How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$ Please help me to find a closed form for this integral:
$$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$
I suspect it might exist because there are similar integrals having closed forms:
$$\begin{align}\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx&=12\zeta(5)-\pi^2\zeta(3)\tag2\\
\int_0^1\frac{\ln^2(1+x)\ln x}x\mathrm dx&=\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\tfrac12\right)\tag3\\
\int_0^1\frac{\ln^3(1+x)\ln x}{x^2}\mathrm dx&=\frac34\zeta(3)-\frac{63}4\zeta(3)\ln2+\frac{23\pi^4}{120}\\&-\frac34\ln^42-2\ln^32+\frac{3\pi^2}4\ln^22-18\operatorname{Li}_4\!\left(\tfrac12\right).\tag4\end{align}$$
Thanks!
 A: Indeed, there is a closed form for this integral:
$$I=\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac{21}4\zeta(3)\ln^22\\-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right).$$
A: An alternate form of the answers given by @Cleo and @Tunk-Fey as sum of $1$ and $1/2$ argumented polylogarithm-products with rational coefficients:
$$I = \frac{99}{16}\operatorname{Li}_5(1)-12\operatorname{Li}_5\left(\frac{1}{2}\right) + 15\operatorname{Li}_1\left( \frac{1}{2} \right)\operatorname{Li}_4(1) - 12\operatorname{Li}_1\left(\frac{1}{2}\right)\operatorname{Li}_4\left(\frac{1}{2}\right) - 15\operatorname{Li}_2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)-\frac{51}{4}\operatorname{Li}_1^2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)+12\operatorname{Li}_2(1)\operatorname{Li}_3\left( \frac{1}{2} \right) - \frac{2}{5}\operatorname{Li}_1^5\left(\frac{1}{2}\right),$$
where $\operatorname{Li}_n$ is the polylogarithm function, and specifically
$$\begin{align}
& \operatorname{Li}_5(1) \ \ \ =  \zeta(5) \\
& \operatorname{Li}_5\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^5} \\
& \operatorname{Li}_4(1) \ \ \ = \zeta(4) = \frac{\pi^4}{90} \\
& \operatorname{Li}_4\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^4} \\
& \operatorname{Li}_3(1) \ \ \ =  \zeta(3) \\
& \operatorname{Li}_3\left(\textstyle\frac{1}{2}\right) = \frac{7}{8} \zeta(3) - \frac{\pi^2}{12} \ln 2 + \frac{1}{6} \ln^3 2 \\
& \operatorname{Li}_2(1) \ \ \ =  \zeta(2) = \frac{\pi^2}{6} \\
& \operatorname{Li}_2\left(\textstyle\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{1}{2} \ln^2 2 \\
& \operatorname{Li}_1\left(\textstyle\frac{1}{2}\right) = \ln2,
\end{align}$$
where $\zeta$ is the Riemann zeta function.
A: UPDATE: The way below may be found in the preprint, A new perspective on the evaluation of the logarithmic integral, $\int_0^1\frac{\log(x)\log^3(1+x)}{x}\textrm{d}x$ by C.I.Valean.

A magical way proposed by Cornel Ioan Valean

We use the powerful form of the Beta function presented in the book, (Almost) Impossible Integrals, Sums, and Series, $\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} \textrm{d}x = \operatorname{B}(a,b)$, (see pages $72$-$73$).
Here is the magic ...
By cleverly differentiating in two different ways to get rid of a nasty integral, we simply get the wonderful result
$$4\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^3 \partial b}\operatorname{B}(a,b)-6\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^2 \partial b^2}\operatorname{B}(a,b)$$
$$=8\int_0^1 \frac{\log(x)\log^3(1+x)}{x}\textrm{d}x-4\int_0^1 \frac{\log^3(x)\log(1+x)}{x}\textrm{d}x-4\int_0^1 \frac{\log^4(1+x)}{x}\textrm{d}x.$$
... and we're wonderfully done!

A first note: a similar strategy has been used in this answer https://math.stackexchange.com/q/3531878.
A BIG BONUS (the extraction of the series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}$):
The extraction of the series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}$ is achieved immediately by observing that using the same Beta function limits, we arrive at

$$\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^3 \partial b}\operatorname{B}(a,b)-\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^2 \partial b^2}\operatorname{B}(a,b)$$
$$=\underbrace{\int_0^1 \frac{\log^2(x)\log^2(1+x)}{x}\textrm{d}x}_{\displaystyle 15/4\zeta(5)-4\sum_{n=1}^{\infty} (-1)^{n-1} H_n/n^4}-\int_0^1 \frac{\log^3(x)\log(1+x)}{x}\textrm{d}x,$$
which assures the desired extraction after turning the second integral into the series we want to calculate.

A: Just a partial answer for now.
We have:
$$ I = -\frac{3}{2}\int_{0}^{1}\frac{\log^2(1+x)\log^2 x}{1+x}\,dx$$
and since:
$$\log(1+z)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}z^n$$
it follows that:
$$ [z^N]\log^2(1+z)=(-1)^{N+1}\sum_{n=1}^{N-1}\frac{1}{n(N-n)}=(-1)^{N+1}\frac{2H_{N-1}}{N},$$
$$\log^2(1+z)=\sum_{n=1}^{+\infty}\frac{2(-1)^{n+1} H_{n-1}}{n}z^{n}.\tag{1}$$
Let we focus now on:
$$J_n = \int_{0}^{1}\frac{x^n\log^2 x}{1+x}\,dx=\frac{\partial^2}{\partial n^2}\int_{0}^{1}\frac{x^n}{1+x}\,dx.$$
We have:
$$ J_n = \frac{1}{4}\left(H_{n/2}^{(3)}-H_{(n-1)/2}^{(3)}\right),$$
hence:
$$ \color{blue}{I = -\frac{3}{4}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}H_{n-1}\left(H_{n/2}^{(3)}-H_{(n-1)/2}^{(3)}\right)}{n}}.\tag{2}$$
or, by partial summation:
$$ \color{purple}{I=-\frac{3}{4}\sum_{n=1}^{+\infty}H_{n/2}^{(3)}(-1)^n\left(\frac{H_n}{n+1}+\frac{H_{n-1}}{n}\right).}\tag{3}$$
Another identity that follows from the Taylor series of $\log^3(1-z)$ is:
$$\color{red}{I=3\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}\left(H_n^2-H_n^{(2)}\right)}{(n+1)^3}.}\tag{4}$$
A: Related problems and techniques: (I), (II). Here is a different form of solution

$$ I = -3\sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{(-1)^k{ n\brack k}k(k-1) }{(n+1)^3n!} ,$$

where $ {n \brack k} $ is the Stirling numbers of the first kind.
