AngouriMath
SolveAHyperbolicOfALinearOverAPowerOfALinear(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A sum of exponentials of the variable, times a polynomial, over a whole power of a
linear:P(x) sum_j c_j e^(r_j x)/(e + f x)^n , which is how a polynomial in
sinh and cosh of a linear arrives, onto the hyperbolic sine and cosine
integrals. Underu = e + f x each term is u^m e^(k u) , the exponential
integral's (APowerTimesAnExponential(System.Int32,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable)), and two of opposite rates pair:
A Ei(k u) + B Ei(-k u) = (A + B) Chi(k u) + (A - B) Shi(k u) , since
Ei(y) + Ei(-y) = 2 Chi(y) and Ei(y) - Ei(-y) = 2 Shi(y) for y > 0 , and
fory < 0 up to a constant. A rate of 0, which an even power leaves, is a power
ofu alone. Rubi's 6.1.1, (c + d x)^m (a + b sinh(e + f x))^n , with
m negative.
https://github.com/asc-community/AngouriMath/issues/1501
linear:
integrals. Under
integral's (APowerTimesAnExponential(System.Int32,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable)), and two of opposite rates pair:
for
of
https://github.com/asc-community/AngouriMath/issues/1501
Remarks
After SolveAnExponentialOfALinearOverAPowerOfALinear(AngouriMath.Entity,AngouriMath.Entity.Variable), which answers one
exponential alone; and only where an integral is left.
exponential alone; and only where an integral is left.
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