# Solving exponential integral

Any idea how to solve this integral, I tried the integration by parts, and it made the things even more difficult. The substitution didn't work either. Here is the integral:

$\displaystyle \int \frac{e^{-\frac{x^2 b c^2}{x^2 + b}}}{\left(b+x^2\right)^2}\,dx$

Or in Mathematica:

Integrate[E^(-((x^2 b c^2)/(x^2 + b)))/(x^2 + b)^2,x]

• As an indefinite integral, this is not likely doable. As a definite integral from $[0,\pi]$, this may be expressed in terms of Bessel functions. – Ron Gordon Feb 2 '14 at 11:30
• If $b \neq 0$, the integrand is analytic (or, at least, I see it's analytic) in $x \in \mathbb{R}$, so you may try a Taylor expansion about $x = 0$ in order to get this: $$\int f(x) \, dx \approx \int \left(\frac{1}{b^2}-\frac{\left(2+b c^2\right) x^2}{b^3}+\frac{\left(6+6 b c^2+b^2 c^4\right) x^4}{2 b^4}+\mathcal{O}(x^6)\right) \, dx.$$ (Output from Mathematica). I hope this may help you. Cheers! – Dmoreno Feb 2 '14 at 11:51
• @Dmoreno Thank you very much.For this problem I need an analytic expression for the solution(numerical results or Taylor expansion would not be much of help). Good thing is that you think that the integrand is analytic if b is not equal to 0. Actually for my problem, all x,b and c are greater than zero. So the problem would be: Integrate[E^(-((x^2 b c^2)/(x^2 + b)))/(x^2 + b)^2,x,Assumptions -> {a > 0, b > 0, c > 0}] So that should be analytic, but I don't know how to solve it... – user113891 Feb 2 '14 at 13:04
• if the integral has a limit i.e. definite integral, it could be solved depending on the limits. – sky-light Aug 18 '15 at 9:47