Let $p$ and $q$ be two positive integers, I wonder if the following integral admits a closed-form and if yeah so in which way it should be worked out.
$$\int_{0}^{1}\frac{\ln(1-x^p)\ln(1+x^q)}{x^{p+q}}dx\,=\sum_{n,k\geq1}\frac{(-1)^n}{nk(np+kq+1-p-q)}$$
Here is a partial answer in terms of an infinite series. After expanding both logarithms into their respective Maclaurin series, integrating term by term and applying a certain series representation of the digamma function, I've obtained this series: $$\int_0^1 \frac{\log(1-x^p)\log(1+x^q)}{x^{p+q}}\,dx=\sum_{k\ge 0} (-1)^{k+1} \frac{\psi^{(0)}\left(\frac{kq+1}{p}\right)+\gamma}{(k+1)(kq-p+1)}$$ Where $\psi^{(0)}$ $(=\Gamma'/\Gamma )$ is the digamma function and $\gamma$ ($=\lim_{n\to\infty} H_n-\log n$) is the Euler-Mascheroni constant. I'm not sure how to simplify this sum, which appears especially challenging because of the appearance of digamma function.
Checking the basic case $q=0$ yields $$\int_0^1 \frac{\log(1-x^p)}{x^p}\,dx=\frac{\psi^{(0)}(1/p)+\gamma}{p-1}$$ Which seems to agree with some of the computational verifications I've done.