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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
Two roots, as in cos(y)^(3/2)/sqrt(a + a sec(y)), are real together where the
cosine is negative too, and there each written apart may have turned its sign: the answer
said provided cos(y) >= 0, half of the integrand's domain, for every one of 165
rows of Rubi's 4.5.4.2. Which turn is known from the constants, and the answer carries
the cosine's sign where they leave an odd number turned. Three, as in
sqrt(a + a sec(y)) sqrt(sec(y)) sqrt(cos(y)), leave a root of 1 - t^2 in
the answer in t, which said provided cos(y) >= 0 where for a negative a the
integrand is real only beyond it; the answer in t is kept beyond where it differentiates
back there.
https://github.com/asc-community/AngouriMath/issues/233

AngouriMath

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AngouriMath

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

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