AngouriMath
SolveAFractionalPowerOfXBesideATrinomialToAWholePowerBelowMinusOne(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A power of x that is not whole beside a whole power below -1 of a trinomial
T = a + b x^n + c x^(2 n) with a symbol in it, x^q T^p , by Rubi's reduction
that raises p by one, down toint x^(q + k n)/T , each asked on its own:
Rubi's1/(x^(3/2) (a + b x^2 + c x^4)^2) .
that raises p by one, down to
int x^m T^p = -x^(m + 1) (b^2 - 2 a c + b c x^n) T^(p + 1)/(a n (p + 1) (b^2 - 4 a c))
+ int x^m T^(p + 1) (b^2 (m + n (p + 1) + 1) - 2 a c (m + 2 n (p + 1) + 1) + b c (m + n (2 p + 3) + 1) x^n)/(a n (p + 1) (b^2 - 4 a c))Rubi's
Remarks
The step is the derivative of its first term, for any m, as for the binomial in
SolveAFractionalPowerOfXBesideABinomialToAWholePowerBelowMinusOne(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean). It was
declined, wherex^(-3/2)/(a + b x^2 + c x^4) and x^(1/2)/(a + b x^2 + c x^4) are
answered. Late, beside the binomial's; the answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233
SolveAFractionalPowerOfXBesideABinomialToAWholePowerBelowMinusOne(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean). It was
declined, where
answered. Late, beside the binomial's; the answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233
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