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SolveByRotatingAwayTheLinearTermInTheTangent(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

A root of a quadratic in tan(x) with a linear term, rotated until it has none:
1/sqrt(a + b tan(x) + c tan(x)^2) is 1/sqrt(A + C tan(y)^2) under
x = y + arctan(m), and the tangent substitution then leaves
1/((1 + t^2) sqrt(A + C t^2)), which is answered in closed form.

Remarks

tan(y + f) is (tan(y) + m)/(1 - m tan(y)) with m = tan(f), which
is the Möbius map that preserves 1 + tan^2 -- the factor the tangent
substitution's dx brings. Under it the quadratic's linear coefficient becomes
b(1 - m^2) + 2(c - a)m, zero for a root of b m^2 + 2(a - c)m - b, whose
discriminant 4((a - c)^2 + b^2) is never negative: the two roots are the
quadratic form's two perpendicular directions and either will do.
The radicand is assembled rather than substituted into. Writing
tan(x) as the quotient inside the root leaves a nested quotient that nothing
downstream reduces, so the substitution never sees the rotated quadratic; here the
root becomes N(S)^p (1 - m S)^(-2p) with N(S) = A + C S^2 computed in
closed form, A = a + b m + c m^2 and C = a m^2 - b m + c. For a half-odd
power the modulus is what the root leaves -- sqrt(N/(1 - mS)^2) is
sqrt(N)/|1 - mS| -- so a sgn(1 - m tan(y)) comes out in front, constant
between its zeros as every such sign in these rules is.
The antiderivative is of the rotated integrand, so the answer is it at
x - arctan(m): a constant shift of the argument, which an antiderivative is
free to have. Rubi's 4.3.9 and the cotangent shapes the tangent substitution rewrites
into tangents on the way in.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

SolveByRotatingAwayTheLinearTermInTheTangent(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)