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SolveASecantBesidePowersOfItsConjugateSums​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

sec(y) (a ± a sec(y))^m (c ∓ c sec(y))^n, with n half-odd and m anything,
integrated in u = a ± a sec(y): the two sums multiply to -a c tan(y)^2, so the
root of the second is tan(y) over the root of u times a constant,
sec(y) tan(y) dy is du over one, and what is left is
u^(m - 1/2) (2 - u/a)^(n - 1/2). The cosecant's are the same with the cotangent.

Remarks

Rubi's sec(e + f x) (a + a sec(e + f x))^m sqrt(c - c sec(e + f x)) and two more of
4.5.2.3 were declined. The constants are not written: the answer is the integrand times
the antiderivative in u over what that differentiates back to, a quotient whose
square is one; and it is kept only where that square is one at the sampled points.
https://github.com/asc-community/AngouriMath/issues/718

























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