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

Method (no overloads)

Summary

(A + B cos(y) + C sin(y))/(a + b cos(y) + c sin(y))^n, for a whole n, by
the denominator, its derivative and a constant: the numerator is
alpha S + beta S' + gamma for S the base, and each power of S comes down to 1/S, which the half-angle tangent answers.

Remarks

With S' = c cos(y) - b sin(y), matching the cosine, the sine and the constant
gives alpha = (B b + C c)/(b^2 + c^2), beta = (B c - C b)/(b^2 + c^2) and
gamma = A - a alpha; beta S'/S^n integrates to a logarithm or a power of
S. For the powers, with T = b sin(y) - c cos(y) = -S',
T^2 = b^2 + c^2 - (S - a)^2, from which

d/dy (T S^m) = (1 + m) S^(m+1) - a (1 + 2m) S^m + m (a^2 - b^2 - c^2) S^(m-1)

which for m = 1 - k writes int S^(-k) through int S^(1-k) and
int S^(2-k), down to int 1/S and y. Rubi's
(A + B cos(x) + C sin(x))/(a + b cos(x) + c sin(x)) and its kin were declined:
under the half-angle tangent each is a rational function over a quadratic with every
coefficient a symbol, beside 1 + t^2, and the square of the denominator was
answered with coefficients past the fiftieth degree in them.
Exact, as the answer for 1/S it is built on: a linear combination and an
identity of derivatives. a^2 = b^2 + c^2, where the base is a square of the
half angle and the recurrence divides by zero, is declined here, as is
b^2 + c^2 = 0, where the base is a + b e^(±i y); both decided. A constant
numerator over the first power is 1/S itself, left to the rules that answer it.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

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