AngouriMath
SolveAFractionalPowerOfXBesideABinomialToAWholePowerBelowMinusOne(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A power of x that is not whole beside a polynomial in x^n over a whole power
below -1 of a binomial inx^n with a symbol in it, x^q P(x^n) (a + b x^n)^p ,
each term of P by Rubi's reduction that raises p by one, down to
int x^(q + k n)/(a + b x^n) , which is asked on its own:
Rubi's1/(x^(3/2) (a + c x^4)^2) and (A + B x^3)/(x^(3/2) (a + b x^3)^3) .
below -1 of a binomial in
each term of P by Rubi's reduction that raises p by one, down to
int x^m B^p = -x^(m + 1) B^(p + 1)/(a n (p + 1)) + (m + 1 + n (p + 1))/(a n (p + 1)) int x^m B^(p + 1)Rubi's
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)) for any m, which is the step. These were declined: the substitution under the root makes
2/(u^2 (a + c u^8)^2) of the first, and the partial fractions over the eighth roots of
a symbolic binomial gave out past the first power, which is answered, as is
x^(-3/2)/(a + c x^4) itself in a tenth of a second. 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
a symbolic binomial gave out past the first power, which is answered, as is
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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