product of gaussian and algebraic function I know the results for $n=3$ but not for any general value of $n$ for the following integral.
$$\int_{-\infty}^{\infty} x^{n}\exp\left[-ax^{2}+bx+c\right]\, dx.$$
Is there  any result for this?
In case there is none, how must I proceed to calculate it?
Thanks 
 A: Of course you need $a > 0$ for the integral to converge.  Let $F_n(a,b,c)$ be your integral.  The exponential generating function of these is
$$ \eqalign{g(z) &= \sum_{n=0}^\infty F_n(a,b,c) \frac{z^n}{n!}\cr & = \int_{-\infty}^\infty \exp(-a x^2 + (b+z) x + c)\; dx \cr &= F_0(a,b+z,c)  = \sqrt{\frac{\pi}{a}} \exp\left(c+\frac{(b+z)^2}{4a}\right)}$$
and $F_n(a,b,c)$ is $n!$ times the coefficient of $z^n$ in the Maclaurin series of that.
A: Actually, after doing the integral I've decided to expand on my comment.
First note
$$
\int_{-\infty}^{\infty} x^ne^{-ax^2+bx + c}dx = e^{-b^2/a + c}\int_{-\infty}^{\infty} x^ne^{-a(x-b/a)^2}dx. 
$$
Putting $u = x - b/a$ we have 
$$
e^{-b^2/a + c}\int_{-\infty}^{\infty} (u+b/a)^ne^{-au^2}du,
$$
so applying the binomial theorem to $(u+b/a)^n$, 
$$
e^{-b^2/a + c}\int_{-\infty}^{\infty} \sum_{j = 1}^n \begin{pmatrix}
n\\ 
j\\
\end{pmatrix}u^j(b/a)^{n-j}e^{-au^2}du
=e^{-b^2/a + c}\sum_{j = 1}^n \begin{pmatrix}
n\\ 
j\\
\end{pmatrix}(b/a)^{n-j}\int_{-\infty}^{\infty} u^je^{-au^2}dx.
$$
Now you need only evaluate the moments of $e^{-au^2}$, which can be done through successive differentiation by $a$ in the even case, and which vanish in the odd. 
