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I am trying to solve an integral given below:

Integrate[r^2/(-α r^3 + r^2 - 2 m r + Q^2), r]

but since coefficients of this cubic polynomial are as parameters, so i could not get any reasonable answer from mathematica, output: 

(* -RootSum[Q^2 - 2 m #1 + #1^2 - α #1^3 &, (Log[r - #1] #1^2)/(2 m - 2 #1 + 3 α #1^2) &] *)

Please help me in finding this integral, what information i have for these coefficients is as: \alpha is a positive number, and m and Q are reals, with m positive real.

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  • $\begingroup$ RootSum[...] is a reasonable answer, see e.g. a related question How do I work with Root objects? $\endgroup$ Commented Dec 29, 2014 at 19:03
  • $\begingroup$ @ Artes thanks for your comment, I am sorry that I cant understand the logic of root sum from the link you mentioned above, could you please briefly describe it here? $\endgroup$ Commented Dec 29, 2014 at 19:11
  • $\begingroup$ @BushraMajeed You will not understand RootSum unless you learn about Root. You have to try a bit harder. You can compute that expression for special values of prarameters e.g. With[{Q = 1, m = 2, α = 0}, -RootSum[ Q^2 - 2 m #1 + #1^2 - α #1^3 &, (Log[r - #1] #1^2)/(2 m - 2 #1 + 3 α #1^2) &]]. $\endgroup$ Commented Dec 29, 2014 at 19:22
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    $\begingroup$ how about Integrate[r^2/(-\[Alpha] r^3 + r^2 - 2 m r + Q^2), r] //Normal//ToRadicals? $\endgroup$ Commented Dec 29, 2014 at 19:30
  • $\begingroup$ @chris I believe that makes a good answer, probably better if you explain a bit about RootSum[]. I'll surely upvote it. $\endgroup$ Commented Dec 29, 2014 at 20:22

2 Answers 2

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or using Rubi integration package (Rubi does not use Root or RootSum in its answers)

(to use Rubi, just change Integrate with Int, everything remains the same:

 Int[r^2/(-a r^3 + r^2 - 2 m r + q^2), r]

Mathematica graphics

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Let's look at this in steps:

Integrate[r^2/(-α r^3 + r^2 - 2 m r + Q^2), r]

returns the formal solution as a sum of the roots of the polynomial of the first argument of RootSum applied to the slots of the second argument.

(* -RootSum[#1^3 (-α)+#1^2-2 #1 m+Q^2&,(#1^2 log(r-#1))/(3 #1^2 α-2 #1+2 m)&] *)

Indeed if we ask Mathematica for an explicit expression

Normal[Integrate[r^2/(-α r^3 + r^2 - 2 m r + Q^2), r]]

Mathematica graphics

Involving the 3 Root

Or even more explicitly 

ToRadicals@Normal[Integrate[r^2/(-α r^3 + r^2 - 2 m r + Q^2), r]]

Mathematica graphics

which I think illustrates nicely why Mathematica (and possibly you) prefers the original expression.

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    $\begingroup$ @ Chris, these useful commands are really helpful for a reasonable output, though the expression for this case is very long and complex numbers involved there, due to roots, and i am considering the case when this polynomial has two complex roots and one real, so I cant avoid these complex numbers. Thanks for your suggestions. $\endgroup$ Commented Dec 30, 2014 at 11:34

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