Chebyshev's other two cases for a binomial differential, a term of a top-level sum asked at the top, and a bare inverse function by parts

[Rafael-SOWNet]

sqrt(1 + x^3)/x had no antiderivative. x^2 sqrt(1 + x^3) and x^5 sqrt(1 + x^3)
did, by the same substitution, and the rule for it (#1275) stopped where Chebyshev's
first case does.

The second case, with a rational function left over

Under u = (a + b x^n)^(1/q), with s = (m + 1)/n:

int x^m (a + b x^n)^(p/q) dx = (q/(n b^s)) int u^(p+q-1) (u^q - a)^(s-1) du

which is a polynomial in u when s >= 1 -- what the rule expanded -- and a
rational function of u when s <= 0, which it read as "not a polynomial" and
declined. sqrt(1 + x^3)/x is (2/3) int u^2/(u^2 - 1) du; 1/(x sqrt(1 - x^3)) is
(2/3) int 1/(u^2 - 1) du. The rational integrator answers those, and the rule now
hands them to it.

The third case, through the reciprocal

Where s + p/q is whole instead, x = 1/y turns x^m (a + b x^n)^(p/q) dx into
-y^m' (b + a y^n)^(p/q) dy with m' = -m - 2 - n p/q, a whole number, and
(m' + 1)/n = -(s + p/q), whole -- the second case in y, with a and b
exchanged, and y = 1/x put back afterwards. x^6 (3 + 4x^4)^(1/4) is
-9 int u^4/(u^4 - 4)^3 du that way, which is the Hermite reduction's (#1287) and
the binomial rule's. Chebyshev proved there is no fourth case, so a binomial
differential outside these two is declined for a reason.

A term of a top-level sum is asked at the top

(x^3 - 1)/(2 + x^3)^(1/3) is two binomial differentials, and was declined although
each term alone is answered. Linearity splits the integrand before anything else,
and every term it produced sat one level down -- where the five rules scoped to the
question asked (#1265, #1280) decline. A term of a top-level sum is strictly smaller
than the sum and is not a continuation of any rule's search, which is what the
scope was measured to stop; it is asked at the top now. Only from the top: one level
down the terms stay one level down, as before. sec(x)^3 + x is the same story with
the secant reduction.

A bare inverse function, by parts against 1

arctan(x sqrt(1 - x^2)) had no antiderivative. It is one step of parts against 1
-- x f(g) - int x g' f'(g) dx, with an algebraic remainder -- and that step was never
taken: the by-parts rule runs on a product, a single node is not one, and a linear
argument is the table's. The remainder is answered where the radical rules reach it,
which is three of Charlwood's and Bondarenko's ln(1/x^4 + x^4); the rest of that
family wants an Euler substitution the library does not have, and declines in about a
second, which a test pins. Asked, not volunteered, like the regrouping beside it.

Measured

Every answer differentiated back.

Rubi corpus, 463-problem sample, against #1287's branch it was cut from, both timed
on the same evening:

#1287       312/463, 0 wrong, 0 timeouts, 76 s
with this   325/463, 0 wrong, 0 timeouts, 80 s

Thirteen more and nothing lost: arctan(x sqrt(1 - x^2)), arctan(x sqrt(1 + x^2))
(Charlwood) and ln(1/x^4 + x^4) (Bondarenko) by parts against 1; and by the binomial
rule sqrt(1 + x^3)/x (Charlwood), 1/(x sqrt(1 - x^3)) and
sqrt(1 + x^8)/(x (1 + x^8)) (Bronstein), x^6 (3 + 4x^4)^(1/4) and
(x^3 - 1)/(2 + x^3)^(1/3) (Timofeev), and five of Welz's: (1 - x^3)^(1/3)/x,
(1 - x^3)^(2/3)/x, 1/(x (1 - x^2)^(1/3)), 1/(x (1 - x^2)^(2/3)),
1/(x (1 - x^3)^(1/3)). The sum of per-problem time over the sample is 75 s to 79 s,
which is the answers.

BinomialDifferentialTest.OutsideTheFirstCase had 1/(x (1 + x^3)^(1/3)) as a case
that may be declined; it is the second case with s = 0 and moved to the answered
theory, with the three that Chebyshev says have no antiderivative kept where they
were.

The five test suites are green.

Part of #718.

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