Compute integral of exponential with polynomial division in argument I'm trying to calculate this integral
$$\int^{\infty}_{-\infty} \frac{1}{|x|}\exp\left(-\frac{a + bx^4}{cx^2}\right)dx $$
As a start, I thought maybe integration by parts was my best bet, so I looked into finding just the integral of the exponent part:
$$\int^{\infty}_{-\infty}\exp\left(-\frac{a + bx^4}{cx^2}\right)dx$$
But even that is looking quite difficult. I've tried dividing the polynomial, as suggested  in this similar question but doesn't really help in this problem.
Is it hopeless to find a closed-form solution for this?
 A: Here's a way to calculate the second integral in the case that $a, b>0$.
Next we will focus on the special case that $b=1$. Note that by letting $x=\sqrt{b}y$ and change the variable, we can easily reduce to this special case.
Note that when $a=0$, the integral is easy to compute (that's the Gauss integral). So we can attempt to differentiate over $a$, and see if we can get a differential equation.
Since the integrand is an even function, we only need to consider the integral over $[0, \infty)$. Let
$$
I(a) = \int_0^\infty \exp \left(-\frac{a+bx^4}{cx^2}\right)\,dx = \int_0^\infty \exp \left[-\frac{1}{c}\left(\frac{a}{x^2} + x^2\right)\right]\,dx.
$$
By taking derivative over $a$ and changing the variable by $(-1/x^2)\,dx = d(1/x)$, we get
$$
I'(a) = \int_0^\infty -\frac{1}{cx^2}\exp \left[-\frac{1}{c}\left(\frac{a}{x^2} + x^2\right)\right]\,dx = -\frac{1}{c}\int_0^\infty\exp\left[-\frac{1}{c}\left(ax^2+\frac{1}{x^2}\right)\right]\,dx.
$$
Change the variable again by letting $y=\sqrt{a}x$, then we get
$$
I'(a) = -\frac{1}{c}\int_0^\infty\exp\left[-\frac{1}{c}\left(y^2+\frac{a}{y^2}\right)\right]\sqrt{a}\,dy = -\frac{\sqrt{a}}{c}I(a).
$$
Next we only need to solve the differential equation with the known initial value $I(0)$.
