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

 Method (no overloads)

Summary

N B^r, for a fractional r and an N and a B that are
polynomials in x and the functions of it in them, answered as
P(x) B^(r + 1) for a polynomial P where there is one: the derivative of
that is B^r (P' B + (r + 1) P B'), and P' B + (r + 1) P B' = N is a linear
system in the coefficients of P once the functions of x are taken for
indeterminates.

Remarks

Bronstein's (5x^2 + 3 (e^x + x)^(1/3) + e^x (3x + 2x^2))/(x (e^x + x)^(1/3)) is
3/x + (5x + e^x (2x + 3))/(e^x + x)^(1/3), and the second is
3x (e^x + x)^(2/3): with P = c_0 + c_1 x and r = -1/3,
c_1 (e^x + x) + (2/3)(c_0 + c_1 x)(e^x + 1) = 5x + e^x (2x + 3) is
c_1 = 3, c_0 = 0. Nothing else read it: the substitution
u = e^x + x wants 1 + e^x beside the root and finds 5x + e^x (2x + 3).
 The degree of P is the degree of N in x, the functions of x held constant, and one more; the system is exact, over the rationals, and its
solution is checked by differentiating back at sampled points, since an identity
between two functions taken for independent indeterminates -- e^x and
e^(2x), say -- can fail to be found and cannot be found wrongly, but a
derivative the reader did not expect can. Closed, and volunteered at any depth.
 https://github.com/asc-community/AngouriMath/issues/718

























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