AngouriMath
SolveAnExponentialBelowTheBarBesideAPowerOfALinear(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
Exponentials of one linear below the bar, beside polynomials of x with one below
the bar: inw , the exponential every other is a whole power of, the exponentials
are a rational function ofw , and where that is a sum of whole powers of w -- its denominator in lowest terms a power of w alone -- each term is an
exponential of a linear over the polynomials, which
SolveAnExponentialOfALinearOverAPowerOfALinear(AngouriMath.Entity,AngouriMath.Entity.Variable) answers onto the exponential
integral.1 + tanh(z) is 2 e^(2z)/(e^(2z) + 1) , so
1/((c + d x)(a + a tanh(e + f x))) is (1 + e^(-2(e + f x)))/(2 a (c + d x)) .
Rubi's 6.3.1 and 6.4.1,(c + d x)^m (a + a tanh(e + f x))^n and the same with
coth , for a negative m .
https://github.com/asc-community/AngouriMath/issues/718
the bar: in
are a rational function of
exponential of a linear over the polynomials, which
SolveAnExponentialOfALinearOverAPowerOfALinear(AngouriMath.Entity,AngouriMath.Entity.Variable) answers onto the exponential
integral.
Rubi's 6.3.1 and 6.4.1,
https://github.com/asc-community/AngouriMath/issues/718
Remarks
The rules for an exponential over a linear read it above the bar: e^(-2x)/x was
answered and1/(x e^(2x)) declined. Where the denominator keeps a factor other
thanw after cancelling, 1/(x (1 + e^x)) , the integral is neither elementary
nor an exponential integral, and this declines.
answered and
than
nor an exponential integral, and this declines.
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