A group of important generating functions involving harmonic number. How to prove the following identities:
$$\small{\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)}\tag1$$
$$\small{\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{n}x^{n}=\operatorname{Li}_3(x)+2\operatorname{Li}_3(1-x)-\ln(1-x)\operatorname{Li}_2(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)}\tag2$$
$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac{\ln^2(1-x)}{1-x}\tag3$$
$$\sum_{n=1}^\infty\frac{H_{n}^2}{n}x^{n}=\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x)-\frac13\ln^3(1-x)\tag4$$
$$\small{\sum_{n=1}^\infty H_n^3x^n=
\frac{\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)+\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)-3\zeta(3)}{1-x}}\tag5$$
$$\small{\sum_{n=1}^\infty H_nH_n^{(2)}x^n=
\frac{\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\frac12\ln x\ln^2(1-x)-\zeta(2)\ln(1-x)-\zeta(3)}{1-x}}\tag6$$
$$\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)x^n=-\frac{\ln^3(1-x)}{1-x}\tag7$$

Edit:
Here is some extra identities and proofs in the answer sections
$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}x^n=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)\tag8$$
$$\sum_{n=1}^\infty\frac{ H_n^{(2)}}{n+1}x^{n}=\frac{2\operatorname{Li}_3(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)}{x}\tag{9}$$
$$\small{\sum_{n=1}^\infty\frac{ H_n^{2}}{n+1}x^{n}=\frac{6\operatorname{Li}_3(1-x)-3\operatorname{Li}_2(1-x)\ln(1-x)-\ln^3(1-x)-3\zeta(2)\ln(1-x)-6\zeta(3)}{3x}}\tag{10}$$

Note: 
Proofs for (3) and (7) should be done without using the formula of   the sterling number of the first kind : $\frac{\ln^k(1+x)}{k!}=\sum_{n=k}^\infty(-1)^{n-k} \begin{bmatrix} n \\  k \end{bmatrix}\frac{x^n}{n!}$.
 A: All the generating functions may be found in the book, (Almost) Impossible Integrals, Sums, and Series.
The versions $\displaystyle \sum_{n=1}^{\infty} x^{n+1} \frac{H_n}{(n+1)^2}$ and $\displaystyle \sum_{n=1}^{\infty} x^{n+1} \frac{H_n^{(2)}}{n+1}$ may be found calculated in $(6.18)$ and $(6.19)$, page $351$ from which we extract $(1)$ and $(2)$ in the post statement by simply readjusting.
The results in $(3)$ and $(7)$ in the post statement may be found on page $355$, and a generalization of $\displaystyle \frac{\log^n(1-x)}{1-x}$ expressed with the help of symmetric polynomials may be found on pages $354$-$355$. Also, $(3)$ in the post statement may be obtained by combining the generating functions in $(4.6)$ and $(4.7)$.
The result in $(4)$ in the post statement may be extracted by using the result in $(4.7)$, page $284$.
Also, the results in $(5)$ and $(6)$ are found on page $284$. 
A: Proof of $(9)$ and $(10)$:
We proved above
$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac{\ln^2(1-x)}{1-x}$$
Integrating both sides gives
$$\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n+1}x^{n+1}=\int\frac{\ln^2(1-x)}{1-x}dx=-\frac13\ln^3(1-x)+C$$
where $C=0$ if we set $x=0$
Then
$$\sum_{n=1}^\infty \frac{H_n^2}{n+1}x^{n+1}=\sum_{n=1}^\infty \frac{H_n^{(2)}}{n+1}x^{n+1}-\frac13\ln^3(1-x)\tag1$$

From
$$\sum_{n=1}^\infty H_n^{(2)}x^n=\frac{\operatorname{Li}_2(x)}{1-x}$$
It follows that
$$\sum_{n=1}^\infty\frac{ H_n^{(2)}}{n+1}x^{n+1}=\int \frac{\operatorname{Li}_2(x)}{1-x}dx=f(x)\tag2$$
now let $1-x=y$ then use the reflection formula $\operatorname{Li}_2(1-y)=\zeta(2)-\ln(y)\ln(1-y)-\operatorname{Li}_2(y)$ 
$$f(x)=-\zeta(2)\int\frac{dy}{y}+\int\frac{\ln(y)\ln(1-y)}{y}dy+\int\frac{\operatorname{Li}_2(y)}{y}dx\\=-\zeta(2)\ln(y)+\left[-\operatorname{Li}_2(y)\ln(y)+\int\frac{\operatorname{Li}_2(y)}{y}dy\right]+\operatorname{Li}_3(y)\\=2\operatorname{Li}_3(y)-\operatorname{Li}_2(y)\ln(y)-\zeta(2)\ln(y)\\=2\operatorname{Li}_3(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)-\zeta(2)\ln(1-x)+C$$ 
Set $x=0$ $\Longrightarrow C=-2\zeta(3)$
Then 
$$f(x)=2\operatorname{Li}_3(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)\tag3$$
Plugging $(3)$ in $(2)$ yields
$$\sum_{n=1}^\infty\frac{ H_n^{(2)}}{n+1}x^{n+1}=2\operatorname{Li}_3(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)\tag4$$
Finally plug $(4)$ in $(1)$ we get
$$\small{\sum_{n=1}^\infty\frac{ H_n^{2}}{n+1}x^{n+1}=2\operatorname{Li}_3(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)-\frac13\ln^3(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)}$$
Or
$$\small{\sum_{n=1}^\infty\frac{ H_n^{2}}{n+1}x^{n}=\frac{6\operatorname{Li}_3(1-x)-3\operatorname{Li}_2(1-x)\ln(1-x)-\ln^3(1-x)-3\zeta(2)\ln(1-x)-6\zeta(3)}{3x}}$$
A: Proof of $(8)$:
From integrating the generating function after dividing by $x$
$$\sum_{n=1}^\infty H_n^{(3)}x^n=\frac{\operatorname{Li}_3(x)}{1-x}$$
it follows that 
$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}x^n=\int\frac{\operatorname{Li}_3(x)}{x(1-x)}dx=\int\frac{\operatorname{Li}_3(x)}{x}dx+\int\frac{\operatorname{Li}_3(x)}{1-x}dx$$
$$=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)+\int\frac{\ln(1-x)\operatorname{Li}_2(x)}{x}dx$$
$$=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)+C$$
Set $x=0\Longrightarrow C=0$
then
$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}x^n=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)$$
A: Different way to prove 
$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac{\ln^2(1-x)}{1-x}$$

Using the proved-above identity
$$\sum_{n=1}^\infty a_nx^n=\frac1{1-x}\sum_{n=1}^\infty (a_n-a_{n-1})x^n,\quad a_{0}=0$$
Set $$a_n=H_n^2-H_n^{(2)}$$
we get
$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac1{1-x}\sum_{n=1}^\infty (H_n^2-H_n^{(2)}-H_{n-1}^2+H_{n-1}^{(2)})x^{n}$$
$$=\frac{1}{1-x}\sum_{n=1}^\infty\left(\frac{2H_n}{n}-\frac{2}{n^2}\right)x^n=\frac{2}{1-x}\left(\color{blue}{\sum_{n=1}^\infty\frac{H_n}{n}x^n-\operatorname{Li}_2(x)}\right)$$
$$=\frac{2}{1-x}\left(\color{blue}{\frac12\ln^2(1-x)}\right)=\frac{\ln^2(1-x)}{1-x}$$
where the blue result follows from dividing both sides of $\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$ by $x$ then integrating from $x=0$ to $x=x$
$$\sum_{n=1}^\infty \frac{H_n}{n}x^n=-\int_0^x\frac{\ln(1-x)}{x(1-x)}\ dx=\operatorname{Li}_2(x)+\frac12\ln^2(1-x)$$
or $$\sum_{n=1}^\infty\frac{H_n}{n}x^n-\operatorname{Li}_2(x)=\frac12\ln^2(1-x)$$
