After finding the exact value of the integral, $$\displaystyle \int_0^1 \int_0^1\left(\frac{\ln (1+x y)}{1+x y}\right)^2 d x d y=-\frac{\zeta(3)}{4}+\frac{1}{3} \ln ^3 2-\frac{\pi^2}{6}+\ln ^2 2+2 \ln 2\,\tag*{} $$
I started to generalise it to $$ I(m, n)=\int_0^1 \int_0^1 \frac{\ln^m (1+x y)}{(1+x y)^n} d x d y. $$ Firstly I derived its reduction formula for integers $m,n \ge 1$ by integration by parts again. $$ \begin{aligned} I(m, n)&=\int_0^1 \int_0^1 \frac{\ln^m (1+x y)}{(1+x y)^n} d x d y \\ & =\int_0^1-\frac{1}{(n-1) y}\left[\int_0^1 \ln ^m(1+x y) d\left(\frac{1}{(1+x y)^{n-1}}\right)\right] d y \\ & =-\frac{1}{n-1} \int_0^1 \frac{1}{y}\left[\frac{\ln ^m(1+y)}{(1+y)^{n-1}}-m \int_0^1 \frac{\ln ^{m-1}(1+x y)}{(1+x y)} d x\right] dy \\ \end{aligned} $$ So $$ \boxed{I(m, n) =-\frac{1}{n-1} \int_0^1 \frac{\ln ^m(1+y)}{y(1+y)^{n-1}} d y+\frac{m}{n-1} I(m-1, n)} \cdots(*) $$ $$ \begin{aligned} I(2,2) & =-\int_0^1 \frac{\ln ^2(1+y)}{y(1+y)} d y+2 I(1,2) \\ & =-\int_0^1 \frac{\ln ^2(1+y)}{y(1+y)} d y+2\left[-\int_0^1 \frac{\ln (1+y)}{y(1+y)}+2 I(0, 2)\right] \\ & =-\int_0^1 \frac{\ln ^2(1+y)}{y(1+y)} d y-2 \int_0^1 \frac{\ln (1+y)}{y(1+y)}+4 \int_0^1 \int_0^1 \frac{1}{(1+x y)^2} d y\\&= -\int_0^1 \frac{\ln ^2(1+y)}{y} d y+\int_0^1 \frac{\ln ^2(1+y)}{1+y} d y-2 \int_0^1 \frac{\ln (1+y)}{y} d y-2 \int_0^1 \frac{\ln (1+y)}{1+y}dy+2\ln 2\\&= -\frac{\zeta(3)}{4}+\frac{\ln ^3 2}{3}-\frac{\pi^2}{6}+\ln ^2 2+2 \ln 2 \end{aligned} $$ where the first integral uses the result :How to evaluate $\int_0^1\frac{\log^2(1+x)}x\mathrm dx$?.
Similarly, $$ \begin{aligned} I(2,3)=&-\frac{1}{2} \int_0^1 \frac{\ln ^2(1+y)}{y(1+y)^2} d y+I(1,3) \\ = & -\frac{1}{2}\left[\int_0^1 \frac{\ln ^2(1+y)}{y} d y-\int_0^1 \frac{\ln ^2(1+y)}{1+y}-\int_0^1 \frac{\ln ^2(1+y)}{(1+y)^2 }dy\right] +\frac{1}{24}\left(9-\pi^2+6 \ln ^2 2\right)\\=& -\frac{\zeta(3)}{8}-\frac{\pi^2}{24}-\frac{1}{3} \ln 2+\frac{7}{8} \end{aligned} $$ Now let’s go further. $$ \begin{aligned} I(3,3) =&-\frac{1}{2} \int_0^1 \frac{\ln ^3(1+y)}{y(1+y)^2} d y+\frac{3}{2} I(2,3) \\ = & -\frac{1}{2}\left[\int_0^1 \frac{\ln ^3(1+y)}{y} d y-\int_0^1 \frac{\ln ^3(1+y)}{1+y} d y-\int_0^1 \frac{\ln ^3(1+y)}{(1+y)^2}dy\right] +\frac{3}{2}\left(-\frac{\zeta(3)}{8}-\frac{\pi^2}{24}-\frac{1}{3} \ln 2+\frac{7}{8}\right)\\=& -\frac{1}{2}\left[\int_0^1 \frac{\ln ^3(1+y)}{y} d y-\frac{\ln ^4 2}{4}-3+\frac{\ln ^3 2}{2}(3+\ln 2)+3\ln 2\right]+ \frac{3}{2}\left( -\frac{\zeta(3)}{8}-\frac{\pi^2}{24}-\frac{1}{3} \ln 2+\frac{7}{8}\right) \end{aligned} $$ Using the result, we have $$ \int_0^1 \frac{\ln ^3(1+y)}{y} d y= -6 \operatorname{Li_4}\left(\frac{1}{2}\right)-\frac{21}{4} \zeta(3) \ln 2 +\frac{\pi^4}{15}-\frac{\ln ^4 2}{4}+\frac{\pi^2}{4} \ln ^2 2 $$
Plugging back yields
$$ \begin{aligned} & \quad I(3,3)\\&=-\frac{1}{2}\left[ -6 \operatorname{Li_4}\left(\frac{1}{2}\right)-\frac{21}{4} \zeta(3) \ln 2+\frac{\pi^4}{15}-\frac{\ln ^4 2}{4}+\frac{\pi^2}{4} \ln ^2 2 -\frac{\ln ^4 2}{4}-3+\frac{\ln ^3 2}{2}(3+\ln 2)+3 \ln 2\right] +\frac{3}{2}\left(-\frac{\zeta(3)}{8}-\frac{\pi^2}{24}-\frac{\ln 2}{3}+\frac{7}{8}\right)\\&= 3 \operatorname{Li_4}\left(\frac{1}{2}\right)+\frac{45}{16} \zeta(3)-\frac{\pi^4}{30}+\frac{\ln ^4 2}{4}-\frac{\pi^2}{8} \ln ^2 2+\frac{27}{32}-\ln ^3 2-\frac{3}{4} \ln 2+\frac{\pi^2}{32} \end{aligned} $$
Theoretically, we can use the result, we have and the reduction formula $(*)$ to evaluate tediously $$\int_0^1 \int_0^1\frac{\ln^m (1+x y)}{(1+x y)^n}d x d y,$$ where $m\ge 3$.
My question is whether there are other elegant methods.
Can you help me? Your comments and solutions are highly appreciated.
