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SolveAPowerOfTheSecantBesideAPowerOfAnImaginaryTangentSumInTheSum(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

A power of the secant, or of the cosine, of z beside a power of A + i A tan(z),
whatever the powers, integrated in that sum: under u = A + i A tan(z),
du = i A sec(z)^2 dz and sec(z)^2 = (u/A) ((2 A - u)/A), so
sec(z)^s u^n dz is u^n ((u/A) ((2 A - u)/A))^r du/(i A) with
r = (s - 2)/2, up to a factor constant wherever it is continuous; and
(u/A)^r ((2 A - u)/A)^r is (sec(z)^2)^r exactly, the two being
1 + i tan(z) and 1 - i tan(z), whose arguments are opposite and less than a
right angle.

Remarks

sec(x)^3 sqrt(a + i a tan(x)) was declined, with the rest of Rubi's 4.3.1.2 whose
powers do not add up to a whole number, which
SolveAPowerOfAnImaginaryTangentBesideAPowerOfTheSecant(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) integrates as an
exponential and this does not: sec^5/(a + i a tan)^(3/2),
(e cos)^(3/2) sqrt(a + i a tan). In u each is a power of u beside a
power of 2 A - u.
The constant is not written, as there: the answer is the integrand times the
antiderivative in u over what that antiderivative differentiates back to, a
quotient constant wherever it is continuous.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

SolveAPowerOfTheSecantBesideAPowerOfAnImaginaryTangentSumInTheSum(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)