Closed-forms for several tough integrals These integrals came up in the process of finding solution to Vladimir Reshetnikov's problem. I wonder if there are closed-forms for the following integrals:
\begin{array}{1,1}
&[\text{1}] &\quad\int_0^1\frac{\operatorname{Li}_3(ax)}{1+2x}\ dx\\[12pt]
&[\text{2}] &\quad\int_0^1\frac{\operatorname{Li}_2(ax)\ln x}{1+2x}\ dx\\[12pt]
&[\text{3}] &\quad\int_0^1\frac{\ln(1-ax)\ln^2 x}{1+2x}\ dx
\end{array}
I have tried many substitutions, integration by parts, or differentiation under integral sign method, but without success so far. I do not need a complete or rigorous answer and your answer can be only Mathematica's or Maple's output since I don't have those software packages in my computer or links of related papers. I'd be grateful for any help you are able to provide.
 A: $\def\Li{{\rm Li}}$I think I know how Mr. Tunk-Fey's approach to find solution to Mr. Vladimir Reshetnikov's problem:
\begin{equation}
I=\int_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx
\end{equation}
He used integration by parts method. Here might be the way he evaluated $I$:
Let 
\begin{equation}
u=\ln(1+2x)\quad\Rightarrow\quad du=\dfrac{2\ dx}{1+2x}
\end{equation}
and
\begin{equation}
v=\int\frac{\ln(1-x)\ln(1+x)}{1+2x}\,dx
\end{equation}
In order to evaluate $v$, he might be using his technique in Mr. Vladimir Reshetnikov's another problem:
\begin{equation}
\int_0^1\frac{\ln^2x}{\sqrt{x^2-x+1}}dx
\end{equation}
See also David H's answer. I will just use that technique to evaluate $I$ without providing the complete steps. After several substitutions, we will arrive at the following equation
\begin{align}
v&=\frac{1}{6}\int\frac{\ln^2x}{1-\frac{2}{3}x}\ dx-\frac{1}{2}\int\frac{\ln^2x}{1-2x}\,dx-\frac{1}{2}\int\frac{\ln^2x}{1+x}\ dx-\frac{1}{6}\int\frac{\ln^2x}{1-\frac{1}{3}x}\,dx\\
&=\frac{1}{6}J_1-\frac{1}{2}J_2-\frac{1}{2}J_3-\frac{1}{6}J_4
\end{align}
where
\begin{align}
J_n=\int\frac{\ln^2x}{1-ax}\ dx=\frac{2}{a}\Li_3(ax)-\frac{2}{a}\Li_2(ax)\ln x-\frac{1}{a}\ln(1-ax)\ln^2x+C
\end{align}
Using $J_n$ we get
\begin{align}
J_1&=3\Li_3\left(\frac{2}{3}x\right)-3\Li_2\left(\frac{2}{3}x\right)\ln x-\frac{3}{2}\ln\left(1-\frac{2}{3}x\right)\ln^2x\\
J_2&=\Li_3\left(2x\right)-\Li_2\left(2x\right)\ln x-\frac12\ln\left(1-2x\right)\ln^2x\\
J_3&=-2\Li_3(-x)+2\Li_2(-x)\ln x+\ln(1+x)\ln^2x\\
J_4&=6\Li_3\left(\frac{1}{3}x\right)-6\Li_2\left(\frac{1}{3}x\right)\ln x-3\ln\left(1-\frac{1}{3}x\right)\ln^2x\\
\end{align}
Hence, we have
\begin{align}
uv&=\left.\left[\frac{1}{6}J_1-\frac{1}{2}J_2-\frac{1}{2}J_3-\frac{1}{6}J_4\right]\ln(1+2x)\right|_{x=0}^1\\
&=\left[\frac{1}{2}\Li_3\left(\frac23\right)-\frac{1}{2}\Li_3\left(2\right)+\Li_3\left(-1\right)-\Li_3\left(\frac{1}{3}\right)\right]\ln3\\
&=\left[\frac{1}{2}\Li_3\left(\frac{2}{3}\right)-\frac{1}{2}\Li_3\left(2\right)-\frac{3}{4}\zeta(3)-\Li_3\left(\frac{1}{3}\right)\right]\ln3
\end{align}
The next step is to evaluate $\displaystyle\int v\,du$ of which consists of three general form of integrals:
\begin{align}
\frac{\Li_3(ax)}{1+2x}\,dx\tag1\\[10pt]
\int_0^1\frac{\Li_2(ax)\ln x}{1+2x}\,dx\tag2\\[10pt]
\int_0^1\frac{\ln(1-ax)\ln^2 x}{1+2x}\,dx\tag3
\end{align}
from which the OP follows. To evaluate $I$, the corresponding values of $a$ are $\displaystyle\frac{2}{3},\,2,\,-1,\,$and $\displaystyle\frac{1}{3}$.
I wish I could evaluate $(1),\,(2),\,$and $(3)$ or, at least, evaluate the integrals with the corresponding values of $a$, but I couldn't (Sorry...). Luckily, Mr. Kirill has provided the results for $a=\displaystyle\frac{2}{3}$ and $a=-1$. I hope he will be generous to provide the results for $a=\displaystyle\frac{1}{3}$ and $a=2$.
A: For $(3)$ you can have the form

$$ I = - \rm Li_3( -2 ) \ln\left( {\frac {a+2}{2}} \right) - {\rm Li_4}( a )+ \int _{0}^{a}\,{\frac {{\rm Li_3}(t) }{t+2}}{dt},$$

where $\rm Li_s(z)$ is the polylogarithm function. For instance when $a=1/2$ we have

$$ I \sim - 0.08900930960. $$

A: Here are values of the integral $[3]$ for some specific values of the parameter $a$:
$$\int_0^1\frac{\ln(1-x)\ln^2x}{1+2x}dx=2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_4\left(\frac13\right)-\operatorname{Li}_4\left(\frac23\right)-\frac14\operatorname{Li}_4\left(\frac14\right)\\-\frac{\ln^42}{12}-\frac{\ln^43}{12}+\frac16\ln2\cdot\ln^33+\frac{\pi^2}6\left(\ln2\cdot\ln3-\ln^22-\operatorname{Li}_2\left(\frac13\right)\right).\tag1$$

$$\int_0^1\frac{\ln(1+x)\ln^2x}{1+2x}dx=3\operatorname{Li}_4\left(\frac12\right)-\frac34\operatorname{Li}_4\left(\frac14\right)\\+\left(7\zeta(3)-\operatorname{Li}_3\left(\frac14\right)\right)\frac{\ln2}4-\frac{3\pi^4}{160}-\frac{\ln^42}{24}-\frac{\pi^2}6\ln^22.\tag2$$

$$\int_0^1\frac{\ln(1+2x)\ln^2x}{1+2x}dx=\operatorname{Li}_4\left(\frac12\right)+\operatorname{Li}_4\left(\frac13\right)+\operatorname{Li}_4\left(\frac23\right)-\frac18\operatorname{Li}_4\left(\frac14\right)\\+\left(\operatorname{Li}_3\left(\frac13\right)+\operatorname{Li}_3\left(\frac23\right)\right)\ln3-\frac{11\pi^4}{360}-\frac{\ln^42}{24}-\frac{\ln^43}4\\+\frac{\pi^2}{12}\left(\ln^23-\ln^22\right)+\frac13\ln2\cdot\ln^33.\tag3$$
