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I've tried to integrate by part, but it seems that Mathematica is still not able to integrate. 

Integrate[(24(1-z)^a)/((-1+(1-z)^a)^2z)+(24a(1-z)^(3a) Hypergeometric2F1[1, 1, 2 - a, 1/z])/((-1+a)(-1+(1-z)^a)^3 (-1+z)z)-(24 a Hypergeometric2F1[1, 1,2+a,1/z])/((1+a)(-1+(1-z)^a)^3 (-1+z)z),z]

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We actually know the result exists but we want to get it from this integral.

Is there any commands in Mathematica that are usually helpful in doing integral?

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  • $\begingroup$ In general, it can be very useful to provide Assumptions to Integrate. For instance the range of a, or whether it is real and the like. By default, Mathematica assumes the most general scenario which is often not necessary, so that assumptions help to find a (simpler) solution $\endgroup$ – Lukas May 31 '16 at 20:52
  • $\begingroup$ One can calculate such an integral for several special values of a e.g. -1/2, 1/2, 3/2. For a integer it is ComplexInfinity. What are assumptions on a? $\endgroup$ – Artes May 31 '16 at 21:08
  • $\begingroup$ @Artes Even more generally, at half-integer "a" $\endgroup$ – yarchik May 31 '16 at 21:11
  • $\begingroup$ @Artes I think 0<a<1 $\endgroup$ – Nahc May 31 '16 at 21:13
  • $\begingroup$ Even Integrate[(24 (1 - z)^a)/((-1 + (1 - z)^a)^2 z), z, Assumptions -> 0 < a < 1] returns unevaluated, so it is not surprising that the full integral does as well. In fact, this simpler integral returns an answer only for a a rational number, and even then typically in terms of RootSum. The only instance of a solution to the full integral that I found was a -> 1/2. $\endgroup$ – bbgodfrey Jun 1 '16 at 4:27

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