An exponential of a multiple of a logarithm is integrated as the power it is

[Rafael-SOWNet]

e^(k ln(q)) is q^k — the definition of the principal power, for every complex q other than zero — and that is the spelling the parser gives every inverse hyperbolic function: acoth(a x) is 1/2 ln((a x + 1)/(a x - 1)). So Rubi's 7.4.2 rows, exponentials of the inverse hyperbolic cotangent, arrived at the integrator as exponentials of a logarithm, which no exponential rule reads — e^acoth(a x) x^3 was declined in 86 ms while x^3 sqrt((a x + 1)/(a x - 1)), the same integrand, is answered by the radical substitution. The simplifier folds the shape since #1430, but the integrand is not simplified before the rules see it (by design), so the fold is a route of the integrator: SolveByFoldingAnExponentialOfALogarithm, right after the flattening of a power of an exponential, asked as a question of its own so the closed rules meet the folded integrand at the top. The multiplier is gathered from the whole product in the exponent, since e^(3 acoth(a x)) arrives as e^(3 (1/2 ln q)).

Of fifteen 7.4.2 rows the probe declined, seven are answered and verified (e^(2 acoth(a x)) (c - a² c x²)² → a polynomial, (c - c/(a² x²))³/e^(2 acoth(a x)), e^acoth(a x)/(c - c/(a x)), e^acoth(a x) (c - c/(a x))^(7/2), e^(1/3 acoth(x)) x², …). Family 7 on the corpus: 159/270, 0 wrong (153 before; 49 timeouts, 46 before — three rows that were declined at once are searched to the budget now). 1774 suite unchanged: 1706, 0 wrong.

ExponentialOfALogarithmIntegralTest: four 7.4.2 shapes with numeric parameters and three plain e^(k ln q) shapes, each differentiated back at five points beyond a x = 1. Full suite 12117 passed, 0 failed. BREAKING-CHANGES.md rows measured on a v2.5.0 build.

Part of #718.

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