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SolveAPowerOfAHyperbolicCosineAndSinePlusTheirAmplitude(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

(a + b cosh(y) + c sinh(y))^n for a^2 = b^2 - c^2, y linear in
x and n half-odd, in closed form:
SolveAPowerOfACosineAndASinePlusTheirAmplitude(AngouriMath.Entity,AngouriMath.Entity.Variable)'s recurrence for the
hyperbolic functions. With S the base and T = b sinh(y) + c cosh(y),
T' = S - a, S' = T and T^2 = S (S - 2a), the last of which is
a^2 = b^2 - c^2, and from them

d/dy (T S^(n-1)) = n S^n - a (2n - 1) S^(n-1)

which steps n down to 1/2, where the integral is 2T/sqrt(S), or up
to -1/2, where it is (ln(T - i q sqrt(S)) - ln(T + i q sqrt(S)))/(i q) for
q = sqrt(2a).
https://github.com/asc-community/AngouriMath/issues/233

Remarks

Rubi's sqrt(b cosh(x) + c sinh(x) + sqrt(b^2 - c^2)) and the rest of the twelve in
its 6.7.1 were declined or past the budget: b cosh(y) + c sinh(y) is
sqrt(b^2 - c^2) cosh(y + phi), so the base is a square of the half angle turned by a
phase, and SolveByTheHalfAngleWhereOnePlusAHyperbolicCosineIsASquare(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) reads
a ± a cosh(y) without one. The recurrence needs no phase.
The base case is two logarithms rather than the arctangent of T/(q sqrt(S)): for
real b > |c|, S is not negative whichever sign a has, and for a
negative a that argument is imaginary and past i, on the arctangent's cut.
The two logarithms are of complex conjugates for a positive a and of two numbers of
one sign for a negative one, real and continuous between the zeros of S either way.
The derivations use only S^(3/2) = S sqrt(S) and (q sqrt(S))^2 = 2a S, which
hold for the principal powers. a^2 = b^2 - c^2 is decided, not assumed, and all three
are there: b = 0 is a ± i a sinh(y) and c = 0 is a ± a cosh(y),
which the half-angle rules answer.
Half-odd powers only. The whole ones are the exponential substitution's, which answers
1/S^2, 1/S^3 and 1/S^4 in 182, 272 and 416 characters, where the
same recurrence stepped up to T/(a S) gives 350, 539 and 728.

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AngouriMath

SolveAPowerOfAHyperbolicCosineAndSinePlusTheirAmplitude​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

SolveAPowerOfAHyperbolicCosineAndSinePlusTheirAmplitude(AngouriMath.Entity,AngouriMath.Entity.Variable)