An exponential times a rational function by Liouville's ansatz, the Hermite reduction first for a repeated factor, and a cache that had been remembering a scoped decline

[Rafael-SOWNet]

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.

The ansatz

Liouville's theorem says the elementary antiderivative of e^h R, with h and R
rational, is e^h times a rational function where it exists at all. So the rule
writes F = e^h N/D, with D read off the denominator of R -- every written power
lowered by one, then the denominator itself, and every level of a single written
power, since with an exponential in front the antiderivative's denominator need not
be one below the integrand's (e^(1/x)(1 + x)/x^4 has x^2 under it) -- and N a
polynomial of unknown coefficients. F' = e^h R is then a polynomial identity linear
in those coefficients, solved by the elimination the symbolic partial-fraction split
uses and checked at sampled points with every symbol pinned before anything is
returned. An ansatz that finds no N has, by Liouville, shown there is no
elementary antiderivative of that shape: e^x/x, e^(x^2) and e^x/(1 + x) are
declined in a millisecond each, and a test pins that they stay declined.

A sum whose terms share one exponential is read whole: e^(x^2) + 2x^2 e^(x^2) is
(x e^(x^2))', and neither term is elementary on its own, so splitting it first --
which linearity does -- loses it.

The identity is built one column at a time as a numeric polynomial rather than
once with the unknowns in it. With fourteen unknowns over a degree-eleven denominator
the single symbolic expansion did not return; the fourteen numeric ones take a moment.

The Hermite reduction, and first

With no exponential the same ansatz is the rational part of the Hermite reduction,
and the full form is taken: R = (N/D)' + M/D_1, with D_1 the product of the
distinct factors and M a second unknown polynomial, so that a logarithmic part
beside the rational one does not defeat it. (1 + x^2 + x^4)/((1 + x^2)(4 + x^2)^2)
has both. What is left, M/D_1, is a proper fraction over a squarefree denominator,
and it is handed to the splits and not to the whole integrator: handing it to the
whole integrator sent one such remainder through every substitution and by-parts
attempt there is, thirty seconds to decline what the splits decline in milliseconds.

A denominator with a written repeated factor takes this first, ahead of the
coprime split and the root peeling. (1 + x^2)/(x (1 + x^3)^2) reached the same
answer through those after seventeen seconds of peeling and re-factoring; this way
it is under a tenth of one. x^7, (x^3 + 2)^2 and (1 + x^4)^3 are the same story.

Integration by parts with neither factor worth differentiating

The classic "neither is a polynomial, try both orderings" case of by parts ran on any
product. Once the Hermite reduction answered ∫u for a small rational piece of a
radical trigonometric integrand, by parts carried on with v' ∫u, three hundred
nodes of mixed arguments against forty, and every substitution below was tried on it:
thirty seconds to decline where it had taken a tenth of one. The case now runs only
with a factor worth differentiating -- a logarithm or an inverse function, whose
derivative is algebraic, or an exponential, whose integral is itself and which the
cyclic cases need. Measured on the Rubi sample with the restriction and without: the
same answers, eight seconds less.

The cache

That restriction surfaced a defect that by parts had been hiding. Five rules answer
only the question asked (#1265) and decline the same integrand one level down, and
the integrator's cache held such a null without the scope it was made in.
cos(x)^(-3), tried inside the search for 1/cos(x)^3 and declined there by the
scoped secant reduction, was then declined from the cache in two milliseconds when
asked for directly -- unseen until now because by parts used to answer it at depth
two regardless. The key carries the scope; a lookup takes a decline only from its own
scope and an answer from either, since an antiderivative that was found is right
wherever it is asked for. A test asks the two integrals in that order.

Measured

Every answer differentiated back, with parameters pinned.

Rubi corpus, 463-problem sample, against the master it was cut from (#1286), both
timed on the same evening:

master      300/463, 0 wrong, 0 timeouts, 102 s
with this   312/463, 0 wrong, 0 timeouts,  76 s

Twelve more and nothing lost: e^x x/(1 + x)^2, e^(x^2)(1 + 2x^2) twice and once as
a sum (Moses), e^(1/x)(1 + x)/x^4 (Hearn), Hebisch's e^(1/(x^2 - 1)) times a
rational function, (4x^5 - 1)/(1 + x + x^5)^2 (Apostol), Stewart's
(1 + x^2 + x^4)/((1 + x^2)(4 + x^2)^2), Timofeev's x^5/(1 + x^4)^3 and
(1 + x^2)/(x (1 + x^3)^2), and two the reduction reached on its own:
ln(x)/(1 + ln(x))^2 and (3x - 1)^(4/3)/x^2. The sum of per-problem time over the
sample is 100 s to 75 s.

Two pinned declines moved as their tests said they would:
BinomialDenominatorIntegralTest had 1/(x^3 + 2)^2 as outside the binomial rule's
shape, and it is the reduction's first and then the rule's;
ExpandedPowerIntegralTest had (a - b x^2)^3/x^7 as unreached, and x^7 is a power
of a linear.

The five test suites are green.

Part of #718.

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