How to obtain a approximate closed-form expression for $\int_0^1 \frac{e^{-a x-\frac{1}{b x}}}{c x} \, dx$?

Question

How to obtain an approximate closed-form expression for $$\int_0^1 \frac{e^{-a x-\frac{1}{b x}}}{c x} \, dx,\quad a>0\land b>0\land c>0,$$ I know there may be a tradeoff between complexity and accuracy and hence multiple solutions exist. The preferred result is expressed with elementary functions. Thank you for your help.

Answer 1

Let $\mu=\sqrt{\frac{a}{b}}$ and $\nu=\sqrt{ab}$: $$\begin{eqnarray*}J(a,b)=\int_{0}^{1}\exp\left(-ax-\frac{b}{x}\right)\frac{dx}{x}&=&\int_{0}^{1}\exp\left[-\nu\left(\mu x+\frac{1}{\mu x}\right)\right]\frac{dx}{x}\\&=&e^{-2\nu}\int_{0}^{\mu}\exp\left[-\nu\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\right]\frac{dx}{x}\\&=&e^{-2\nu}\int_{-\infty}^{\sqrt{\mu}-\frac{1}{\sqrt{\mu}}}\frac{2 e^{-\nu z^2}\,dz}{\sqrt{4+z^2}}\\&=&2e^{-2\nu}\int_{-\infty}^{\frac{1}{2}\left(\sqrt{\mu}-\frac{1}{\sqrt{\mu}}\right)}\frac{e^{-4\nu z^2}\,dz}{\sqrt{1+z^2}}\end{eqnarray*}$$ and $ \int_{-\infty}^{0}\frac{e^{-4\nu z^2}}{\sqrt{1+z^2}}=\frac{1}{2}e^{2\nu}K_0(2\nu)$, so that: $$ J(a,b) = K_0(2\nu)+2e^{-2\nu}\int_{0}^{\frac{1}{2}\left(\sqrt{\mu}-\frac{1}{\sqrt{\mu}}\right)}\frac{e^{-4\nu z^2}\,dz}{\sqrt{1+z^2}}$$ where $K$ is a modified Bessel function of the second kind and the remaining integral is easy to approximate through the Cauchy-Schwarz inequality.