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.

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:
$$
\boxed{\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx}} \, \mathrm{d}x = 2\sqrt{\frac{c}{a}}e^{-\frac{b}{g}}K_{1}\left(\frac{2}{|g|}\sqrt{ac}\right)}
$$
