AngouriMath

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AngouriMath

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

 Method (no overloads)

Summary

A power of a square written out, (A + B w + C w^2)^p with B^2 = 4 A C and
w = x^k, as the power of its root, L = w + B/(2 C): the integral is
F times the integral with L^(2p) in its place, where
F = (A + B w + C w^2)^p / L^(2p).

Remarks

The square is C L^2, and F is constant on every interval where L is
not zero: its derivative is 2 p w' L^(2p-1) (C L^2 Q^(p-1) - Q^p) / L^(4p), and
Q^(p-1) Q = Q^p wherever Q is not zero. So F goes in front of the
integral, whatever p is: a symbol, 3/4, or half an odd number with a
leading coefficient of unknown sign. For a whole p it is C^p. The square
is a factor of the integrand or nothing is read: below the bar 1/F comes out,
and out of a sum nothing does.
 After SolveByTakingARootOfAPerfectSquare(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean), which answers half an odd power
of a square in a whole power of x with the sign written out, and before which nothing
read k as anything but a whole number. This reads k as any exponent, a
fraction or a symbol: Rubi's 1.2.3.2 x^2 (a^2 + 2 a b x^3 + b^2 x^6)^p,
x sqrt(a^2 + 2 a b x^n + b^2 x^(2n)) and (a^2 + 2 a b x^2 + b^2 x^4)^(3/4) were declined. At the top only, where the answer is the caller's, as for the sign the
rule above writes.
https://github.com/asc-community/AngouriMath/issues/718

























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