What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$? Some time ago I asked How to find $\displaystyle{\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$.
Thanks to great effort of several MSE users, we now know that
\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln x}xdx=&\,\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)\\&\,-\frac{21}4\zeta(3)\ln^22-12\operatorname{Li}_4\!\left(\tfrac12\right)\ln2-12\operatorname{Li}_5\!\left(\tfrac12\right)\tag1
\end{align}
Now, a natural follow-up to that question is to bump the power of the logarithm and to ask:

Question: What is a closed form for the next integral?
  $$I=\int_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx\tag2$$

I think it is likely that $I$ has a closed form, because there are several very similar integrals having known closed forms:
$$\int_0^1\frac{\ln^2(1+x)\,\ln^2x}xdx=\frac{\pi^2\,\zeta(3)}3-\frac{29\,\zeta(5)}8\tag3$$

$$\int_0^1\frac{\ln^3(1-x)\,\ln^2x}xdx=12\zeta^2(3)-\frac{23\pi^6}{1260}\tag4$$

\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln^2x}{x^2}dx=&\,\frac{3\zeta(3)}2+2\pi^2\zeta(3)+\frac{3\zeta(5)}2-\frac{21\zeta(3)}2\ln^22\\&\,-\frac{63\zeta(3)}2\ln2+\frac{23\pi^4}{60}-\frac{4\ln^52}5-\frac{3\ln^42}2\\&\,-4\ln^32+\frac{2\pi^2}3\ln^32+\frac{3\pi^2}2\ln^22-24\operatorname{Li}_5\!\left(\tfrac12\right)\\&\,-36\operatorname{Li}_4\!\left(\tfrac12\right)-24\operatorname{Li}_4\!\left(\tfrac12\right)\ln2\tag5
\end{align}
 A: This is not going to be a complete answer but since this kind of approach has not been presented here yet and since I believe that it can be brought to a successful completion given enough time under disposal( which I lack now) I present the approach now.
Denote:
\begin{eqnarray}
{\mathcal I}^{(2,3)} := \int\limits_0^1 \frac{\log(\xi)^2 \log(1+\xi)^3}{\xi} d\xi
\end{eqnarray}
Then we have:
\begin{eqnarray}
&&{\mathcal I}^{(2,3)} = \left. \frac{\partial^2}{\partial \theta_1^2}  \frac{\partial^3}{\partial \theta_2^3} \int\limits_0^1 \xi^{\theta_1-1} (1+\xi)^{\theta_2} d\xi \right|_{\theta_1=\theta_2=0} \\
&&= \left. \frac{\partial^2}{\partial \theta_1^2}  \frac{\partial^3}{\partial \theta_2^3}
\left[\sum\limits_{l=0}^\infty \frac{ (\theta_2)_{(l)} }{ \theta_1^{(l+1)} } \cdot 2^{\theta_2-l} (-1)^l \right] 
\right|_{\theta_1=\theta_2=0}\\
&&= \sum\limits_{l=1}^\infty 
\left(\log(2)^2 + \frac{\log(4)}{l} + \frac{2}{l^2} + [H_l]^2 - H_l^{(2)}-\frac{2}{l} H_l - 2 \log(2) H_l \right)\cdot \\
&&\left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l\cdot  2^l} \right)
\end{eqnarray}
The first line is straightforward. In the second line we computed the integral in question by integrating by parts. Finally in the last line we computed the partial derivatives using Higher order derivatives of the binomial factor and the chain rule. Now, the sums look scary but it appears that those sums have actually a much simpler integrals representation than the original integral we want to compute. As a matter of fact the following holds:
\begin{eqnarray}
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) x^l = -\int\limits_0^1 \frac{x}{1-\xi x} \cdot [\log(1-\xi)]^3 d\xi
\end{eqnarray}
Using the generation function above we compute the harmonic sums in question. We have:
\begin{eqnarray}
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) \cdot \frac{1}{2^l} &=& \frac{21}{4} \zeta(4)\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} &=& -\frac{3 \pi ^2 \zeta (3)}{8}+12 \zeta (5)-\frac{7}{120} \pi ^4 \log (2)\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^3} \right) \cdot \frac{1}{2^l} &=& -\int\limits_0^1 \frac{1}{\xi} Li_2(\frac{\xi}{2}) \log(1-\xi)^3 d\xi\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) \cdot \frac{1}{2^l} \cdot H_l &=& -\frac{7}{8} \pi^2 \zeta(3) + \frac{279}{16} \zeta(5)\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} \cdot H_l &=& \int\limits_{0}^1 \frac{Li_2(-\xi)}{\xi(1+\xi)} \cdot [\log(\frac{1-\xi}{1+\xi})]^3 d\xi\\
\sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} \cdot \left([H_l]^2-H_l^{(2)}\right) &=&
-12 \left( \zeta(-4,1,1)-\zeta(4,-1,1)\right)- \frac{1}{8} \left(\pi^4 \log(2) + 14 \pi^2 \zeta(3) - 279 \zeta(5)\right) 
\end{eqnarray}
It is clear that the remaining sums are more complicated and more time is required to bring this thread to completion. We will finish this work as soon as possible.
A: partial solution
using the following identity: ( I can provide the proof if needed)
$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
replace $x$ with $-x$ , then multiply both sides by $\ln^3x$ and integrate from $0$ to $1$, we have
\begin{align}
I&=\int_0^1\frac{\ln^2(1+x)\ln^3x}{1+x}\ dx=\sum_{n=1}^\infty (-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^n\ln^3x\ dx\\
&=-6\sum_{n=1}^\infty \frac{(-1)^n}{(n+1)^4}\left(H_n^2-H_n^{(2)}\right)=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n-1}^2-H_{n-1}^{(2)}\right)\\
&=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n}^2-H_{n}^{(2)}-2\frac{H_n}{n}+\frac2{n^2}\right)\\
&=6\left(\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^4}-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}-\frac{31}{16}\zeta(6)\right)
\end{align}
