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

Method (no overloads)

Summary

A product of powers of quotients of polynomials in x, a symbolic
exponent among them, where one root stands in two of the bases, written over each
linear once: K (x - r_1)^(e_1) ... (x - r_k)^(e_k), with e_i the sum of the
exponents r_i has in the bases, symbols and all. K is the integrand over
that product, and its logarithmic derivative is zero -- that of q^s is
s q'/q for any constant s and on every branch, and of e^(s ln(q)) the same -- so it is constant wherever it is defined and stands in front of the answer,
the way Rubi writes sqrt(c - a^2 c x^2)/sqrt(1 - a^2 x^2) in front of its.

Remarks

e^atanh(a x) x^m sqrt(c - a^2 c x^2) is sqrt((1 + a x)/(1 - a x)) x^m sqrt(c - a^2 c x^2):
the root 1/a stands in both bases with exponents -1/2 and 1/2, and
-1/a with 1/2 twice, so the integrand is K x^m (a x + 1); and
e^(2 p atanh(a x)) (c - a^2 c x^2)^p is K (a x + 1)^(2 p). Each is a sum of
powers to the rules, answered in milliseconds. SolveByWritingEachLinearOnce(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) reads rational exponents only, and declined both: the first ran out of time behind it
and the second was left unevaluated. Rubi's 7.3.6 and 7.4.2.
https://github.com/asc-community/AngouriMath/issues/233
Only where it helps: a root stands in two sums to powers that are not positive whole
numbers and keeps a power once they are added, and the product is one power that is not
whole, of x or of a linear, beside whole powers of the other linears above the bar. Its
antiderivative is then a sum of powers of that one, which holds on every interval
between the roots whatever the exponent, and K is constant on each. Two powers
that are not whole -- e^(n atanh(a x))/(c - a^2 c x^2) is
K (a x + 1)^(n/2 - 1) (a x - 1)^(-n/2 - 1) -- are left to the rules behind this one,
which answer them as a power of the quotient, and one beside a whole power below the bar
is hypergeometric. Rational exponents alone are the other rule's. The answer is checked
at sampled points before it is given.

AngouriMath

Table of contents

AngouriMath

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

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