In the following link here I found the integral & the evaluation of
$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
I'll also include a simpler version together with the question: is it possible to find some easy
ways of computing both integrals without using complicated sums that require multiple zeta
formulae and "never-ending long" generating functions?
$$i). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx$$
$$ii). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
On
\begin{align*} \int_0^1 \frac{\text{Li}_2(x)^3}{x}dx &= \sum_{m,n=1}^\infty \frac{1}{m^2 n^2}\int_0^1 x^{m+n-1}\text{Li}_2(x)\; dx \\ &= \sum_{m,n=1}^\infty \frac{1}{m^2 n^2} \left(\frac{\zeta(2)}{m+n}-\frac{H_{n+m}}{(n+m)^2} \right) \\ &= \zeta(2) \sum_{m,n=1}^\infty \frac{1}{m^2 n^2 (m+n)}-\sum_{m,n=1}^\infty \frac{1}{m^2 n^2}\frac{H_{n+m}}{(n+m)^2} \tag{1} \end{align*}
Notice that with the substitution $r=m+n$, the double Euler sum may be transformed into a single sum as
\begin{align*} \sum_{m,n=1}^\infty \frac{1}{m^2 n^2 (m+n)} &= \sum_{r=2}^\infty \frac{1}{r}\sum_{k=1}^{r-1}\frac{1}{k^2 (r-k)^2} \\ &= \sum_{r=2}^\infty \frac{1}{r}\sum_{k=1}^{r-1}\left(\frac{1}{r^2 k^2}+\frac{1}{r^2 (r-k)^2} +\frac{2}{k r^3}+\frac{2}{r^3(r-k)}\right) \\ &= \sum_{r=2}^\infty \frac{1}{r}\left(2 \frac{H_{r}^{(2)}}{r^2}+4\frac{H_r}{r^3}-6\frac{1}{r^4} \right) \\ &= -6\zeta(5)+2\color{blue}{\sum_{r=1}^\infty \frac{H_{r}^{(2)}}{r^3}}+4\color{orange}{\sum_{r=1}^\infty \frac{H_r}{r^4}} \\ &= -6\zeta(5)+2\color{blue}{\left(3\zeta(2)\zeta(3)-\frac{9 \zeta (5)}{2} \right)}+4\color{orange}{\left(3 \zeta (5)-\zeta(2)\zeta(3) \right)} \\ &= -3\zeta(5)+2\zeta(2)\zeta(3) \tag{2} \end{align*}
Repeating the first 3 steps as above,
\begin{align*} \sum_{m,n=1}^\infty \frac{H_{m+n}}{m^2 n^2 (m+n)^2}&= \sum_{r=1}^\infty \frac{H_r}{r^2}\left(2 \frac{H_{r}^{(2)}}{r^2}+4\frac{H_r}{r^3}-6\frac{1}{r^4} \right) \\ &= -6 \color{green}{\sum_{r=1}^\infty \frac{H_r}{r^6}}+4\color{maroon}{\sum_{r=1}^\infty \frac{(H_r)^2}{r^5}}+2\sum_{r=1}^\infty \frac{H_r H_r^{(2)}}{r^4} \\ &= -6\color{green}{\left(-\zeta(4)\zeta(3)-\zeta(2)\zeta(5)+4 \zeta (7) \right)}+4\color{maroon}{\left( -\frac{5}{2}\zeta(4)\zeta(3)-\zeta(2)\zeta(5)+6 \zeta (7)\right)}+2\sum_{r=1}^\infty \frac{H_r H_r^{(2)}}{r^4} \\ &= -4\zeta(4)\zeta(3)+2\zeta(2)\zeta(5)+2\sum_{r=1}^\infty \frac{H_r H_r^{(2)}}{r^4} \end{align*}
Substituting every thing in equation (1), we get
$$\displaystyle \int_0^1 \frac{\text{Li}_2(x)^3}{x}dx =\boxed{ \displaystyle 9\zeta(4)\zeta(3)-5\zeta(2)\zeta(5)-2\sum_{r=1}^\infty \frac{H_r H_r^{(2)}}{r^4}} \tag{3}$$
Using the methods shown in this paper, I obtained
$$\displaystyle \sum_{r=1}^\infty \frac{H_r H_r^{(2)}}{r^4}=\frac{3}{4}\zeta(4)\zeta(3)+2\zeta(2)\zeta(5)-\frac{51 \zeta (7)}{16} \tag{4}$$
\begin{align*} \int_0^1 \frac{\text{Li}_2(x)^3}{x}dx &=9\zeta(4)\zeta(3)-5\zeta(2)\zeta(5)-2\color{brown}{\sum_{r=1}^\infty \frac{H_r H_r^{(2)}}{r^4}}\\ &= 9\zeta(4)\zeta(3)-5\zeta(2)\zeta(5)-2\color{brown}{\left(\frac{3}{4}\zeta(4)\zeta(3)+2\zeta(2)\zeta(5)-\frac{51 \zeta (7)}{16} \right)} \\ &= \boxed{\displaystyle \frac{15}{2}\zeta (3)\zeta(4)-9\zeta(2)\zeta(5)+\frac{51 \zeta (7)}{8}} \tag{5} \end{align*}
References
On
Here is a slightly different approach to evaluating $$\int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx,$$ which, like the answer given by @Zaid Alyafeai, makes use of the result $$\sum^\infty_{n = 1} \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3).$$
We start by writing $$I = \int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx = \int^1_0 \frac{\text{Li}_2 (x)}{x} \cdot \text{Li}_2 (x) \, dx.$$ Integrating by parts we have $$I = \text{Li}_2 (1) \text{Li}_3 (1) + \int^1_0 \frac{\text{Li}_3 (x) \ln (1 - x)}{x} \, dx = \zeta (2) \zeta (3) + \int^1_0 \frac{\text{Li}_3 (x) \ln (1 - x)}{x} \, dx,$$ where we have made use of the well-known results of $\text{Li}_s (0) = 0$ and $\text{Li}_s (1) = \zeta (s)$ respectively.
Integrating by parts again $$I = \zeta (2) \zeta (3) - \int^1_0 \frac{\zeta (4) - \text{Li}_4 (x)}{1 - x} \, dx.$$ Recalling $$\text{Li}_s (x) = \sum^\infty_{n = 1} \frac{x^n}{n^s} \quad \text{and} \quad \zeta (4) = \sum^\infty_{n = 1} \frac{1}{n^4},$$ we can write \begin{align*} I &= \zeta (2) \zeta (3) - \int^1_0 \frac{1}{1 - x} \left [\sum^\infty_{n = 1} \frac{1}{n^4} - \sum^\infty_{n = 1} \frac{x^n}{n^4} \right ] \, dx\\ &= \zeta (2) \zeta (3) - \sum^{\infty}_{n = 1} \frac{1}{n^4} \int^1_0 \frac{1 - x^n}{1 - x} \, dx. \end{align*}
From the integral representation for the harmonic number $H_n$, namely $$H_n = \int^1_0 \frac{1 - x^n}{1 - x} \, dx,$$ we can rewrite our integral as $$I = \zeta (2) \zeta (3) - \sum^\infty_{n = 1} \frac{H_n}{n^4} = \zeta (2) \zeta (3) - \left \{3 \zeta (5) - \zeta (2) \zeta (3) \right \},$$ or $$\int^1_0 \frac{\text{Li}_2^2 (x)}{x} \, dx = 2 \zeta (2) \zeta (3) - 3 \zeta (5).$$
By series expansion
$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx=\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)}$$
By some manipulations
$$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n(n+k)}$$
Now use that
$$\frac{H_k}{k} = \sum_{n\geq 1}\frac{1}{n(n+k)}$$
Hence we conclude that
$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = \zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_k}{k^4} $$
The euler some is known
$$\sum_{k\geq 1}\frac{H_k}{k^4} = 3\zeta(5)-\zeta(2)\zeta(3)$$
Finally we get
$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$
The other integral is very complicated to evaluate. I obtained formula using non-linear euler some here.
$$ \int^1_0\frac{\mathrm{Li}_{2}(x)^3}{x}\, dx = \zeta(3)\zeta(2)^2- \zeta(2) S_{3,2} +\sum_{k\geq 1} \frac{H_k^{(3)} H_k}{k^3}\\-\mathscr{H}(3,3)+\zeta(3) \zeta(4)-\zeta(3)\mathscr{H}(2,1)$$
where
$$ S_{p \, , \, q} = \sum_{n\geq 1} \frac{H^{(p)}}{n^q}$$
$$\begin{align}\mathscr{H}(p,q) = \int^1_0 \frac{\mathrm{Li}_p(x)\mathrm{Li}_q(x)}{x}\,dx \end{align}$$