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A polynomial times a fractional power of a base holding a function of x is answered as a polynomial times the next power of the base, by a linear system with the functions of x as indeterminates - #1321

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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). Nothing read it: the substitution u = e^x + x wants
1 + e^x beside the root and finds 5x + e^x (2x + 3), and no other rule
reads a root of an exponential beside a polynomial at all.

For N B^r with a fractional r and N and B polynomials in x and
the functions of x in them, the ansatz is P(x) B^(r + 1): its derivative
is B^r (P' B + (r + 1) P B'), so 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 -- 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. The identity is written as one polynomial in x,
the functions and the unknowns, a derivative of a logarithm bringing a
1/x that is cleared; each monomial in x and the functions is one
equation, and the system is solved exactly over the rationals. The degree
of P is that of N in x, the functions held constant, and one more.
The solution is checked by differentiating back at sampled points, since
two functions taken for independent indeterminates -- e^x and e^(2x)
-- can make an identity fail to be found, never found wrongly, but a
derivative the reader did not expect could. Closed, and volunteered at
any depth.

The split takes Bronstein's apart first, 3/x and the rest, and gathers
the rest's three terms over their shared root, which is where this is
asked. The multivariate polynomial exposes its terms for the rows, which
it did not before.

Rubi sample (463 problems, 5 s budget), against the #1319 master this branch is cut from:

master at #1319     410/463, 0 wrong, 0 timeouts, ~97 s
with this           411/463, 0 wrong, 0 timeouts, 97 s

One more, nothing lost: Bronstein 19, in 0.8 s. Moses 136 goes from 0.9 s to 0.2 s; nothing else moved by more than noise. Independent of #1320, merged meanwhile.

The five test suites are green.

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

… x is answered as a polynomial times the next power of the base, by a linear system with the functions of x as indeterminates

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)`. Nothing read it: the substitution `u = e^x + x` wants
`1 + e^x` beside the root and finds `5x + e^x (2x + 3)`, and no other rule
reads a root of an exponential beside a polynomial at all.

For `N B^r` with a fractional `r` and `N` and `B` polynomials in `x` and
the functions of `x` in them, the ansatz is `P(x) B^(r + 1)`: its derivative
is `B^r (P' B + (r + 1) P B')`, so `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 -- 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`. The identity is written as one polynomial in `x`,
the functions and the unknowns, a derivative of a logarithm bringing a
`1/x` that is cleared; each monomial in `x` and the functions is one
equation, and the system is solved exactly over the rationals. The degree
of `P` is that of `N` in `x`, the functions held constant, and one more.
The solution is checked by differentiating back at sampled points, since
two functions taken for independent indeterminates -- `e^x` and `e^(2x)`
-- can make an identity fail to be found, never found wrongly, but a
derivative the reader did not expect could. Closed, and volunteered at
any depth.

The split takes Bronstein's apart first, `3/x` and the rest, and gathers
the rest's three terms over their shared root, which is where this is
asked. The multivariate polynomial exposes its terms for the rows, which
it did not before.

Rubi sample: +1 (Bronstein 19), 0 wrong, 0 timeouts.

Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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