Write each linear once under a symbolic power (#1974)

A product of powers of quotients of linears whose exponents carry a symbol,
e^atanh(a x) x^m sqrt(c - a^2 c x^2) or e^(2 p atanh(a x)) (c - a^2 c x^2)^p,
was declined or ran out of time: the rule that writes each linear once reads
rational exponents only. Each linear's exponent is now the symbolic sum of
those it has in the bases, and K, the integrand over that product, has a
logarithmic derivative of zero on every branch, so it stands in front. Only
where the product is one power that is not whole beside whole ones; the
answer is checked at sampled points.

Rubi pocket (1,818 rows of 7.3.x and 7.4.x): 1726 -> 1748 solved, none lost;
the moves re-run alone gain 22 rows of 7.3.6 and 7.4.2. Samples of every
family unchanged.

Part of #233.


Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>

Author: Rafael-SOWNet

BREAKING-CHANGES.md

@@ -4763,6 +4763,26 @@ and this is the same answer reached through the even root
 |---|---|---|
 | `"sqrt(x)/(sqrt(1/x + 1)*sqrt(1 - x^2))".Integrate("x")` | `integral(...)` | an antiderivative in `sqrt(x)`, `provided x >= 0` |
 
+### A product of powers of quotients of linears with a symbolic exponent is written over each linear once
+
+**Answers where there were none.** `e^atanh(a x) x^m sqrt(c - a^2 c x^2)`, `e^(2 p atanh(a x)) (c - a^2 c x^2)^p` and
+seventeen more of Rubi's 7.3.6 and 7.4.2 were declined. The first is `sqrt((1 + a x)/(1 - a x)) x^m sqrt(c - a^2 c x^2)`,
+in which the root `1/a` stands in both bases with exponents `-1/2` and `1/2`, and the second is
+`K (a x + 1)^(2 p)`; the rule below reads rational exponents only. Each exponent of a linear is now the sum, symbols
+and all, of those it has in the bases, and `K`, the integrand over that product, has a logarithmic derivative of zero
+on every branch, so it stands in front of the answer. Only where the product comes to one power that is not whole, of
+x or of a linear, beside whole powers of the other linears above the bar: the antiderivative is then a sum of powers
+of it on every interval between the roots, and is checked at sampled points. Two powers that are not whole are left
+to the rules that answer them as a power of the quotient
+([#233](https://github.com/asc-community/AngouriMath/issues/233)).
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"e^atanh(a*x)*x^m*sqrt(1-a^2*x^2)".ToEntity().Integrate("x")` | `integral(...)` | `K (x^(m + 1)/(m + 1) + a x^(m + 2)/(m + 2))`, 151 characters |
+| `"e^atanh(a*x)*x^m*(c-a^2*c*x^2)^(5/2)".ToEntity().Integrate("x")` | `integral(...)` | 619 characters |
+| `"e^(2*p*atanh(a*x))*(c-a^2*c*x^2)^p".ToEntity().Integrate("x")` | `integral(...)` | `K (a x + 1)^(2 p + 1)/(a (2 p + 1))`, 153 characters |
+| `"e^(n*acoth(a*x))*(c-a*c*x)^(2+1/2*n)".ToEntity().Integrate("x")` | `integral(...)` | 585 characters |
+
 ### A product of powers of quotients of linears is written over each linear once
 
 **Improvement, not silent.** `e^atanh(a x) sqrt(c - c/(a x))` is