While dealing with algebraic integrals composited of logarithms and polynomials, I learned about using the derivatives of the Beta Function in order to evaluate them. Applying this knowledge I was able to show that
$$\int_0^1 \log^n(x^a)\log^m(1-x^b)x^c(1-x^b)^t\mathrm dx~=~\frac{a^n}{b^{n+1}}\left.\frac{\partial^{n+m}}{\partial x^n\partial y^m}B(x,y)\right|^{x=\frac{c+1}{b}}_{y=t+1}\tag1$$
Which is not difficult at all since it is just the application of the substitution $u=x^b$. The explicit evaluation gets more and more complicated while $n$ and $m$ are increasing. Furthermore I can only apply this formula for the case $n,m\in\mathbb{N}$. Anyway now I thought about the following integral
$$\int_0^1 \log^n(x^a)\log^m(1-x^{\color{red}{\alpha}})x^b(1-x^{\color{red}{\beta}})^t~dx\tag2$$
With $\alpha\ne\beta$. For this case the simple substitution $u=x^{\alpha}$ or $u=x^{\beta}$, respectively, does not work out since it produces a term of the type $(1-x^{\alpha/\beta})$ which does not fit within the integral representation of the Beta Function and its derivatives. Therefore, I think it is maybe not the right approach at all but I could not figure out a different way to get started with $(2)$. IBP or a different subsititution seem pointless to me as well as a series expansion of the logarithm because of the powers $m$ and $n$ respectively.
How can one tackle $(2)$? I would be interested in a general formula similar to $(1)$ if possible. I would be glad about an attempt concerning small values of $n$ and $m$. My priority lies within the evaluation of the composition of $\log(1-x^{\alpha})$ and $(1-x^{\beta})^t$ therefore the powers of the logarithm are of minor matter.
Thanks in advance!
For m=1, $$ \int_{0}^{1} \log(x^a)^n \log(1-x^\alpha) x^b (1-x^\beta)^t\, dx = \frac{a^n}{\beta^{n+1}} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \frac{\partial^n}{\partial Q^n} B(Q, t+1)\Bigr|_{\substack{Q=\frac{1+b+k\alpha}{\beta}}} $$
Where B is the Beta function.