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

 Method (no overloads)

Summary

Half-odd powers of a ± a sec(y), beside powers of d sec(y) or
d cos(y) and anything rational in the sine and cosine of y, by the
half-angle tangent t = tan(y/2): 1 + sec(y) is 2/(1 - t^2),
1 - sec(y) is -2 t^2/(1 - t^2) and sec(y) is
(1 + t^2)/(1 - t^2), so the whole is a rational function of t beside a root of
1 - t^2 or of 1 + t^2 -- or of both, which is elliptic and declined. The
cosecant's the same way through the complement, csc(y) = sec(pi/2 - y).

Remarks

Rubi's (a + b sec)^m (d sec)^n files with a^2 = b^2 hold about a thousand
problems with a half-odd m, and none was answered, sqrt(1 + sec(x)) included: the half angle at which a + a cos(y) is a square, the rule before this
one, leaves a root of the cosine below the bar here, since 1 + sec(y) is that
square over cos(y).
 Exact where the integrand is real. a (1 + sec(y)) is not negative with
t inside (-1, 1) for a positive a and outside it for a negative
one, and either way (2a/(1 - t^2))^p is (2a)^p (1 - t^2)^(-p) for the
principal powers; so for the secant's power beside it. For 1 - sec(y) the square
t^2 comes out of the root as |t|, a sign constant between the zeros of
tan(y/2) in front. At the question asked or one below it, since it lands on the
chain in t.
https://github.com/asc-community/AngouriMath/issues/718

























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