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

 Method (no overloads)

Summary

A quotient of two homogeneous polynomials in sin(u) and cos(u),
integrated by t = tan(u) — which turns it into a rational function of t whenever the two degrees differ by an even number.

Remarks

1/(cos(x) + sin(x))^6 took 231 seconds to be declined and
1/(b^2 cos(x)^2 + a^2 sin(x)^2) was one of the Rubi sample's three timeouts.
Both are this shape, and so is sin(x)/(cos(x)^3 + sin(x)^3).
Why the degrees decide it. Writing t = tan(u) and c = cos(u), a
homogeneous polynomial of degree n in the pair is c^n times a polynomial
in t alone. So a quotient of degrees n over d is
c^(n-d) N(t)/D(t), and with c^2 = 1/(1 + t^2) and
dx = dt/(a(1 + t^2)):
f dx = [N(t)/D(t)] (1 + t^2)^((d - n)/2 - 1) dt / a
which is rational exactly when d - n is even. Odd is a genuine boundary rather
than a first cut: there the substitution leaves a square root of 1 + t^2 behind,
which is a different problem and not a rational one.
 This is the third of Bioche's rules, and it is the one the other two do not cover:
SolveByTangentSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) answers an integrand that is a function of
tanalone, and 1/(b^2 cos^2 + a^2 sin^2) is not — it is a function
of tan times sec^2, which is what the degree difference of two is saying.
Where the answer holds.t = tan(u) is a bijection on each interval
between the poles of the tangent, so the answer is an antiderivative on each of them —
the standing caveat on this substitution, shared with the half-angle one, and not
something this writes as a condition.
 https://github.com/asc-community/AngouriMath/issues/718

























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