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

 Method (no overloads)

Summary

A half-odd power of f = cos(y) beside sqrt(a - a f) or its reciprocal,
y linear in x, in closed form, and the sine's the same. With g the
other function and Q = a - a f, Rubi's substitution
v = g/(sqrt(Q) sqrt(f)) makes it rational: a f v^2 = 1 + f, so
f = -1/(1 - a v^2) and Q = a (2 - a v^2)/(1 - a v^2), and
dv/dy = ± sqrt(Q)/(2a f^(3/2)), plus for the cosine, so that
f^p Q^q dy = ± 2a f^(p + 3/2) Q^(q - 1/2) dv is a whole power of
1 - a v^2 over one of 2 - a v^2 or none. Its integral, back in y,
is a sum of g f^k/sqrt(Q) for half-odd k and of artanh(1/z)/sqrt(a) and artanh(sqrt(2)/z)/sqrt(2a), z = sqrt(a) v.

Remarks

Rubi's cos(c + d x)^(5/2)/sqrt(a - a cos(c + d x)) and the ten beside it in 4.2.2.1,
and 1/(sqrt(sin(x)) sqrt(a - a sin(x))) in 4.1.2.1, were answered by the sign of the
complement, u = cos(y) with sin(y) = sgn(sin(y)) sqrt(1 - u^2): right, but
489 to 1,567 characters with a sgn(sqrt(cos(y)) - 1) in them whose derivative is left
unevaluated, so that the corpus could not check them, owing a > 0 and
cos(y) >= 0, and up to 25 GB to build.
 Exact for any a of either sign, at every real y where the integrand is
defined, a negative cosine included: the step uses only v^2 = (1 + f)/(a f), a
quotient of squares, and the chain rule for principal roots of real quantities. Rubi
writes artanh(z), whose argument's square (1 + f)/f exceeds one where the
integrand is real for a positive a, a value on the cut and complex by a constant
there. artanh(1/z) has the same derivative and is ln(f (1 + z)^2)/2, since
z^2 - 1 = 1/f, and artanh(sqrt(2)/z) is ln(f (sqrt(2) + z)^2/(1 - f))/2;
each is written with ln(f) apart, because for a negative a the product
meets the cut where g changes sign inside the real domain. z is real or
imaginary, so (1 + z)^2 and (sqrt(2) + z)^2 are never negative numbers, and
f keeps its sign between the zeros of the cosine: the answer is continuous on each
interval between the zeros of f and of Q, and real where the integrand is.
 Not for a + a f, whose arguments are real beyond one for a negative a and
which the rules after this answer. At the question asked, the two powers alone, before
the polynomial term, which hands the reciprocal of their product down a level, and
q = ±1/2.
https://github.com/asc-community/AngouriMath/issues/233

























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