Integrate a function of the hyperbolic tangent alone under u = tanh

[Rafael-SOWNet]

Part of #233.

The parser writes tanh(L) as (e^(2 L) - 1)/(e^(2 L) + 1) and coth(L) as its reciprocal, so an integrand built from those alone is R(u) with u = tanh(L), and dx = du/(b (1 - u^2)):

• tanh(x)/sqrt(a + b tanh(x)^2 + c tanh(x)^4) is u/((1 - u^2) sqrt(a + b u^2 + c u^4)).
• sin(tanh(x)) is sin(u)/(1 - u^2), onto the sine and cosine integrals.

The trigonometric counterparts are answered under u = tan(x). The hyperbolic functions arrive as exponentials, and nothing read them as a tangent. The new rule, SolveAFunctionOfTheHyperbolicTangent, applies only where R is not rational in u. Rational functions of tanh keep the answers the exponential substitution gives them; tanh(x)^2 and tanh(x)/(1 + tanh(x)) are pinned as unchanged.

Measured against master 8a1ef454:

	master	this branch
Rubi 6.3 and 6.4, all 728 rows	689	722
moves, each re-run alone	3 of 41	36 of 41, 0 wrong
families 0 (whole) and 1 (40 a file), samples of 2–8		identical
performance gate		passed
suite		15,603 passed, 0 failed

The gains include tanh/coth over the root of a quartic in them, tanh/(a + b tanh^4)^k, sin(tanh(a + b x))^k with its cos and coth twins, and (b coth(c + d x)^4)^(k/3).

The rebase onto #1950–#1955 conflicted at the shared anchors. I resolved it by keeping every rule. The combined tests of #1952, #1953, #1955 and this branch pass, 21/21. All gained rows were declined (35) or timed out (1) on 2.5.0, and BREAKING-CHANGES.md has the entry.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura