I would like to evaluate the integral:
\begin{equation} \int_0^\infty e^{-\frac{(a+bx)^2}{c+dx}} dx. \end{equation}
$ a,b,c,d\in \mathbb{R}$ and the denominator in the exponent has no zeros over the range of integration, i.e. $c$ and $d$ are strictly positive (if $c$ and $d$ are both negative there would be no zeros but then the integral is not convergent).
=====Edit=====
It might be useful to provide some context, and actually directly ask a question. The obvious question is:
Is there a closed-form solution? If so, what is it. If not, how do you know (proof? something less rigorous)?
If there is an answer that can be evaluated for given $a,b,c,d \in \mathbb{R}$ then that is nearly a perfect answer for me too--i.e. I am not going to quibble about whether various special functions or series cannot be evaluated (exactly) in a finite number of calculations, as long as they can be computed (like Erf).
I don't demand the answer is complete, but concrete and/or verifiable information is appreciated.
I am also open to any information about series solutions, systematic expansions, or other ways to express the integral.
It is obvious to me that the integral has at least a calculable answer in terms of Erf if $d=0$. I am also interested in concrete information about systematic expansions in a ratio of one or more of the parameters $a,b,c,d$ if some condition holds. I am working on this now, but struggling to find plausible conditions.
===What I have tried===
I've tried simple analytical tricks like substitution to absorb the denominator into one variable--didn't work because I still have a power of the integration variable in the denominator.
I've submitted the integral to Mathematica which also didn't work because it returned the same integral form. This sort of got me interested in algorithms for symbolic integration and I have learned a small amount from the this thesis, and it was good because it provides a mathematical grounding--so I am open to learning more on that front as well.
Partial answer that I'll add to later if I have time.
Using polynomial division, you get:
$$\frac{(a+bx)^2}{c+dx} = \frac{b^2}{d}x + \frac{b}{d}\left(2a - \frac{bc}{d}\right) + \frac{\left(a - \frac{bc}{d}\right)^2}{c+dx}$$
So for $a=\frac{bc}{d}$ we get: \begin{align} f(\tfrac{bc}{d},b,c,d) &= \int_0^\infty \exp\bigg(- \underbrace{\tfrac{b^2}{d}x}_{u} - \tfrac{b^2c}{d^2}\bigg)\;dx = e^{- \tfrac{b^2c}{d^2} } \cdot \int_0^\infty \exp(-u)\; \cdot \frac{d}{b^2}du\\\\ &= \frac{d}{b^2} e^{- \tfrac{b^2c}{d^2} } \end{align}
For $d=0$ we get:
\begin{align} f(a,b,c,0) &= \int_0^\infty \exp\big(- \big(\underbrace{\frac{a+bx}{\sqrt{c}}}_{u}\big)\big)^2\;dx = \int_0^\infty \exp(-u^2) \frac{\sqrt{c}}{b} \;du = \frac{\sqrt{c}}{b} \left(\frac{\sqrt{\pi}}{2} - \frac{\sqrt{\pi}}{2}\operatorname{erf}(\tfrac{a}{\sqrt{c}})\right) \\\\ &= \frac{\sqrt{\pi c}}{2b}\left( 1 - \operatorname{erf}\big(\tfrac{a}{\sqrt{c}}\big) \right) \end{align}