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SolveByLogarithmTowerAnsatz(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

An exponential times a rational function of x and ln(x), with the
logarithm allowed in the exponent too, closed by an ansatz
F = e^h P(x, L)/(D(x) L^k) with L = ln(x).

Remarks

Hearn's (-1 + (1 - x) ln(x))/(e^x ln(x)^2) is (x e^(-x)/ln(x))' and Hebisch's
e^(x + 1/ln(x)) (-1 + (1 + x) ln(x)^2)/ln(x)^2 is (x e^(x + 1/ln(x)))'; neither
had an antiderivative, and neither is reached by parts or by any substitution,
since the logarithm is not a whole subtree to replace. They are the exponential
ansatz one level up the tower: x and ln(x) are algebraically
independent, so with L standing for ln(x) and L' = 1/x, the
derivative of the ansatz divided by e^h is a rational function of x and
L, and asking it to equal the integrand's is one polynomial identity in the
two -- linear in the coefficients of P, one equation per monomial
x^i L^j. The identity is exact, so a solution is an answer and its absence a
decline; the derivative of what comes out is checked against the integrand at
sampled points all the same.
The denominator is tried from the integrand's: its x-part as the exponential
ansatz tries its denominators, times L^k for every k from zero up to
the written power, smallest first so that the answer carries no common factor. Degrees are bounded like the other
ansatz's, and the columns are built one monomial at a time by differentiating the
candidate and clearing one common denominator, so nothing here is expanded with
unknowns in it.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

SolveByLogarithmTowerAnsatz(AngouriMath.Entity,AngouriMath.Entity.Variable)