An odd power of the complement left by the substitution of a sine or a cosine is its sign times a root, and the sign is a constant between the complement's zeros

[Rafael-SOWNet]

Charlwood's ln(sin(x)) sqrt(1 + sin(x)) by parts leaves
-2 cos(x)^2/(sin(x) sqrt(1 + sin(x))), and under u = sin(x) that is
-2 cos(x)/(u sqrt(1 + u)) du, a cosine over: the substitution reads even
powers of the complement as powers of 1 - u^2 and refuses an odd one,
since cos(x) is not a function of sin(x). It is one on every interval
between the zeros of the cosine, sgn(cos(x)) sqrt(1 - u^2), and the sign
is a constant there. So the integrand is that constant times an algebraic
function of u -- -2 sgn(cos(x)) sqrt(1 - u)/u -- which is integrated,
-4 sqrt(1 - u) + 4 atanh(sqrt(1 - u)), and the sign goes back in as
sgn(cos(x)). Exact wherever the cosine is not zero, the generic case,
and checked on both signs of it. Charlwood's
cos(x)^2/sqrt(1 + cos(x)^2 + cos(x)^4) is the same under the cosine:
-u^2/(sin(x) sqrt(1 + u^2 + u^4)), which is -sgn(sin(x)) u^2/sqrt(1 - u^6)
once the roots combine, and -sgn(sin(x)) arcsin(u^3)/3 -- Rubi's
x/3 + arctan(...)/3 written the other way.

A rule of its own, and late, after the half-angle substitution: inside
the general substitution it answered 1/(1 + sin(x)) as a sign times a
root where the half-angle substitution gives the tangent of the half
angle, and the remainder by parts left beside that root was nine seconds
of search for ln(sin(x))/(1 + sin(x)), which the tangent answers in a
moment. Only for an integrand with a root in it, since a rational
function of the sine and cosine is the half-angle substitution's and
Jeffrey's paid half a second here on the way to it; only where the
complement is a factor of the product, since inside a sum,
1/(cos(x) + sin(x)), the sign beside a root of 1 - u^2 is a search that
timed out; and only where what is left is algebraic in u, since beside a
logarithm of u it ended nowhere.

Rubi sample (463 problems, 5 s budget), against the #1320 master this branch is cut from:

master at #1320     411/463, 0 wrong, 0 timeouts, ~97 s
with this           413/463, 0 wrong, 0 timeouts, 97 s

Two more, nothing lost: Charlwood 128 in 1.4 s and Charlwood 20 in 0.3 s. Charlwood 131 (sec(x)/sqrt(sec(x)^4 - 1), answered) takes 0.7 s where it took 0.4; nothing else moved by more than noise. Three things were measured on the way and are not in this branch: the sign inside the general substitution (Apostol 192 to a timeout, Charlwood 125 and 47 lost), the rule without the algebraic condition, and the rule without the root condition (half a second on each of Jeffrey 20 and 23).

Independent of #1321 (merged meanwhile) and #1322 (open), which touch other rules.

The five test suites are green.

Part of #718.

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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura