# Integration of a rational function

I am trying to solve the following rational integral (I got assistance here):

$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$ where $\{a, b, c\}\in \mathbb{R}$. I would like also to find an analytical solution using Mathematica. I have tried Integrate,

Integrate[(a + x)/(b^2 + (a + x)^2)*1/(
1 + c*(a - x)^2), {x, -Infinity, Infinity}]


but this gives me an imaginary result based on $a$, $b$ and $c$. I believe the result should be 100% real.

Am I using Integrate the correct way?

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"I believe the result should be 100% real." - no doubt, but you should know that sometimes Mathematica, out of convenience, expresses real results using complex numbers, somewhat similar to expressing the sine as complex exponentials. Anyway, have you seen Assuming[]? – J. M. May 15 '13 at 11:42
@J.M. I'll take a look at it and let you know how it turns out, thanks. – BillyJean May 15 '13 at 11:45
...or the Assumptions option to Integrate? Assumptions -> {a ∈ Reals, b ∈ Reals, c ∈ Reals} would probably work. – gpap May 15 '13 at 11:46
Alternatively, use the Residue Theorem. Implicitly assuming $b\gt 0$ (WLG) and $c\gt 0$, we may (very quickly) obtain the solution as 2 \[Pi] I Sum[Residue[(a + x)/((b^2 + (a + x)^2) (1 + c (a - x)^2)), {x, z}], {z, {b I - a, I/Sqrt[c] + a}}] // Simplify, yielding $\frac{2 a \sqrt{c} \pi }{1+2 b \sqrt{c}+4 a^2 c+b^2 c}$. – whuber May 15 '13 at 12:56

Try this:

Integrate[(a + x)/((b^2 + (a + x)^2) (1 + c*(a - x)^2)), {x, -Infinity, Infinity},
Assumptions -> {a ∈ Reals, b ∈ Reals, c ∈ Reals, b != 0, c > 0}]


Which gives:

(-4 a b^2 c π + 2 a Sqrt[c] (1 + 4 a^2 c + b^2 c) π Abs[b])/
(Abs[b] + c (8 a^2 - 2 b^2 + (4 a^2 + b^2)^2 c) Abs[b])


Note that I've added additional assumptions so that Mathematica doesn't return the result as a ConditionalExpression.

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