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

Method (no overloads)

Summary

An integrand that is a rational function of tanh(y) -- of e^(2y), that
is, even as a function of v = e^y -- with a symbol among its coefficients,
integrated by u = tanh(y): e^(2y) is (1 + u)/(1 - u) exactly and
dx is du/((1 - u^2) d), so the integrand is a rational function of
u with the written factors kept -- a + b coth(y)^2 is
(b + a u^2)/u^2 -- and roots of such functions are admitted as they are.
Failing evenness, the half of y: every e^(k y) is v^(2k) for
v = e^(y/2), and the substitution is u = tanh(y/2), the hyperbolic
half-angle.

Remarks

Under u = e^x, Rubi's 1/(a + b coth(x)^2)^2 is a rational function of the
palindromic quartic (a + b) u^4 + 2(b - a) u^2 + (a + b) squared, whose roots
are what the budget went on -- fifty-seven of the four hundred and seventeen in the
hyperbolic sample were this; here it is u^4/((b + a u^2)^2 (1 - u^2)), which
the symbolic partial fractions answer. Rubi's route for the hyperbolic tangent,
cotangent, secant and cosecant families. With rational coefficients the exponential
substitution's quartics are factored over the rationals and answered, and are left
to it. Before the substitution search, which with a symbolic slope spends the whole
budget on u = c + d x and never returns here.
How the written factors survive. Each maximal rational function of v among the factors, powers and radicands -- and not the whole quotient -- is read as
v^m R(s) with s = v^2 and m either 0 or 1, which every rational
function even or odd in v is, and homogenised on its own: P(s)/Q(s) as
the polynomials sum p_i (1 + u)^i (1 - u)^(d - i) over the same for Q,
with the powers of 1 - u and 1 + u both share divided out; the
conjugate that SolveByHyperbolicTangentSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) clears its root
with squares every degree, and left this integrand a quotient of degree sixteen
whose common factor only a gcd over the symbols would find. The stray v's
cancel in an even integrand -- sech(x)^4 is (2v/(s + 1))^4 -- and one
left over at the top is an odd integrand, which the half of y takes.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

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