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

 Method (no overloads)

Summary

A power of the variable times an odd half-power of a quadratic without a linear term —
x^m (a + b x^2)^(k/2) with k odd — turned into a power of the sine times
a power of the cosine, which
IntegrateAPowerOfSineTimesAPowerOfCosine(AngouriMath.Entity,PeterO.Numbers.ERational,PeterO.Numbers.ERational,AngouriMath.Entity,AngouriMath.Entity) answers in closed form.

Remarks

1/(1 + x^2)^(3/2), x^5/sqrt(5 + x^2), 1/(x^2 sqrt(1 - x^2)) and
their kin had no antiderivative. A square root of a quadratic is the largest single
class the Rubi suites leave unanswered here, and this is the part of it that comes out
closed.
Why the trigonometric substitution rather than Euler's. Euler's rationalises the
radical and lands on a rational function, which then has to be integrated by the whole
chain — a speculative hand-off that
#1244 measured
at tens of seconds on integrands it never answers. This lands on sin^p cos^q,
which is a Laurent polynomial after one more substitution and is integrated term by
term without asking the integrator anything. Same class of integrand; one of the two
routes is closed and the other is a search.
Three substitutions, chosen by the two signs. With r = sqrt(|a/b|):
a > 0, b > 0: x = r tan(t), under which a + b x^2 = a sec(t)^2 and dx = r sec(t)^2 dt, leaving sin(t)^m cos(t)^(-m-k-2).
a > 0, b < 0: x = r sin(t), under which a + b x^2 = a cos(t)^2 and dx = r cos(t) dt, leaving sin(t)^m cos(t)^(k+1).
a < 0, b > 0: x = r sec(t), added in
#1277. Its domain
is two intervals rather than one, and the sign that makes the second one right is
written out in IntegrateAPowerTimesARadicalQuadratic(AngouriMath.Entity,AngouriMath.Entity,System.Int32,System.Int32,PeterO.Numbers.ERational,PeterO.Numbers.ERational).
 Both signs negative is a radicand negative everywhere, with no real integrand to
integrate, and there is no fourth case.
Coming back. The answer arrives in sin(t), cos(t) and
tan(t), each of which is algebraic in x under the substitution, and in
t itself wherever the recursion bottoms out on int 1 dt — which is where
an arcsin or an arctan enters an otherwise algebraic answer.
Volunteered, and deliberately so — this is the one rule of the five that is not
scoped to AnsweringTheQuestionAsked. Scope is right for a
rule whose answers only ever let some other search carry on into ground that was
doomed, which is what
#1265 is about.
It is wrong for this one, because the sub-integrals it answers are the ones integration
by parts asks for and then uses: int x arcsin(x) dx leaves
int x^2/sqrt(1 - x^2) dx, which is squarely this rule's and was refused for
sitting one level down. Ungating it is worth ten more of the Rubi sample and
measures free — same wall clock, same timeouts, and the same 28.8 s over a probe of
integrands that are declined either way.
 https://github.com/asc-community/AngouriMath/issues/718

























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