Integrate[1/(a + b Sech[z])^2, z] == ((b + a Cosh[z]) Sech[z] (-((1/(a^2 - b^2)^(3/2)) (2 b (-2 a^2 + b^2) ArcTan[((-a + b) Tanh[z/2])/Sqrt[a^2 - b^2]] (b + a Cosh[z]) Sech[z])) + z (a + b Sech[z]) + (a b^2 Tanh[z])/(a^2 - b^2)))/ (a^2 (a + b Sech[z])^2)

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2002-12-18