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This question is a continuation of another question of mine, in an attempt to gain more understanding on how to use Mathematica to simplify special functions.

Consider the expression:

expr = Hypergeometric2F1[1, 3 - z, 1 + z, -1];

If I try

FunctionExpand[expr]

Hypergeometric2F1[1, 3 - z, 1 + z, -1]

Mathematica seems to suggest that no simpler expression for this hypergeometric function can be written down. However if we dig deeper, we can find an identity on the wolfram functions webpage that reduces expr to a simple expression involving just two gamma functions in this case:

Hypergeometric2F1[a, b, a - b + 3, -1] == ((Sqrt[Pi] Gamma[3 + a - b])/(2^a ((b - 1) (b - 2)))) (1/(Gamma[(a + 1)/2] Gamma[a/2 + 1 - b]) - a/(Gamma[a/2 + 1] Gamma[(a + 3)/2 - b]) + ((a (a + 1))/4) (1/(Gamma[(a + 3)/2] Gamma[a/2 + 2 - b])))/.a->1/.b->3-z//FullSimplify

enter image description here

Considering that the identity can be found on the wolfram functions webpage, I presume that Mathematica should be aware of it. Therefore my question: Is it possible to obtain this simplification directly in Mathematica, without having to look it up manually on a webpage? If yes, what is the appropriate command?

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    $\begingroup$ "Considering that the identity can be found on the Wolfram Functions webpage, I presume that Mathematica should be aware of it." - not necessarily. Note that adding FullSimplify[] in your last snippet did not yield True. $\endgroup$ – J. M.'s technical difficulties Apr 11 '17 at 16:01

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