Computing $\int_0^\infty \exp\left( -\frac{x^2}{A} -\frac{B}{x} -Cx \right){\rm d}x$ Exactly as the title suggests, I want to compute or approximate the integral of the form
$$\int_0^\infty \exp\left( -\frac{x^2}{A} -\frac{B}{x} -Cx \right){\rm d}x$$
where $A,B,C>0.$ I got the question from a friend who asked me how to evaluate it.
I was told that an asymptotic formula or a good approximation for large values of $t$ (if we integrate from $0$ to $t$ instead) is good enough, as long as it's written in terms of $A,B,C.$ I feel like contour integration might do the trick but I'm a bit too rusty on that. Thank you in advance.
 A: If we consider
$$f(a,b,c)=\int_0^\infty \exp\left( -a x^2 -\frac{b}{x} -c x \right)\,dx$$
(with $a \geq 0$, $b\geq 0$, $c\geq 0$), exploring all possibilities
$$f(a,0,0)=\frac{\sqrt{\pi }}{2 \sqrt{a}}$$
$$f(0,b,0)\qquad\qquad \text{does not converge}$$
$$f(0,0,c)=\frac 1c$$
$$f(a,b,0)=\frac b{4\sqrt \pi}G_{0,3}^{3,0}\left(\frac{a b^2}{4}|
\begin{array}{c}
 -\frac{1}{2},0,0
\end{array}
\right)$$ where appears Meijer G function
$$f(a,0,c)=\frac{\sqrt{\pi }}{2 \sqrt{a}}\,e^{\frac{c^2}{4 a}}\,\text{erfc}\left(\frac{c}{2 \sqrt{a}}\right)$$
where appears the complementary error function
$$f(0,b,c)=2 \sqrt{\frac{b}{c}} K_1\left(2 \sqrt{b c}\right)$$
where appears the modified Bessel function of the second kind.
Unfortunately, for the general case, none of the CAS I used gives an answer.
$$f(a,b,c)= \text{   } \large ???$$
Edit
If $a$ or $c$ are "small", we can expand the integrand as Taylor series and obtain results for $f(a,b,c)$. For $a$, it will be a linear combination of Bessel functions; for $c$ it will be a linear combination of Meijer G functions.
A: By rescaling $x$, we may set one constant to unity, so we can consider the integral
$$
I=\int\limits_0^\infty dx \ \exp \left( -ax^2-bx^{-1}-x\right)
$$
For large $a$
The integrand has a moveable maxima at $x_0(a,b)$. Let $x=ux_0$, then the maxima is fixed at $u=1$, and we have
$$
I=\int\limits_0^\infty du \ x_0 \exp \left( -ax_0^2u^2-bx_0^{-1}u^{-1}-x_0u \right)
$$
If you want to expand any of the exponentials, expanding this equation is a good idea. Expanding the whole exponent around $u=1$, we get a Gaussian integral, resulting in a complementary error function (for fixed $b$)
$$
I(a) \sim \frac{x_0}{2} \sqrt{\frac{\pi x_0}{b+ax_0^3}} e^{f(x_0)} \operatorname{erfc}\left[ g(x_0)\right] \quad ; \quad a \to \infty
$$
$x_0$, $f$, and $g$ are given at the end of the post. It turns out this is also good when $b \to \infty$ for fixed $a$. Here are plots together with the numerical integration:

Now consider the integral
$$
J(t)=\int\limits_t^\infty dx \ \exp \left( -ax^2-bx^{-1}-x\right)
$$
So that the integral in the comments is given by
$$
\int\limits_0^t dx \ \exp \left( -ax^2-bx^{-1}-x\right) = I-J(t)
$$
Where $a$ and $b$ are fixed. We develop the asymptotic series for $J$ by integration by parts.
$$
J(t)=\int\limits_t^\infty dx \ \frac{1}{-2ax+bx^{-2}-1} \frac{d}{dx} \left[ \exp \left( -ax^2-bx^{-1}-x\right)\right]
$$
$$
J(t)=\frac{\exp \left(-at^2-bt^{-1}-t \right)}{2at-bt^{-2}+1}-\int\limits_t^\infty dx \ \dots
$$
For the integral on the right, we have $\int dx \dots << J(t)$ for $t \to \infty$, so long as $a$ and $b$ are fixed. Neglecting this term, we have
$$
J(t) \sim \frac{\exp \left(-at^2-bt^{-1}-t \right)}{2at-bt^{-2}+1} \quad ; \quad t \to \infty
$$
Here is a plot with $a=b=3/2$:

Here are the functions (thanks Mathematica!)
$$
x_0(a,b)=\frac{1}{24a}\left[ -4 + \frac{2+2i\sqrt{3}}{(1-54a^2 b+6 \sqrt{3} \sqrt{a^2 b (27a^2 b-1})^{1/3}} +(2-2i\sqrt{3})(1-54a^2+6 \sqrt{3} \sqrt{a^2 b (27a^2 b-1)})^{1/3}\right]
$$
$$
f(x)=\frac{-3b^2+x^4-6bx^2(1+2ax)}{4x(b+ax^3)}
$$
$$
g(x)=\frac{x^2-3b}{2 \sqrt{bx+ax^4}}
$$
$x_0$ is real, despite the imaginary units floating around. Probably, it can be simplified if one was so inclined.
