Analytical evaluation of a definite integral

I want to evaluate analytically the integral $\int _ {-1} ^ {1} \mathrm{d} x \frac{1}{ax+b} e^{cx} \sqrt{1-x^2}$, where $a$, $b$ and $c$ are real numbers.

I tried Mathematica, but with no success. Any ideas?

• This isn't worth being an answer, but I would start by rewriting a bit and differentating in the exponent to get rid of the fraction. – user111187 May 12 '14 at 20:45
• Substitution x=sinϕ may also be useful. Then the integrand can be expressed as a function of sinϕ only. – Urgje May 12 '14 at 21:15
• Could user111187 be a bit more specific? – user127054 May 12 '14 at 21:21
• @user127054 My idea was to write the integrand as $\frac{1}{a} \frac{dx}{x+b/a} e^{c(x+b/a)-bc/a}$, take the factor $e^{-bc/a}$ out of the integral, and then differentiate with respect to $c$ in order to remove the $\frac{1}{x+b/a}$ factor. The problem is that you need to integrate back in order to get the original integral. – user111187 May 13 '14 at 6:50

If the quantity $$\beta(x) := \frac{\sqrt{1-x^2}}{ax + b}$$ is such that $\beta(x) \in [-1,1]$ you can define $$\sin(\alpha x) = \beta \,.$$ The integral then reduces to $$I = \int_{-1}^1 e^{cx}\,\sin(\alpha x)\,{\rm d}x \,.$$ From Wikipedia, $$I = \left.\frac{e^{cx}}{\sqrt{c^2+\alpha^2}}\,\sin(\alpha x - \phi)\right|_{-1}^1$$ where $$\phi := \frac{c}{\sqrt{c^2+\alpha^2}} \,.$$
• What is $\alpha$? – user111187 May 13 '14 at 6:45
• I was thinking of expressing $\beta(x)$ in terms of a Fourier series expansion in sines and cosines. I'll try to correct the answer and make it more explicit when I get some free time. – Biswajit Banerjee May 13 '14 at 22:16