Closed-form solution for $\int_m^\infty x^{-a} \exp(-b x) \exp(-c x^2) dx$ Apologies if this is a duplicate. After quite a while searching I couldn't quite find this integral solved in here.
I'm currently working on an application in Probability/Statistics where I need to compute 
$$
\int_m^\infty x^{-a} \exp(-b x) \exp(-c x^2) dx,
$$
where $m >0$, $a>1$, $b \in \mathbb{R}$ and $c >0$.
I'd love to not have to compute this by quadrature, although I'd be happy with a principled approximation (i.e., one for which we could control the error $\epsilon$). So let me first ask for a closed-form solution and we'll see if that's possible.
I threw the usual arsenal at it: integration by parts, differentiation under the integral sign, etc. Not claiming these don't work; just that when I tried them it was not fruitful.  
Edits
Following the answer by @martycohen, we can complete the square and, with the change of variables $u = x + \frac{b}{2c}$, arrive at$^\ast$
$$
\int_{m + \frac{b}{2c}}^\infty \left(2cu - b\right)^{-a}\exp(-cu^2)du.
$$
Am not sure how to get out of this quagmire, though.
$^\ast$ ignoring constants.
 A: This is a job for
completing the square!
$\begin{array}\\
I(a, b, c)
&=\int_m^\infty x^{-a} \exp(-b x) \exp(-c x^2) dx\\
&=\int_m^\infty x^{-a} \exp(-(cx^2+b x)) dx\\
&=\int_m^\infty x^{-a} \exp(-c(x^2+(b/c) x)) dx\\
&=\int_m^\infty x^{-a} \exp(-c(x^2+(b/c) x+(b^2/(4c^2))-(b^2/(4c^2)))dx\\
&=\int_m^\infty x^{-a} \exp(-c(x+(b/(2c)))^2+c(b^2/(4c^2))) dx\\
&=\exp(c(b^2/(4c^2)))\int_m^\infty x^{-a} \exp(-c(x+(b/(2c)))^2) dx\\
&=\exp(b^2/(4c))\int_{m+b/(2c)}^\infty (y-b/(2c))^{-a} \exp(-cy^2) dy
\quad\text{(error here fixed)}\\
\end{array}
$
This can be integrated
for some values of $a$,
such as the negative of
an odd integer.
If $a$ is the negative of an even integer,
the result involves the
error function.
A: Concerning the antiderivative$$I=\int\left(2cu - b\right)^{-a}\exp(-cu^2)du$$ let use first change variable
$u=\frac t{2c}$ to make 
$$2cI=J=\int (t-b)^{-a} \exp(-k t^2) \qquad \text{where} \qquad k=\frac 1{4c} >0$$
what you can notice if that, if $b=0$, the solution is given by 
$$J=-\frac{1}{2} k^{\frac{a-1}{2}} \Gamma \left(\frac{1-a}{2},k t^2\right)$$ which seem to be valid $\forall a$.
Now, just as an idea, why not to write
$$(t-b)^{-a}=\sum_{n=0}^\infty (-1)^n \binom{-a}{n}\,b^n\, t^{-(a+n)}$$ making
$$J=-\frac 12\sum_{n=0}^\infty (-1)^n \binom{-a}{n}\,b^n\,k^{\frac{a+n-1}{2} } \Gamma \left(-\frac{a+n-1}{2} ,k t^2\right)$$
