# Evaluating $\int e^{ax} x^b (1-x)^c \mathrm{dx}$

Edit: clarify question

The integrand looks kind of like a gamma density function, and kind of like a beta density function, so maybe it has a somewhat nice solution?

$$\int e^{ax} x^b (1-x)^c \mathrm{dx}$$

Wolfram alpha does not want to do it.

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It is definitely a thing. –  copper.hat Aug 21 '12 at 15:34
What are you asking exactly? –  gt6989b Aug 21 '12 at 15:39
It's not polite to make someone do something if they do not want to do it. –  Graphth Aug 21 '12 at 16:33
You can expand out the $(1-x)^c$ to get terms of the form $\int e^{ax}x^n dx$. Wolfram Alpha then gives a solution in terms of the incomplete Gamma function. This is a form that can be integrated by parts-set $dv=e^{ax}dx, u=x^n$ and step down the exponents, giving $\int e^{ax}x^n dx=\frac {x^n e^{ax}}a -\frac na \int x^{n-1}e^{ax}dx$
This only works if $b$ and $c$ are both integral; I'm presuming (with admittedly no evidence either way) that the OP wasn't assuming integer values... –  Steven Stadnicki Aug 21 '12 at 16:26
You could try a binomial series, but that would only work if $x$ was sufficiently small. –  Calvin McPhail-Snyder Aug 21 '12 at 17:16