Alternative ways to evaluate $\int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$ 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?
$$i). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx$$
$$ii). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
 A: 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 
$$ S_{p \, , \, q} = \sum_{n\geq 1} \frac{H^{(p)}}{n^q}$$
$$\begin{align}\mathscr{H}(p,q)  = \int^1_0 \frac{\mathrm{Li}_p(x)\mathrm{Li}_q(x)}{x}\,dx  \end{align}$$
A: 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
$$\int^1_0 \frac{\text{Li}_2^2 (x)}{x} \, dx = 2 \zeta (2) \zeta (3) - 3 \zeta (5).$$
