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

 Method (no overloads)

Summary

An integrand e^h R, with R rational in x and h rational
in x or absent, integrated by the ansatz F = e^h N/D: D read
off the denominator of R, N a polynomial of unknown coefficients, and
F' = e^h R a linear system in them.

Remarks

e^x x/(1 + x)^2 is (e^x/(1 + x))'; e^(x^2)(1 + 2x^2) is
(x e^(x^2))'; (4x^5 - 1)/(1 + x + x^5)^2 is (-x/(1 + x + x^5))'.
None had an antiderivative: the first two are not a polynomial times an exponential,
which is the shape by parts reads, and the third has a denominator nothing factors.
Each is the derivative of something of the same shape, and that is what is looked
for -- Liouville's theorem says the elementary antiderivative of e^h R, where
there is one, is e^h times a rational function, so an ansatz that fails here
fails because there is no such antiderivative and not because the shape was wrong.
For h = 0 this is the rational part of the Hermite reduction, with the
logarithmic part required to vanish; a denominator with a repeated factor and a
logarithmic part beside it is left to the splits, as before.
Which D. Differentiating raises the multiplicity of every factor of the
denominator by one, so D is the denominator of R with each factor's
written power lowered by one, and then the denominator itself, and 1 where the
denominator is 1. With h = p/q the identity is
((N' D - N D') q^2 + (p' q - p q') N D) D_R  =  N_R D^2 q^2
            
a polynomial identity in x, linear in the coefficients of N, solved by
the same elimination the symbolic partial-fraction split uses and checked the same way
before anything is returned: at sampled points with every symbol pinned. The degree
of N is bounded by the degrees in that identity, with a little to spare;
an unknown the system does not need is zero.
https://github.com/asc-community/AngouriMath/issues/718

























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