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SolveByEulerSubstitution​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

 Method (no overloads)

Summary

A rational function of x and one square root of a quadratic in x,
rationalised by an Euler substitution and handed to the rational integrator.

Remarks

The trigonometric substitution beside this answers x^m sqrt(a + b x^2)^k and
nothing wider; sqrt(2 - x - x^2)/x^2, 1/((4 + x^2) sqrt(1 + 4x^2)) and
x sqrt(2 r x - x^2) had no antiderivative, and neither did any of the
integrands that other rules reduce to a rational function of x and one such
root -- the nested radicals of Bondarenko's suite under u = sqrt(1 + x), the
by-parts remainders of Charlwood's inverse functions.
Euler's three substitutions, for Q = a x^2 + b x + c:
a > 0: sqrt(Q) = t - sqrt(a) x, so x = (t^2 - c)/(2 sqrt(a) t + b).c > 0: sqrt(Q) = x t + sqrt(c), so x = (2 sqrt(c) t - b)/(a - t^2).c = 0: sqrt(Q) = x t, so x = b/(t^2 - a) -- the third substitution at
the root the quadratic has at zero.
Each makes x, sqrt(Q) and dx/dt rational in t, so the
integrand becomes a rational function of t; whichever applies with its radical
a rational number is taken first, and a symbol in a or c takes the first
in the generic case. The result goes to the rational integrator directly --
long division, the splits, the Hermite reduction, the binomial rule -- and not back
into the chain: an earlier Euler rule rewrote and handed on, and its cost was in the
open search below it and in a Simplify it ran before it could tell whether it
applied at all (https://github.com/asc-community/AngouriMath/issues/1265). This one
decides by reading the tree, in microseconds, and finishes in one closed step or not
at all.
Bounded in the degree of the rational function it produces, since a degree the
splits cannot factor is declined by them at a cost that grows with it. Asked, not
volunteered, like every rule that lands on a search of any size.
https://github.com/asc-community/AngouriMath/issues/718

























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