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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). And so a power
of a product of powers of x and a sum, (x^m (a + b x^n))^p, a whole power where
an exponent of x is not a rational number, 1/(a x + b x^n), and beside either a
power of a monomial, (c x)^m, which is x^m times a constant.

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), and over it for a
factor below the bar. Only the fractional part of p is in it, since a whole
power of x^j T distributes. 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. A symbol in an exponent does not confine the integrand to a
positive x: x^(3 (n - 1)/2)/(a x^(n - 1) + b x^n + c x^(n + 1))^(3/2) is real for
a negative x and any n, the phases of the two principal powers cancelling to a
sign, and K is that sign; so it is written there too, and is 1 for a positive x.
(c x^e)^m for a whole e is x^(e m) times c^q (c x^e)^r/x^(e r),
for the whole part q and the rest r of m, constant on either side
of 0 by the same argument; for a positive x it is c^m, whatever c is. It
is taken apart beside the sum, and alone where e is past 1, (d x^3)^n:
(c x)^m alone is a power of a linear, a substitution's, and for an e that
is not whole it is SolveByDistributingAPowerOfAMonomial(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)'s.
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; and so were 1/(a x + b x^n), sqrt(c x)/(a x + b x^n)^(3/2) and 1/sqrt(x^(2 - n) (a + b x^n)). 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: below
it, integration by parts integrates the answer again, and x^2 sqrt(a x + b x^4),
whose integral is not elementary, was half a minute of that before it was declined.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

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