Integral: how to evaluate $$\int_0^1{\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \text{d}x}$$
Same context I'm not sure of the closed form of the integral, as I haven't evaluated it yet. However, after evaluating the integral $$\int_0^1 \ln (1+x) \ln (1-x) \, dx,$$ I thought about integrating $$\int_0^1 \ln^3(1-x) \ln^2(1+x) \, dx.$$
I am also interested in evaluation of $$\int_0^1 \ln^n(1-x) \ln^m(1+x) \, dx.$$ If possible
My work
replace $x\rightarrow\frac{1-x}{1+x}$, there is $$\int_0^1{\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \text{d}x}=2\underset{I}{\underbrace{\int_0^1{\frac{\ln ^2\left( \dfrac{2}{1+x} \right) \ln ^3\left( \dfrac{2x}{1+x} \right)}{\left( 1+x \right) ^2}\text{d}x}}}$$
Expand the integral
\begin{align*} &\ln ^2\left( \frac{2}{1+x} \right) \ln ^3\left( \frac{2x}{1+x} \right) =-\ln ^2\text{2}\ln ^3\left( 1+x \right) +\text{2}\ln\text{2}\ln ^4\left( 1+x \right) -\ln ^5\left( 1+x \right) \\ &+\text{3}\ln ^2\text{2}\ln ^2\left( 1+x \right) \ln \left( 2x \right) -\text{6}\ln\text{2}\ln ^3\left( 1+x \right) \ln \left( 2x \right) +\text{3}\ln ^4\left( 1+x \right) \ln \left( 2x \right) \\ &-\text{3}\ln ^2\text{2}\ln \left( 1+x \right) \ln ^2\left( 2x \right) +\text{6}\ln\text{2}\ln ^2\left( 1+x \right) \ln ^2\left( 2x \right) -\text{3}\ln ^3\left( 1+x \right) \ln ^2\left( 2x \right) \\ &+\ln ^2\text{2}\ln ^3\left( 2x \right) -\text{2}\ln\text{2}\ln \left( 1+x \right) \ln ^3\left( 2x \right) +\ln ^2\left( 1+x \right) \ln ^3\left( 2x \right) \end{align*}
Evaluation of $I_1$
$$I_1=-\ln ^22\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=\frac{\ln ^52}{2}+\frac{\text{3}\ln ^42}{2}+\text{3}\ln ^32-\text{3}\ln ^22$$
Evaluation of $I_2$
$$I_2=\text{2}\ln 2\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-\ln ^52-\text{4}\ln ^42-\text{12}\ln ^32-\text{24}\ln ^22+\text{24}\ln 2$$
Evaluation of $I_3$
$$I_3=-\int_0^1{\frac{\ln ^5\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-60+\frac{\ln ^52}{2}+\frac{\text{5}\ln ^42}{2}+\text{10}\ln ^32+\text{30}\ln ^22+\text{60}\ln 2$$
Evaluation of $I_4$
\begin{align*} I_4&=\text{3}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln \left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x\\&=\text{3}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x+\text{3}\ln ^32\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x \end{align*}
$$I_{41}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2$$
$$I_{42}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2$$
Expanding the integral does not seem like a great way to evaluate it. I did not provide explanations for some smaller integrals due to the length of the post.
Closed form by Mathematica
\begin{align*} &-120 + 6 \pi^2 + \frac{\pi^4}{30} + 96 \log(2) - 6 \pi^2 \log(2) - \frac{1}{30} \pi^4 \log(2) \\ &\quad - 48 \log^2(2) + 2 \pi^2 \log^2(2) + 16 \log^3(2) - \frac{2}{3} \pi^2 \log^3(2) \\ &\quad - 3 \log^4(2) + \frac{3}{5} \log^5(2) + 24 \text{Li}_4 \left( \frac{1}{2} \right) + 24 \text{Li}_5 \left( \frac{1}{2} \right) \\ &\quad + 45 \zeta(3) - 2 \pi^2 \zeta(3) - 24 \log(2) \zeta(3) + 12 \log^2(2) \zeta(3) + \frac{3}{5} \zeta(5) \end{align*}