calculus - How to find ${largeint}_0^1frac{ln^3(1+x)ln x}xmathrm dx$

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!

Answer

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

$$\begin{align} I=&\ \frac{\pi^4}{120}\ln2-\frac{33}4\zeta(3)\ln^22+\frac{\pi^2}2\ln^32-\frac{11}{20}\ln^52+6\zeta(5)+\pi^2\zeta(3)\\ &\ +6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}-18\operatorname{Li}_4\left(\frac12\right)\ln2-24\operatorname{Li}_5\left(\frac12\right). \end{align}$$

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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
$$\begin{align} \color{blue}{\mathbf{H}_{4}\left(\frac12\right)}=&\ \frac{\ln^52}{40}-\frac{\pi^2}{36}\ln^32+\frac{\zeta(3)}{2}\ln^22-\frac{\pi^2}{12}\zeta(3)\\&+\frac{\zeta(5)}{32}-\frac{\pi^4}{720}\ln2+\operatorname{Li}_4\left(\frac12\right)\ln2+2\operatorname{Li}_5\left(\frac12\right).\tag5 \end{align}$$

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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.

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References :

$[1]\ $ Harmonic number

$[2]\ $ Polylogarithm