<s>[INST] To determine the integration of $ \int_{0}^{+\infty} \exp\!\Big(-\Big(\frac{ax^2+bx+c}{gx+h}\Big)\Big) dx$. What is the integration of the following function:
$$  \int_\nolimits{0}^{+\infty} \exp\!\bigg(-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg) \bigg)dx.$$
What I have done is as follows:

Here, $\kappa=c-\Big(\frac{bg-ah}{g^2}\Big)h$.
\begin{align*}\implies & \int_{0}^{+\infty} \exp\!\bigg(-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg)\bigg) dx \\ =& \int_{0}^{+\infty}\exp\!\bigg(-\bigg(\frac{a}{g}x+\frac{b g-a h}{g^2}\bigg)\bigg)\exp\!\bigg(-\frac{\kappa}{gx+h}\bigg)dx.\end{align*}
I am finding it difficult to proceed further from here. Is this approach correct or is there any other intuitive way to solve this problem? Thanks in advance for the help.
 [/INST] Defining the incomplete Bessel function as
$$
K_\nu(x,y) = \int_1^{\infty} t^{-\left(\nu+1 \right)}e^{-\left(xt + \frac{y}{t} \right)}\, \mathrm{d}t
$$
We get
\begin{align}
\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx+h}} \, \mathrm{d}x & \overset{\color{purple}{x = \frac{h}{g}(t - 1)}}{=}\frac{h}{g}e^{\frac{2ah}{g^2}-\frac{b}{g}} \int_{1}^{\infty}e^{-\left[\left( \frac{ah}{g^2}\right)t + \left(\frac{ah}{g^2}- \frac{b}{g} + \frac{c}{h} \right)\frac{1}{t} \right]}\, \mathrm{d}t\\
& =\boxed{\frac{h}{g}e^{\frac{2ah}{g^2}-\frac{b}{g}} K_{-1}\left(\frac{ah}{g^2},\frac{ah}{g^2}- \frac{b}{g} + \frac{c}{h}\right)}
\end{align}

For the special case of $h=0$ you can obtain a closed-form in terms of a Modified Bessel function.
Notice that
$$
\frac{ax^2+bx+c}{gx} = \frac{a}{g}x + \frac{c}{gx} + \frac{b}{g}
$$
Now, for the integral to converge we require that the $\frac{ax^2+bx+c}{gx} \to + \infty$ when $x \to \infty$. This condition is met when $\frac{a}{g}>0$, so we'll do the rest of the analysis assuming this condition holds true. Additionally, we don't want $\frac{ax^2+bx+c}{gx}\to -\infty$  when $x \to 0^+$ since this would also make the integral divergent. To avoid this we also require that $\color{Purple}{\frac{c}{g}>0}$. With this we see that
$$
\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx}} \, \mathrm{d}x = e^{-\frac{b}{
g}}\int_{0}^{\infty} e^{-\left(\frac{a}{g}x + \frac{c}{gx} \right) } \, \mathrm{d}x  \overset{\color{Purple}{u = \frac{g}{c}x}}{=}\frac{c}{g}e^{-\frac{b}{g}} \int_{0}^{\color{Purple}{+\infty}} e^{-\left(\frac{\left(\color{green}{\frac{2}{|g|}\sqrt{ac}}\right)^2}{4}u + \frac{1}{u}\right)} \, \mathrm{d}u
$$
Now, from the Digital Library of Mathematical Functions we know that
$$
K_{\nu}(x) = \frac{1}{2}\left( \frac{1}{2}x\right)^{\nu}\int_{0}^{\infty} e^{-\left( \frac{x^2}{4t} + t\right)} \frac{1}{t^{\nu+1}} \, \mathrm{d}t, \qquad \forall x \in \mathbb{R}
$$
which means that
$$
\frac{4}{x} K_{1}(x)=\int_{0}^{\infty} e^{-\left( \frac{x^2}{4t} + t\right)} \frac{1}{t^{2}} \, \mathrm{d}t \overset{\color{blue}{u = \frac{1}{t}}}{=}  \int_{0}^{\infty} e^{-\left(\frac{\color{green}{x}^2}{4} u + \frac{1}{u} \right)} \, \mathrm{d}u
$$
So combining everything, if $\frac{a}{g}, \frac{c}{g} >0$ then:
$$
\boxed{\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx}} \, \mathrm{d}x = 2\sqrt{\frac{c}{a}}e^{-\frac{b}{g}}K_{1}\left(\frac{2}{|g|}\sqrt{ac}\right)}
$$
</s>