I came across a quite difficult integral, which I can not solve. The parameters a and b are both > 0. For fixed b I can find solutions, but no formula with b as a variable.
The function that I want to integrate: $$ \int_0^1 \log((1-x^a)^{-b} -1) \ dx,\\ a,b > 0 $$
$$\begin{eqnarray*}\int_{0}^{1}\log\left((1-x^a)^{-b}-1\right)\,dx &=& \frac{1}{a}\int_{0}^{1} x^{1/a}\log((1-x)^{-b}-1)\frac{dx}{x}\\&=&\frac{1}{a}\int_{0}^{1} (1-x)^{1/a}\log(x^{-b}-1)\frac{dx}{1-x}\\&=&\frac{1}{a}\int_{1}^{+\infty} \left(1-\frac{1}{x}\right)^{1/a}\log(x^{b}-1)\frac{dx}{x(x-1)}\\&=&\frac{1}{ab}\int_{1}^{+\infty} \left(\frac{x^{1/b}-1}{x^{1/b}}\right)^{1/a}\log(x-1)\frac{dx}{x(x^{1/b}-1)}\\&=&\frac{1}{ab}\int_{0}^{1} \left(1-x^{1/b}\right)^{1/a-1}x^{1/b-1}\log\left(\frac{1-x}{x}\right)\,dx\end{eqnarray*}$$ where by differentiating Euler's Beta function $$ \frac{1}{ab}\int_{0}^{1} \left(1-x^{1/b}\right)^{1/a-1}x^{1/b-1}\left(-\log x\right)\,dx = -b H_{1/a} $$ and by exploiting Beta directly $$ \frac{1}{ab}\int_{0}^{1} \left(1-x^{1/b}\right)^{1/a-1}x^{1/b-1}\left(-\frac{x^n}{n}\right)\,dx = -\frac{b\,\Gamma\left(\frac{1}{a}\right)\,\Gamma(bn+1)}{n\,\Gamma\left(1+\frac{1}{a}+b n\right)} $$ so the original integral can be written in terms of the digamma function and hypergeometric functions.