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

 Method (no overloads)

Summary

A half-odd power of a ± a sin(y) times whole or half-odd powers of other linears
in the same sine, c + d sin(y), and nothing else of x: by
SolveByTheHalfAngleWhereOnePlusASineIsASquare(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean,System.Boolean), asked first. The cosine's
the same.

Remarks

Rubi's 1/(sqrt(a + a sin(e + f x)) (c + d sin(e + f x))^2),
1/((a + a sin(e + f x))^2 sqrt(c - c sin(e + f x))) and their kin in 4.1.2.1
ran past the budget. The half angle answers each, its question in u taking
tens of milliseconds, but it was not reached: the polynomial term above it wrote the
reciprocal as a power, took the constant out of the root, and asked the substitution
search below that, whose simplification of what it found took the whole budget.
 The answer is that rule's, so exact on the same terms. Read narrowly: a product of
powers of linears in one sine or one cosine, one of them half-odd and of
a ± a sin(y), and at least two of them, since one alone is
SolveAHalfPowerOfOnePlusASine(AngouriMath.Entity,AngouriMath.Entity.Variable)'s.
https://github.com/asc-community/AngouriMath/issues/233

























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