AngouriMath
SolveABinomialToAWholePowerBelowMinusOneByTheRecurrence(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A power of x beside a whole power below -1 of a binomial with a symbol in it,
x^m (a + b x^n)^p for whole m, n and p <= -2 , by Rubi's reduction that
raises p by one, down toint x^m/(a + b x^n) , which is asked on its own:
raises p by one, down to
J(p) = -x^(m + 1) (a + b x^n)^(p + 1)/(a n (p + 1)) + (m + 1 + n (p + 1))/(a n (p + 1)) J(p + 1)Remarks
The derivative of x^(m + 1) B^(p + 1) is x^m B^p ((m + 1 + n (p + 1)) B - a n (p + 1)) ,
which is the step, exact for whole powers. Rubi's1/(x^2 (a + b x^3)^5) , which the
root of its perfect square leaves from1/(x^2 (a^2 + 2 a b x^3 + b^2 x^6)^(5/2)) , was
declined: the partial fractions over the cube roots of a symbolic binomial gave out past the
first power, which is answered. Late, so that every rule answering the power as written goes
first; the answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233
which is the step, exact for whole powers. Rubi's
root of its perfect square leaves from
declined: the partial fractions over the cube roots of a symbolic binomial gave out past the
first power, which is answered. Late, so that every rule answering the power as written goes
first; the answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233
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