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

 Method (no overloads)

Summary

A rational function of e^(k x), turned into a rational function of one variable
by u = e^(k x).

Remarks

With u = e^(k x) we have dx = du/(k u), and every exponential in the
integrand becomes a whole power of u, so a quotient built from them is a
quotient of polynomials — which the rational integrator answers.
What was missing.1/(1 + e^x) had no antiderivative, and neither did
1/(e^x + e^(-x)). e^x/(1 + e^x) did, which is the shape of the gap: the
general substitution answers an exponential integrand only when the numerator happens
to be the derivative of something in it, and declines the rest for want of anything to
substitute for.
It is also how the hyperbolic functions get integrated, since they are not nodes
here — tanh(x) is built as (e^(2x) - 1)/(e^(2x) + 1) and sech(x) as 2/(e^x + e^(-x)), so both are rational functions of an exponential, and
tanh, coth, sech and csch had no antiderivative at all.
 The answers come out in the exponential rather than as ln(cosh(x)) or
2 arctan(e^x), and unfolded — the rational integrator writes a logarithm as
ln((2ax + b - D)/(2ax + b + D)) and leaves the arithmetic in the coefficients
standing. That is its shape on plain rational integrands too, not something this
rewrite introduces; 1/(x(x+1)) comes out as
ln((2x + 1 - 1)/(2x + 1 + 1)) with nothing exponential in sight.
One k for the whole integrand. The exponents are read as linear in the
variable and their slopes taken together by greatest common divisor, so e^x beside e^(2x) gives u and u^2 under one substitution. A slope that
is not a whole number, or an exponent that is not linear — e^(x^2), whose
integral is not elementary at all — is declined.
No condition is owed.e^(k x) is positive for every real x and
never zero, so the substitution is invertible on the whole line and introduces no
interval of its own; u is a genuine change of variable rather than a branch.
 https://github.com/asc-community/AngouriMath/issues/718

Summary

A quotient by an exponential, handed on as the product with its reciprocal:
N/b^(f(x)) as N * b^(-f(x)).

Remarks

sin(x) * e^(-x) had an antiderivative and sin(x)/e^x did not. They are
the same integrand, and the difference is only which node is on top: integration by
parts matches a Mulf and nothing else, so a quotient never reached it. The
simplifier does not turn one into the other — it keeps sin(x)/e^x as written —
so nothing upstream closed the gap either.
 The family is wider than the trigonometric case that exposed it: x/e^x,
ln(x)/e^x and sin(x)/2^x were all declined for the same reason, and each
is answered once written as a product.
Only an exponential denominator — a base free of the variable with the variable
in the exponent. x^n in a denominator is a rational function and belongs to
partial fractions, which reads it as written; turning that into a negative power would
take it away from the rules that answer it. A numerator free of the variable is left
alone too, since SolveAsPolynomialTerm(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) already takes that one.
 https://github.com/asc-community/AngouriMath/issues/718

























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