Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to know the underlying reason for different outcomes for the two integration operation below. One of them includes a few assumptions, otherwise both have the same integrand:

Clear[a, b, c, d]
Integrate[(1/(a q^2 + b q Cos[θ] - c) - 1/(a q^2 + b q Cos[θ] + c)), {θ, 0, 2 π}, 
  Assumptions -> {a, c, b} ∈ Reals]
ConditionalExpression[-((2 Log[-((c - b q + a q^2)/Sqrt[-c^2 + b^2 q^2 - 2 a c q^2 - a^2 q^4])])/ Sqrt[-c^2 + b^2 q^2 - 2 a c q^2 - a^2 q^4]) + (2 Log[(c - b q + a q^2)/ Sqrt[-c^2 + b^2 q^2 - 2 a c q^2 - a^2 q^4]])/Sqrt[-c^2 + b^2 q^2 - 2 a c q^2 - a^2 q^4] + (2 (-Log[-((c + q (b - a q))/Sqrt[-c^2 + b^2 q^2 + 2 a c q^2 - a^2 q^4])] ....
Integrate[(1/(a q^2 + b q Cos[θ] - c) - 1/(a q^2 + b q Cos[θ] + c)), {θ, 0, 2 π}]
-((2 π Sqrt[1 - (2 b q)/(-c + q (b + a q))])/( c + q (b - a q))) - 
 (2 π)/((c + q (b + a q)) Sqrt[1 - (2 b q)/(c + q (b + a q))])
share|improve this question
Well, in the first one you tell Mathematica that a, b and c are real whereas in the second Mathematica treats them as complex as that's the default assumption. – RunnyKine Aug 20 '14 at 4:58
@RunnyKine So the second outcome should reduce to the first (more lengthy) output when a,b,c real? – thils Aug 20 '14 at 5:08
Well, the first one appears lengthy because it's a ConditionalExpression. So it's giving you various answers that correspond to different combinations of a, b and c in the Reals. – RunnyKine Aug 20 '14 at 5:10
The second outcome is applicable to more general conditions, and should reduce to the first with application of those a,b,c combinations, how do you show this? – thils Aug 20 '14 at 5:47
Probably not as simple as it seems, running Assuming[{a, c, b} \[Element] Reals, FullSimplify[ eq1 == eq2]] does not simplify to True (where eq1 and eq2 are the respective integral outputs) – rhermans Aug 20 '14 at 10:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.