Integral of two logs and a power: $\int_0^1 u^c \log(1-au)\log(1-bu)\,\mathrm du$ Does the following integral have a closed form in terms of known functions?
$$ f(a,b,c) = \int_0^1 u^c \log(1-au)\log(1-bu)\,\mathrm du.$$
The parameters are possibly complex, and satisfy
$$\Re(c)>-1, \qquad |a|\leq1, \qquad |b|\leq1. $$
The best I could do was
$$ f(a,b,c) = \frac{1}{1+c}\frac{\partial^2}{\partial s \partial t}\Big|_{s=t=0}F_1(1+c; -s,-t; 2+c; a, b),$$
where $F_1$ is an Appell function, but this doesn't help me much. Also, when $a=b$, I know that $f$ is the second derivative of the beta function.
Is it possible to get a nicer expression for $f$?
 A: Maybe differentiate with respect to $a,b$ ?
$$ \frac{\partial}{\partial a} \frac{\partial}{\partial b}f(a,b,c)
= \int_0^1 \frac{u^{c+2}}{(1- au)(1-bu)} \; du  $$
Then partial fractions might work, you can plug in $u = 1/a, 1/b$ to get the coefficients $A,B$.
$$\frac{1}{(1- au)(1-bu)}  = \frac{A}{1-au} + \frac{B}{1-bu}  $$
The result is an integral like, which can be solved with integration by parts.
$$  \int_0^1 \frac{u^{c+2}}{1- au} \; du  $$
Then you have to take the antiderivative with respect to $a$ and $b$.  That result might be complex.

It is interesting to see $a,b$ vary.  There are degenerate cases when $a=0$ or $b=0$.
A: Not a complete answer, but a hint. A series solution might be helpful.
Start with the simple integral and use Taylor series for the logarithm:
$$\int_0^1 u^p ~du=\frac{1}{p+1}$$
$$\int_0^1 u^c u^k u^l ~du=\frac{1}{k+l+c+1}$$

$$\int_0^1 u^c \log(1-au) \log(1-bu)~du=\sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \frac{a^k b^l}{k~l~(k+l+c+1)}$$

We can turn it into a single sum with Lerch Trancendent:
$$ \sum_{l=1}^{\infty} \frac{b^l}{l~(k+l+c+1)}=-\frac{b ~\Phi (b,1,k+c+2)+\log(1-b)}{k+c+1} $$
