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I was trying to (re)calculate a problem of an older Wolfram blog post (Problem 11457, by M. L. Glasser) with Mathematica 9.0.0.0 (on OS X 10.8.2).

Assuming[0 < a < b, Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}]]

Instead of the expected solution, it just returns the integral unevaluated. Is this a regression?

More details: As pointed out in the commentes, the indefinite integral

Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], x]

still gives the same result in Mathematica 8 and 9.

The next two each returned ConditionalExpression in Mathematica 8 but return unevaluated in Mathematica 9:

Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}]
Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}, Assumptions -> 0 <= a <= b]

The actual problem

Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}, Assumptions -> 0 < a < b]

computes correctly to ((a - b)^2 \[Pi])/(4 (a + b)) in Mathematica 8 but still returns unevaulated in Mathematica 9.

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    $\begingroup$ I can confirm that it does not work in MMA 9 win 7 64 bit, but works in MMA 8.0.1. In MMA8 I get ((a - b)^2 \[Pi])/(4 (a + b)) $\endgroup$
    – Ajasja
    Jan 24, 2013 at 12:16
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    $\begingroup$ The topic is misleading. I thought it was about Regression in Statistics $\endgroup$
    – asim
    Jan 24, 2013 at 14:35
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    $\begingroup$ Mathematica 8 and 9 give the same correct indefinite integral. The difference is in the calculation for the limits of integration. $\endgroup$
    – Searke
    Jan 24, 2013 at 15:45
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    $\begingroup$ Here are another instances of definite integrals which are unevaluated in ver.9 while they are in ver.8 mathematica.stackexchange.com/questions/18327/… $\endgroup$
    – Artes
    Jan 24, 2013 at 17:18
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    $\begingroup$ [I am NOT putting this into a response.] Yes, this appears to be a regression. Investigating... $\endgroup$
    – Daniel Lichtblau
    Jan 25, 2013 at 1:18

1 Answer 1

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This works in V9.0.1:

Assuming[0 < a < b, 
  Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}, 
   GenerateConditions -> False]] // Timing
(* {3.835651, ((a - b)^2 π)/(4 (a + b))} *)
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  • $\begingroup$ Good catch! Works with v9.0.0.0 also. $\endgroup$
    – Norbert Fabritius
    Oct 18, 2013 at 11:40

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