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SolveByTakingAPowerOfXOutOfAFractionalPower​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

A fractional power of a sum whose every term has x to a power in it,
(a x^j + b x^n)^p with 0 < j < n, as K x^(j p) (a + b x^(n - j))^p:
the integral is K times the integral with x^(j p) (a + b x^(n - j))^p in its
place, where K = (a x^j + b x^n)^p / (x^(j p) (a + b x^(n - j))^p).

Remarks

K is constant on every interval where x and the sum are not zero: with
S = x^j T, its logarithmic derivative is p (S'/S - j/x - T'/T), and
S'/S = j/x + T'/T. So it goes in front of the integral, as the factor of a square
does in SolveByWritingAPowerOfASquareAsAPowerOfItsRoot(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean). For an even whole
j and a whole j p it is sgn(x)^(j p): x^j is not negative for a
real x, so (x^j T)^p is |x|^(j p) T^p.
 Rubi's 1.1.4.2 and 1.1.4.3 write these by the hundred: 1/sqrt(a x^2 + b x^5) and
1/(x sqrt(b x^(2/3) + a x)) were declined, the first since the factorization
over the rationals does not read a symbol, and the second since nothing took the power
of x out of the root. With it out, the binomial is what the rules for
x^m (a + b x^n)^p read. At the top only, where the answer is the caller's.
https://github.com/asc-community/AngouriMath/issues/718

























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