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SolveByWritingAPowerOfAnExponentialAsAMultipleOfOne​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

A power of an exponential the flattening above leaves -- a symbol for the base or the
power, (F^(g (e + f x)))^n -- beside a polynomial below the bar. Its logarithmic
derivative is the constant n g f ln F, so it is K e^(n g f ln(F) x) with
K constant wherever it is differentiable; with a symbol for K the integrand
is one the exponential integral's rules read, and the answer is written back with
K = (F^(g (e + f x)))^n e^(-n g f ln(F) x). A whole power is the exponential of
the product exactly, whatever the base, and is written so. Rubi's 2.2,
(a + b (F^(g (e + f x)))^n)^p/(c + d x)^m.
https://github.com/asc-community/AngouriMath/issues/718

Remarks

(F^u)^n is F^(n u) only where F^u is on the principal branch, which
it is for a positive F and need not be for a symbol; K carries the
difference, and the answer holds for every F. Only beside a polynomial below the
bar, where nothing else reads the power: above it, the rules for a polynomial times an
exponential answer it as written.
 And beside another power of an exponential with the same rate, decided by value, which
nothing reads from the two as written: Rubi's 2.3
(a + b (F^(e (c + d x)))^n)^p (G^(h (f + g x)))^(d e n ln(F)/(g h ln(G))) is
K_2 e^(r x) (a + b K_1 e^(r x))^p, which the substitution u = e^(r x) answers, and was a search past its budget. The constants come back beside the
exponential they multiplied and are cancelled against it, and the answer is Rubi's.
https://github.com/asc-community/AngouriMath/issues/233

























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