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SolveAHalfPowerOfOnePlusASine(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

(a + b sin(c x + d))^n or the same with a cosine, for a^2 = b^2 and
n a positive half-integer, in closed form.

Remarks

sqrt(1 + sin(x)) had no antiderivative. It is -2 cos(x)/sqrt(1 + sin(x)):
differentiating that gives (2 sin (1 + sin) + cos^2)/(1 + sin)^(3/2), and with
cos^2 = 1 - sin^2 = (1 - sin)(1 + sin) the numerator is (1 + sin)^2. That
cancellation is what a^2 = b^2 buys, and it is the whole of the rule.
Above the base case, one step of by parts lowers n by one:

int (a + b sin)^n = -b cos (a + b sin)^(n-1)/(c n) + (a (2n - 1)/n) int (a + b sin)^(n-1)

which is checked by differentiating the first term and using b^2 cos^2 =
(a - b sin)(a + b sin). Applied until n is 1/2, so
(1 - sin(2x/3))^(5/2) is two steps and a base. A cosine is the same rule
with sin replaced by -cos in the first term, since d cos = -sin.
No condition is owed.a + b sin is never negative when a^2 = b^2,
so its square root is real wherever the integrand is, and the base-case answer is a
single antiderivative across the zeros of 1 + sin: both it and the integrand
vanish there. a^2 = b^2 is required decidably, so a symbolic a beside a
numeric b is not this shape. Rubi's rule for the same case is the source.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

SolveAHalfPowerOfOnePlusASine(AngouriMath.Entity,AngouriMath.Entity.Variable)