How to compute $\int_0^1x^a(1-x)^be^{cx}dx$?

Question

How to compute the integral $I(a,b,c) = \int_0^1x^a(1-x)^be^{cx}dx$ ?

I know that, $\int_0^1{x^a(1-x)^b}dx = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$. Using this result, I tried integration by parts; I got the following recurrence relation,

$I(a,b,c) = \frac{b}{c}I(a,b-1,c) - \frac{a}{c}I(a-1,b,c)$.

I don't know how to proceed. Any help is appreciated. Thanks in advance.

Answer 1

This is essentially the the moment generating function of the beta distribution. The result is hypergeometric and cannot be further simplified.