复杂的对数积分（一）
\[\Large\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx
\]
\(\Large\mathbf{Solution:}\)
Start with integration by parts (IBP) by setting \(u=\ln^3(1+x)\) and \(\mathrm{d}v=\dfrac{\ln x}{x}\ \mathrm{d}x\) yields
\[\begin{align*}
I&=-\frac32\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}\ \mathrm{d}x\\
&=-\frac32\int_1^2\frac{\ln^2x\ln^2 (x-1)}{x}\ \mathrm{d}x\quad\Rightarrow\quad\color{red}{x\mapsto1+x}\\
&=-\frac32\int_{\large\frac12}^1\left[\frac{\ln^2x\ln^2 (1-x)}{x}-\frac{2\ln^3x\ln(1-x)}{x}+\frac{\ln^4x}{x}\right]\ \mathrm{d}x\quad\Rightarrow\quad\color{red}{x\mapsto\frac1x}\\
&=-\frac32\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ \mathrm{d}x+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ \mathrm{d}x-\left.\frac3{10}\ln^5x\right|_{\large\frac12}^1\\
&=-\frac32\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ \mathrm{d}x}+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ \mathrm{d}x-\frac3{10}\ln^52
\end{align*}\]
Applying IBP again to evaluate the red integral by setting \(u=\ln^2(1-x)\) and \(\mathrm{d}v=\dfrac{\ln^2 x}{x}\ \mathrm{d}x\) yields
\[\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ \mathrm{d}x}=\frac13\ln^52+\frac23\color{blue}{\int_{\large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ \mathrm{d}x}
\]
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.
\[\begin{align*}
\color{blue}{\int\frac{\ln^3x\ln (1-x)}{1-x}\ \mathrm{d}x}=&-\int\sum_{n=1}^\infty H_nx^n\ln^3x\ \mathrm{d}x\\
=&-\sum_{n=1}^\infty H_n\int x^n\ln^3x\ \mathrm{d}x\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\int x^n\ \mathrm{d}x\right]\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\frac{x^{n+1}}{n+1}\right]\\
=&-\sum_{n=1}^\infty H_n\left[\frac{x^{n+1}\ln^3x}{n+1}-\frac{3x^{n+1}\ln^2x}{(n+1)^2}+\frac{6x^{n+1}\ln x}{(n+1)^3}-\frac{6x^{n+1}}{(n+1)^4}\right]\\
=&-\ln^3x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{n+1}+\ln^3x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^2}+3\ln^2x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^2}\\&-3\ln^2x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^3}-6\ln x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^3}+6\ln x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^4}\\&+6\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^4}-6\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^5}\\
=&\ -\sum_{n=1}^\infty\left[\frac{H_nx^{n}\ln^3x}{n}-\frac{x^{n}\ln^3x}{n^2}-\frac{3H_nx^{n}\ln^2x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}\right.\\& \left.\ +\frac{6H_nx^{n}\ln x}{n^3}-\frac{6x^{n}\ln x}{n^4}-\frac{6H_nx^{n}}{n^4}+\frac{6x^{n}}{n^5}\right]\\
=&\ -\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+\operatorname{Li}_2(x)\ln^3x+3\color{blue}{\mathbf{H}_{2}(x)}\ln^2x-3\operatorname{Li}_3(x)\ln^2x\\&\ -6\color{blue}{\mathbf{H}_{3}(x)}\ln x+6\operatorname{Li}_4(x)\ln x+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x)
\end{align*}\]
Therefore
\[\begin{align*}
\color{blue}{\int_{\Large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ \mathrm{d}x}
=&\ 6\color{blue}{\mathbf{H}_{4}(1)}-6\operatorname{Li}_5(1)-\left[\color{blue}{\mathbf{H}_{1}\left(\frac12\right)}\ln^32-\operatorname{Li}_2\left(\frac12\right)\ln^32\right.\\&\left.\ +3\color{blue}{\mathbf{H}_{2}\left(\frac12\right)}\ln^22-3\operatorname{Li}_3\left(\frac12\right)\ln^22+6\color{blue}{\mathbf{H}_{3}\left(\frac12\right)}\ln 2\right.\\&\ -6\operatorname{Li}_4(x)\ln 2+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x)\bigg]\\
=&\ 12\zeta(5)-\pi^2\zeta(3)+\frac{3}8\zeta(3)\ln^22-\frac{\pi^4}{120}\ln2-\frac{1}
{4}\ln^52\\&\ -6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}+6\operatorname{Li}_4\left(\frac12\right)\ln 2+6\operatorname{Li}_5\left(\frac12\right)
\end{align*}\]
Using the similar approach as calculating the blue integral, then
\[\begin{align*}
\int\frac{\ln^3x\ln (1-x)}{x}\ \mathrm{d}x&=-\int\sum_{n=1}^\infty \frac{x^{n-1}}{n}\ln^3x\ \mathrm{d}x\\
&=-\sum_{n=1}^\infty \frac{1}{n}\int x^{n-1}\ln^3x\ \mathrm{d}x\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\int x^{n-1}\ \mathrm{d}x\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\frac{x^{n}}{n}\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\left[\frac{x^{n}\ln^3x}{n}-\frac{3x^{n}\ln^2x}{n^2}+\frac{6x^{n}\ln x}{n^3}-\frac{6x^{n}}{n^4}\right]\\
&=\sum_{n=1}^\infty \left[-\frac{x^{n}\ln^3x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}-\frac{6x^{n}\ln x}{n^4}+\frac{6x^{n}}{n^5}\right]\\
&=6\operatorname{Li}_5(x)-6\operatorname{Li}_4(x)\ln x+3\operatorname{Li}_3(x)\ln^2x-\operatorname{Li}_2(x)\ln^3x
\end{align*}\]
Hence
\[\int_{\large\frac{1}{2}}^1\frac{\ln^3x\ln (1-x)}{x}\ \mathrm{d}x=\frac{\pi^2}{6}\ln^32-\frac{21}{8}\zeta(3)\ln^22-6\operatorname{Li}_4\left(\frac{1}{2}\right)\ln2-6\operatorname{Li}_5\left(\frac{1}{2}\right)+6\zeta(5)
\]
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 the 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
\[\begin{align*}
\color{blue}{\mathbf{H}_{4}\left(x\right)}=&\frac14\zeta(3)\ln^2 x-\frac18\int\frac{\ln^2x\ln^2(1-x)}x\mathrm dx+\frac12\int\frac{\ln x}x\bigg[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\bigg]\ \mathrm dx\\&+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac12\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ \mathrm dx+\frac{\pi^4}{60}\ln x\\
=&\frac14\zeta(3)\ln^2 x+\frac{\pi^4}{60}\ln x+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac18\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ \mathrm dx}\\&+\frac12\left[\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ \mathrm dx}-\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ \mathrm dx}-\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ \mathrm dx}\right]\tag2
\end{align*}\]
Evaluating the red integral using the same technique as the previous one yields
\[\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ \mathrm dx}=\frac13\ln^3x\ln^2(1-x)-\frac23\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ \mathrm dx}
\]
Evaluating the purple integral yields
\[\begin{align*}
\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ \mathrm dx}&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\frac{\partial}{\partial n}\left[\int x^{n-1}\ \mathrm dx\right]\\
&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\left[\frac{x^n\ln x}{n}-\frac{x^n}{n^2}\right]\\
&=\color{blue}{\mathbf{H}_{3}(x)}\ln x-\color{blue}{\mathbf{H}_{4}(x)}
\end{align*}\]
Evaluating the green integral using IBP by setting \(u=\ln x\) and \(\mathrm dv=\dfrac{\operatorname{Li}_3(x)}{x}\mathrm dx\) yields
\[\begin{align*}
\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ \mathrm dx}&=\operatorname{Li}_4(x)\ln x-\int\frac{\operatorname{Li}_4(x)}x\ \mathrm dx\\
&=\operatorname{Li}_4(x)\ln x-\operatorname{Li}_5(x)
\end{align*}\]
Evaluating the orange integral using IBP by setting \(u=\operatorname{Li}_3(1-x)\) and \(\mathrm dv=\dfrac{\ln x}{x}\ \mathrm dx\) yields
\[\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ \mathrm dx}=\frac12\operatorname{Li}_3(1-x)\ln^2 x+\frac12\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ \mathrm dx}
\]
Applying IBP again to evaluate the maroon integral by setting \(u=\operatorname{Li}_2(1-x)\) and
\[\mathrm dv=\dfrac{\ln^2 x}{1-x}\mathrm dx\quad\Rightarrow\quad
v=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x\]
we have
\[{\begin{align*}
\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ \mathrm dx}=&\left[2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)\\
&-2\int\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ \mathrm dx+2\int\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ \mathrm dx+\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ \mathrm dx}
\end{align*}}\]
We use the generating function for the generalized harmonic numbers evaluate the above integrals involving polylogarithm.
\[\begin{align*} \int\frac{\operatorname{Li}_k(x)\ln x}{1-x}\ \mathrm dx&=\sum_{n=1}^\infty H_{n}^{(k)}\int x^n\ln x\ \mathrm dx\\ &=\sum_{n=1}^\infty H_{n}^{(k)}\frac{\partial}{\partial n}\left[\int x^n\ \mathrm dx\right]\\ &=\sum_{n=1}^\infty H_{n}^{(k)}\left[\frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^2}\right]\\ &=\sum_{n=1}^\infty\left[\frac{H_{n+1}^{(k)}x^{n+1}\ln x}{n+1}-\frac{x^{n+1}\ln x}{(n+1)^{k+1}}-\frac{H_{n+1}^{(k)}x^{n+1}}{(n+1)^2}+\frac{x^{n+1}}{(n+1)^{k+2}}\right]\\ &=\sum_{n=1}^\infty\left[\frac{H_{n}^{(k)}x^{n}\ln x}{n}-\frac{x^{n}\ln x}{n^{k+1}}-\frac{H_{n}^{(k)}x^{n}}{n^2}+\frac{x^{n}}{n^{k+2}}\right]\\ &=\color{blue}{\mathbf{H}_{1}^{(k)}(x)}\ln x-\operatorname{Li}_{k+1}(x)\ln x-\color{blue}{\mathbf{H}_{2}^{(k)}(x)}+\operatorname{Li}_{k+2}(x) \end{align*}\]
Dividing generating function of \(\color{blue}{\mathbf{H}^{(k)}(x)}\) by \(x\) and then integrating yields
\[\begin{align*} \sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n}&=\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ \mathrm dx\\ \color{blue}{\mathbf{H}_{1}^{(k)}(x)}&=\int\frac{\operatorname{Li}_k(x)}{x}\ \mathrm dx+\int\frac{\operatorname{Li}_k(x)}{1-x}\ \mathrm dx\\ &=\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ \mathrm dx \end{align*}\]
Repeating the process above yields
\[\begin{align*} \sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^2} &=\int\frac{\operatorname{Li}_{k+1}(x)}{x}\ \mathrm dx+\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ \mathrm dx\\ \color{blue}{\mathbf{H}_{2}^{(k)}(x)}&=\operatorname{Li}_{k+2}(x)+\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ \mathrm dx \end{align*}\]
where it is easy to show by using IBP that
\[\begin{align*} \int\frac{\operatorname{Li}_2(x)}{1-x}\ \mathrm dx&=-\int\frac{\operatorname{Li}_2(1-x)}{x}\ \mathrm dx\\ &=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln(x)-\operatorname{Li}_2(1-x)\ln x-\ln (1-x)\ln^2x \end{align*}\]
and
\[\int\frac{\operatorname{Li}_3(x)}{1-x}\ \mathrm dx=-\int\frac{\operatorname{Li}_3(1-x)}{x}\ \mathrm dx=-\frac12\operatorname{Li}_2^2(1-x)-\operatorname{Li}_3(1-x)\ln x \]
Now, all unknown terms have been obtained. Putting altogether to \((2)\), we have
\[{\begin{align*}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+C\tag3
\end{align*}}\]
The next step is finding the constant of integration. Setting \(x=1\) to \((3)\) yields
\[{\begin{align*}
\color{blue}{\mathbf{H}_{4}(1)}
&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac65\operatorname{Li}_5(1)-\frac15\operatorname{Li}_4(1)-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(1)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(1)}+C\\
3\zeta(5)+\zeta(2)\zeta(3)&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac{19}{30}\operatorname{Li}_5(1)+\frac{3}{5}\operatorname{Li}_3(1)+C\\
C&=\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)
\end{align*}}\]
Thus
\[{\begin{align*}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\tag4
\end{align*}}\]
and setting \(x=\dfrac12\) 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
\[\boxed{\begin{align*}
\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx=&\ \color{blue}{\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac25\ln^52+\frac{\pi^2}3\ln^32-\frac{21}4\zeta(3)\ln^22}\\&\color{blue}{-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right)}
\end{align*}}\]
\(\mathbf{References :}\)
\([1]\) : Harmonic number
\([2]\) : Polylogarithm
\(\Large\mathbf{Here \,\,\,is \,\,\,another \,\,\,way \,\,\,solving \,\,\,this \,\,\,problem\,\,:}\)
Step 1: Expressing the integral as a sum
It is easy to derive the formula
\[\left(\sum^{\infty}_{n=1}a_nx^n\right)\left(\sum^{\infty}_{n=1}b_nx^n\right)=\sum^\infty_{n=1}\sum^{n}_{k=1}a_kb_{n-k+1}x^{n+1}
\]
We apply this formula to derive the Taylor series of \(\ln^2(1+x)\).
\[\begin{align*}
\ln^2(1+x)
&=\left(\sum^{\infty}_{n=1}\frac{(-1)^{n-1}}{n}x^n\right)\left(\sum^{\infty}_{n=1}\frac{(-1)^{n-1}}{n}x^n\right)\\
&=\sum^\infty_{n=1}\sum^n_{k=1}\frac{(-1)^{k-1}(-1)^{n-k}}{k(n-k+1)}x^{n+1}\\
&=\sum^\infty_{n=1}\frac{(-1)^{n+1}}{n+1}\sum^n_{k=1}\left(\frac{1}{k}+\frac{1}{n-k+1}\right)x^{n+1}\\
&=\sum^\infty_{n=1}\frac{(-1)^{n+1}2H_n}{n+1}x^{n+1}
\end{align*}\]
Apply this formula again to obtain the Taylor series of \(\displaystyle\frac{\ln^2(1+x)}{1+x}\).
\[\begin{align*}
\frac{\ln^2(1+x)}{1+x}
&=\left(\sum^\infty_{n=1}\frac{(-1)^{n+1}2H_n}{n+1}x^{n+1}\right)\left(\sum^{\infty}_{n=1}(-1)^{n-1}x^{n-1}\right)\\
&=\sum^\infty_{n=1}\sum^n_{k=1}\frac{(-1)^{k+1}(-1)^{n-k}2H_k}{k+1}x^{n+1}\\
&=\sum^\infty_{n=1}2(-1)^{n+1}\sum^n_{k=1}\frac{H_k}{k+1}x^{n+1}\\
\end{align*}\]
The inner sum is
\[\begin{align*}
\sum^n_{k=1}\frac{H_k}{k+1}
&=\sum^n_{k=1}\frac{H_{k+1}}{k+1}-\sum^n_{k=1}\frac{1}{(k+1)^2}\\
&=\sum^{n+1}_{k=1}\frac{H_k}{k}-H_{n+1}^{(2)}\\
&=\sum^{n+1}_{k=1}\frac{1}{k}\sum^k_{j=1}\frac{1}{j}-H_{n+1}^{(2)}\\
&=\sum^{n+1}_{j=1}\frac{1}{j}\left(\sum^{n+1}_{k=1}\frac{1}{k}-\sum^{j-1}_{k=1}\frac{1}{k}\right)-H_{n+1}^{(2)}\\
&=H_{n+1}^2-\sum^{n+1}_{j=1}\frac{H_j}{j}\\
&=\frac{H_{n+1}^2-H_{n+1}^{(2)}}{2}
\end{align*}\]
Hence
\[\frac{\ln^2(1+x)}{1+x}=\sum^\infty_{n=1}(-1)^{n+1}\left(H_{n+1}^2-H_{n+1}^{(2)}\right)x^{n+1}
\]
Pluck this into the integral.
\[\begin{align*}
\int^1_0\frac{\ln^3(1+x)\ln{x}}{x}{\rm d}x
&=-\frac{3}{2}\int^1_0\frac{\ln^2(1+x)\ln^2{x}}{1+x}{\rm d}x\\
&=-\frac{3}{2}\sum^\infty_{n=1}(-1)^{n+1}\left(H_{n+1}^2-H_{n+1}^{(2)}\right)\int^1_0x^{n+1}\ln^2{x} \ {\rm d}x\\
&=-3\sum^\infty_{n=1}\frac{(-1)^{n+1}\left(H_{n+1}^2-H_{n+1}^{(2)}\right)}{(n+2)^3}\\
&=3\sum^\infty_{n=1}\frac{(-1)^{n}\left(H_{n}^{(2)}-H_{n}^2\right)}{(n+1)^3}\\
\end{align*}\]
Step 2: Evaluation of \(\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{(n+1)^3}\)
We begin with some simple manipulations of the sum.
\[\begin{align*}
\sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{(n+1)^3}
&=\sum^\infty_{n=1}\frac{(-1)^nH_{n+1}^{(2)}}{(n+1)^3}-\sum^\infty_{n=1}\frac{(-1)^n}{(n+1)^5}\\
&=-\frac{15}{16}\zeta(5)-\underbrace{\sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^3}}_{S}
\end{align*}\]
Consider the function \(\displaystyle f(z)=\frac{\pi\csc(\pi z)\psi_1(-z)}{z^3}\). At the positive integers,
\[\begin{align*}
{\rm Res}(f,n)
&=\operatorname*{Res}_{z=n}\left[\frac{(-1)^n}{z^3(z-n)^3}+\frac{(-1)^n(H_n^{(2)}+2\zeta(2))}{z^3(z-n)}\right]\\
&=\frac{6(-1)^n}{n^5}+\frac{(-1)^nH_n^{(2)}}{n^3}+\frac{2(-1)^n\zeta(2)}{n^3}
\end{align*}\]
Summing them up gives
\[\sum^\infty_{n=1} {\rm Res}(f,n)=-\frac{45}{8}\zeta(5)+S-\frac{3}{2}\zeta(2)\zeta(3)
\]
At the negative integers,
\[\begin{align*}
{\rm Res}(f,-n)
&=-\frac{(-1)^n\psi_1(n)}{n^3}\\
&=\frac{(-1)^nH_n^{(2)}}{n^3}-\frac{(-1)^n\zeta(2)}{n^3}-\frac{(-1)^n}{n^5}
\end{align*}\]
Summing them up gives
\[\sum^\infty_{n=1} {\rm Res}(f,-n)=S+\frac{3}{4}\zeta(2)\zeta(3)+\frac{15}{16}\zeta(5)
\]
At \(z=0\),
\[\begin{align*}
{\rm Res}(f,0)
&=[z^2]\left(\frac{1}{z}+\zeta(2)z\right)\left(\frac{1}{z^2}+\zeta(2)+2\zeta(3)z+3\zeta(4)z^2+4\zeta(5)z^3\right)\\
&=4\zeta(5)+2\zeta(2)\zeta(3)
\end{align*}\]
Since the sum of the \(\mathrm{reisudes} =0\),
\[\sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{(n+1)^3}=-\frac{41}{32}\zeta(5)+\frac{5}{8}\zeta(2)\zeta(3)
\]
Step 3: Evaluation of \(\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{2}}{(n+1)^3}\)
Formula \((45)\) in this page states that this sum is equal to
\[4{\rm Li}_5\left(\frac{1}{2}\right)+4{\rm Li}_4\left(\frac{1}{2}\right)\ln{2}+\frac{2}{15}\ln^5{2}-\frac{107}{32}\zeta(5)+\frac{7}{4}\zeta(3)\ln^2{2}-\frac{2}{3}\zeta(2)\ln^2{2}-\frac{3}{8}\zeta(2)\zeta(3)
\]
Step 4: Obtaining the final result
Combining our previous results, we get
\[\begin{align*}
&{\int^1_0\frac{\ln^3(1+x)\ln{x}}{x}{\rm d}x}\\
&={3\sum^\infty_{n=1}\frac{(-1)^n\left(H_{n}^{(2)}-H_n^2\right)}{(n+1)^3}}\\
&=3\Bigg(\frac{33}{16}\zeta(5)+\zeta(2)\zeta(3)-4{\rm Li}_5\left(\frac{1}{2}\right)-4{\rm Li}_4\left(\frac{1}{2}\right)\ln{2}-\frac{2}{15}\ln^5{2}\\
&~~~-\frac{7}{4}\zeta(3)\ln^2{2}+\frac{2}{3}\zeta(2)\ln^3{2}\Bigg)\\
&=\boxed{\color{blue}{\dfrac{99}{16}\zeta(5)+\dfrac{\pi^2}{2}\zeta(3)-12{\rm Li}_5\left(\dfrac{1}{2}\right)-12{\rm Li}_4\left(\dfrac{1}{2}\right)\ln{2}-\dfrac{2}{5}\ln^5{2}-\dfrac{21}{4}\zeta(3)\ln^2{2}+\dfrac{\pi^2}{3}\ln^3{2}}}
\end{align*}\]
