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

 Method (no overloads)

Summary

Fractional powers of two different linears in the variable, (a x + b)^(p/q) beside (c x + d)^(r/q), rationalised by t = ((a x + b)/(c x + d))^(1/q) where every term of the integrand carries a whole power of the second linear once
the first is written through t.

Remarks

With t^q = (a x + b)/(c x + d), x = (d t^q - b)/(a - c t^q) and
c x + d = (a d - b c)/(a - c t^q), both rational in t; and
(a x + b)^(p/q) = t^p (c x + d)^(p/q). So each term of the integrand is a
power of t times a power of the second linear, and where that power's
numerator is the same modulo q across the terms above the line and across
those below, the powers left are whole and the integrand is rational in t.
1/((x - 1)^4 (x + 1)^2)^(1/3) is (x - 1)^(-4/3) (x + 1)^(-2/3): with
t^3 = (x - 1)/(x + 1) it is t^(-4) (x + 1)^(-2), and
x (1 + x)^(2/3) sqrt(1 - x) over a sum of two such products is the same with
q = 6, every product there carrying (1 - x)^(7/6).
Where it holds: everywhere.t is written back as the quotient of the
two principal roots, (a x + b)^(1/q) / (c x + d)^(1/q), and not as the root of
the quotient: a whole power of a quotient is the quotient of the powers, so
t^p is (a x + b)^(p/q) / (c x + d)^(p/q) for every complex x,
where the root of the quotient agrees with that only where the two linears have the
same sign. Written the first way the answer to Timofeev's 315 held on (-1, 1) and failed above 1, where sqrt(1 - x) is imaginary.
 After SolveByLinearRadicalSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean,System.Boolean), which answers one linear base
and declines two of different roots; this is those.
https://github.com/asc-community/AngouriMath/issues/718

























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