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.
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.
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 $\int e^{ax}x^n dx=\frac {x^n e^{ax}}a -\frac na \int x^{n-1}e^{ax}dx$
$\int e^{ax}x^b(1-x)^c~dx$
$=\int_0^xx^b(1-x)^ce^{ax}~dx+C$
$=\int_0^xt^b(1-t)^ce^{at}~dt+C$
$=\int_0^1(xt)^b(1-xt)^ce^{axt}~d(xt)+C$
$=x^{b+1}\int_0^1t^b(1-xt)^ce^{axt}~dt+C$
$=\dfrac{x^{b+1}\Phi_1(b+1,-c,b+2;x,ax)}{b+1}+C$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)