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IntegrateAPolynomialOverABinomial(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A polynomial over a binomial a x^n + b, n >= 3, decomposed at the
n-th roots of -b/a and integrated term by term in closed form.

Remarks

1/(x^3 + 1) had an antiderivative and 1/(x^3 + 2) did not: the first
factors over the rationals and the coprime split takes it apart, the second does not
and nothing further was tried. The same for 1/(x^5 + 1), 1/(x^6 + 2) and
every 1/(a x^n + b) with a symbol in it, which no split over the rationals can
reach at all. Rubi's test suite has these by the dozen.
The decomposition is a formula, not a computation. Writing the denominator as
a (x^n - c) with c = -b/a and rho^n = c, the roots are
rho e^(i theta_k) and the residue of x^m/(x^n - c) at one of them is
rho^(m+1-n) e^(i (m+1-n) theta_k) / n. A root on the real line gives
rho^(m+1-n) cos((m+1-n) theta) / n over x - rho cos(theta); a conjugate
pair gives, over x^2 - 2 rho cos(theta) x + rho^2,

(2 rho^(m+1-n) / n) (cos((m+1-n) theta) x - rho cos((m-n) theta))

and each of those is a logarithm plus an arctangent, written out here rather than
handed back to the integrator: the quadratic is (x - h)^2 + k^2 with
h = rho cos(theta) and k = rho sin(theta), and
int (P x + Q) / ((x - h)^2 + k^2) dx is
(P/2) ln((x - h)^2 + k^2) + ((Q + P h)/k) arctan((x - h)/k). Handing the pieces
back would have the quadratic rule decide the sign of a discriminant that is
-4 rho^2 sin^2(theta) and cannot be read as negative once rho is a
symbol, and answer with a piecewise for what is one branch.
Which roots, by the sign of c. For c > 0 the real
rho = c^(1/n) puts the roots at 2 pi k / n; for c < 0 it is
rho = (-c)^(1/n) and they sit at pi (2k + 1) / n, so that every
rho and every angle is real and the answer is real on the real line. A symbol
has no sign to read, and takes the first form with rho = c^(1/n): the
factorisation x^n - c = prod (x - rho zeta_k) is an identity for any
rho with rho^n = c, so the antiderivative is correct for every
c and happens to be written through a complex rho where c is
negative -- which is what Rubi's own answer for 1/(a + b x^3) does with its
(a/b)^(1/3), and is the generic case this integrator gives elsewhere.
After every split over the rationals, so that a denominator which factors exactly
keeps the exact answer it had. A proper fraction only; an improper one has been
divided out above.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

IntegrateAPolynomialOverABinomial(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable)