Mixed trigonometric arguments unified, a half-power of 1 + sin closed, a power of an inverse tangent by parts, and a symbolic repeated linear factor

[Rafael-SOWNet]

Four families the Rubi sample was declining, each answered by one rule or one bound,
with the two runaway declines each opened measured and closed before it shipped.

Different multiples of one argument

sin(x)/cos(2x), (cos(x) + sin(x))/sin(2x), cos(x)/(sin(x) tan(x/2)) had no
antiderivative, while each with the double or half angle written out by hand came
out. Every trigonometric rule reads one argument. The new rule rewrites the
integrand to one: down to the greatest common divisor of the slopes through the
Chebyshev recurrences cos(nt) = T_n(cos t), sin(nt) = sin(t) U_(n-1)(cos t); or --
when only an argument and its double are present and the integrand is even in the
smaller, every monomial sin^p cos^q having p + q even -- up to the double, at
half the degree. Moses's sec(2t)/(1 + sec(t)^2 + 3 tan(t)) is answered in a third
of a second the second way and declined after five the first.

What it hands on is one quotient with the common factors cancelled, because
tan(x)/tan(2x) becomes (s/c) / (2sc/(2c^2 - 1)) and on that shape, or on the
single quotient with the s still in it, the search ran twenty-five seconds to
decline 1 - 1/(2c^2). Simplify is not the tool: it folds 2sc back into
sin(2x) and hands the rule its own input.

Two bounds, both from the corpus rather than from caution. Rational in the
trigonometric functions only -- a radical or fractional power of one is a different
integrand, and sqrt(cos(x) sin(x)^3) beside sin(2x) and (2 - 3 sin^2)^(3/5) sin(4x)
were each a five-second timeout for a decline that took a tenth of that before. And
a cap on the degree the rewriting produces: 1/(cos(x) + cos(3x))^5 is degree
fifteen in the sine and cosine, and went the same way.

(a + b sin(cx + d))^n for a^2 = b^2

sqrt(1 + sin(x)) had no antiderivative. It is -2 cos(x)/sqrt(1 + sin(x)), and the
cancellation cos^2 = (1 - sin)(1 + sin) that checks it is what a^2 = b^2 buys.
Above the base case, one step of parts lowers n by one -- Rubi's rule for the same
case -- so (1 - sin(2x/3))^(5/2) is two steps and the base. The cosine is the same
rule with sin replaced by -cos in the boundary term. a^2 = b^2 is decided, not
assumed; sqrt(1 + 2 sin(x)) is elliptic and is declined.

A power of an inverse tangent, by two rounds of parts

x arctan(x)^2 leaves x^2 arctan(x)/(1 + x^2) after one step, which has more
nodes and needs one more step to finish; on nodes alone the second step was refused.
The power fell from two to one, and that is the descent the step made. The measure
is now the pair, power first: a step that lowers the power may grow the expression,
and one that keeps it must shrink it, so the pair is well-founded.

Two things were measured on the way. A remainder whose power has fallen to zero is
whatever the derivative and the integral made together -- for asec(x)^2 times a
radical, a hundred-node radical expression -- and admitting it on the power opened
a by-parts search at every level: thirty seconds to decline, one without. So the
power admits a remainder only while a factor is left for the next round. And the
remainder is re-spelled as the product its next step reads, since Simplify writes
it as a quotient and parts runs on a product -- but only where what stands beside
the factor is rational in the variable, for the same measured reason.

A repeated linear factor with a symbol in it

1/(a + b e^(px))^2 is 1/(p u (a + b u)^2) under u = e^(px). The symbolic
partial-fraction split (#1284) read (a + b u)^2 as a quadratic; it is a repeated
linear factor, and is now one block P/(a + b u)^2 with P of degree one -- which
is then a shape nothing read: (c + d x)/(a + b x)^2 came out with numbers and not
with symbols, the rational-root split being the only thing that ever answered it.
Under t = a + b x it is a sum of powers of t, and that rule is here too. A
repeated symbolic quadratic has nothing to land on and stays declined.

Measured

Every answer differentiated back, with parameters pinned.

Rubi corpus, 463-problem sample, against #1284's branch it was cut from, both timed
on the same afternoon:

#1284       290/463, 0 wrong, 0 timeouts, 95 s
with this   300/463, 0 wrong, 0 timeouts, 95 s

Ten more and nothing lost: sin(x)/cos(2x), (cos(x) + sin(x))/sin(2x),
cos(x)/(sin(x) tan(x/2)), sec(2t)/(1 + sec(t)^2 + 3 tan(t)), sqrt(1 + sin(x)),
(1 - sin(2x/3))^(5/2), x arctan(x)^2, arctan(x)^2/x^3, 1/(a + b e^(px))^2, and
ln(sin(x))/(1 + sin(x)), which the by-parts measure reached on its own. The sum of
per-problem time over the sample is 94 s on both sides.

ExpandedPowerIntegralTest.APowerInsideAQuotientIsNotReached pinned
(a - b x^2)^3/x^7 as declined and said it would move rather than be deleted when
something answered it; x^7 is a power of a linear, and it moved.

The five test suites are green.

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura