[\pi,0])\cup[\epsilon,1]\cup C$ is $0$ since $z=1$ is a removable singularity and the indent around $z=0$ vanishes.)
Next, note that \begin{align} I_\nu(\nu^{-1},1) &=\int^\frac{\pi}{2}_0\frac{x^2}{\tan{x}(\cos^2{x}+\nu^2\sin^2{x})}{\rm d}x\\ &=\frac{1}{4}\int^\pi_0\frac{x^2}{\tan\left(\frac{x}{2}\right)(1+(1-\nu^2)\cos{x}+\nu^2)}{\rm d}x\\ &=\frac{1}{8\nu}\int^\pi_0\left(x^2\cot\left(\frac{x}{2}\right)+2\sum^\infty_{n=1}\left(\frac{\nu-1}{\nu+1}\right)^nx^2\cos(nx)\cot\left(\frac{x}{2}\right)\right){\rm d}x\\ &=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\int^1_0\frac{1-x}{(1+x)(1+\xi x)}(\ln^2{x}-\pi^2)+\frac{1+x}{(1-x)(1-\xi x)}\ln^2{x}\ {\rm d}x\right) \end{align} Here $\xi=\dfrac{\nu-1}{\nu+1}$. Utilising the partial fraction decompositions $$\frac{1-x}{(1+x)(1+\xi x)}=\frac{\nu+1}{1+x}-\frac{\nu}{1+\xi x}$$ $$\frac{1+x}{(1-x)(1-\xi x)}=\frac{\nu+1}{1-x}-\frac{\nu}{1-\xi x}$$ in tandem with the easily verifiable fact $$\int^1_0\frac{\ln^2{x}}{1+\lambda x}{\rm d}x=-\frac{2{\rm Li}_3(-\lambda)}{\lambda}$$ yields \begin{align} I_\nu(\nu^{-1},1) &=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\frac{\pi^2\nu}{\xi}\ln(1+\xi)-\pi^2(\nu+1)\ln{2}+\frac{7\zeta(3)}{2}(\nu+1)-\frac{4\nu}{\xi}\chi_3(\xi)\right)\\ &=\chi_3\left(\frac{\nu-1}{\nu+1}\right)-\frac{7\zeta(3)}{8}+\frac{\pi^2}{4}\ln\left(1+\frac{1}{\nu}\right) \end{align} where $\displaystyle\chi_s(z)=\sum_{n\ \text{odd}}\frac{z^n}{n^s}=\frac{1}{2}\left({\rm Li}_s(z)-{\rm Li}_s(-z)\right)$ is the Legendre-chi function.
Integrating back, \begin{align} I_\nu(\nu^{-1},1) &=\underbrace{\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)-\frac{7\zeta(3)}{8}\nu}_{\text{Let this be C}}+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_3(v)}{(1+v)^2}{\rm d}v\\ &=C-\left.\frac{2\chi_3(v)}{1+v}\right|^\frac{1-\nu}{1+\nu}_1+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_2(v)}{v(1+v)}{\rm d}v\\ &=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}-\left.\color{white}{\frac{1}{1}}2\chi_2(v)\ln(1+v)\right|^\frac{1-\nu}{1+\nu}_1+\int^\frac{1-\nu}{1+\nu}_1\frac{\ln(1+v)\ln\left(\frac{1+v}{1-v}\right)}{v}{\rm d}v\\ &=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln{2}\\ &\ \ \ \ +\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)-\ln^2(1-v)+\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v \end{align} Repeatedly integrating by parts, it is not hard to derive that \begin{align} \frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)}{v}{\rm d}v =&\ -\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)+\frac{7\zeta(3)}{8}-{\rm Li}_3\left(\frac{1+\nu}{2}\right)+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)\\ &\ +\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)\\ -\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1-v)}{v}{\rm d}v =&\ {\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)\\ \frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v =&\ -2\chi_3(\nu)+2\chi_2(\nu)\ln{\nu}+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right) \end{align} Therefore, we get, for $I(\nu^{-1},1)$, \begin{align} I(\nu^{-1},1) =&\color{brown}{\ (1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-2\chi_3(\nu)+{\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_3\left(\frac{1+\nu}{2}\right)-\frac{7\zeta(3)}{8}\nu}\\ &\ \color{brown}{+2\chi_2(\nu)\ln{\nu}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)}\\ &\ \color{brown}{+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)+\frac{\pi^2}{4}\ln{2}}\\ &\ \color{brown}{+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right)+\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)-\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)}\\ &\ \color{brown}{-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)} \end{align} If no mistakes were made, this formula should hold for $0<\nu\le1$. Further simplifications to the formula may be possible through some polylogarithm identities.