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

 Method (no overloads)

Summary

A rational function of x and one square root of a palindromic quartica x^4 + d x^3 + b x^2 + s d x + a, integrated by u = x - 1/x or
u = x + 1/x as s is -1 or +1: the quartic is x^2 times a quadratic in u, and the rest becomes a rational function of u where it is one.

Remarks

Charlwood's (1 + x^2)/((1 - x^2) sqrt(1 + x^4)) had no antiderivative. A root of
a quartic is nothing Euler's substitutions or the trigonometric ones read, and it is
not a binomial. But 1 + x^4 = x^2 (x^2 + 1/x^2) = x^2 ((x - 1/x)^2 + 2), and with
u = x - 1/x, du = (1 + 1/x^2) dx, the integrand is -du/(u sqrt(u^2 + 2)) -- a root of a quadratic over a rational function, which Euler answers. Its
companion (1 - x^2)/((1 + x^2) sqrt(1 + x^4)) wants u = x + 1/x, under
which x^2 + 1/x^2 is u^2 - 2. Both are tried.
The algebra. With u = x + s/x for s = -1 or +1:
x^2 + 1/x^2 = u^2 - 2s and x + s/x = u, so
Q = a x^4 + d x^3 + b x^2 + s d x + a = x^2 (a u^2 + d u + b - 2 a s);
and dx = x^2 du/(x^2 - s). For N/(D sqrt(Q)) the integrand is then
[N x/(D (x^2 - s))] du/sqrt(a u^2 + d u + b - 2as), and for N sqrt(Q)/D it
is [N x^3/(D (x^2 - s))] sqrt(a u^2 + d u + b - 2as) du. Odd terms fix the
sign: Timofeev's (1 - x^2)/((1 + 2ax + x^2) sqrt(1 + 2ax + 2bx^2 + 2ax^3 + x^4)) is one of u = x + 1/x only, -du/((u + 2a) sqrt(u^2 + 2au + 2b - 2)), and
its coefficients are symbols. The bracket is a rational
function of x; the rule asks whether it is one of u, by undetermined
coefficients on P(u)/S(u) and a check at sampled points, and declines where
it is not -- which is the only way this can fail, and is exact.
The sign of x.sqrt(Q) = |x| sqrt(a u^2 + d u + b - 2as), and the
rule takes |x| = x: what comes out is an antiderivative for x > 0. It
is made one everywhere by parity where the integrand has one, which is exact: an
odd integrand has an even antiderivative, so F(|x|) serves on both sides, and
an even one has an odd antiderivative, sgn(x) F(|x|). An integrand of neither
parity -- every one with odd terms under the root -- is answered by the sign of
x instead: the integrand in u is sgn(x) R(u) du/sqrt(q(u)) on both
sides, so sgn(x) G(x + s/x) is an antiderivative on both, and is what is
returned. A caller that knows its variable is positive, as the exponential
substitution does of u = e^x, says so and gets G(x + s/x) bare.
 A half-odd power of the quartic, Q^(3/2), is Q sqrt(Q), a whole power
beside the root, and is read as that: sinh(x)^2 sinh(2x)/(1 - sinh(x)^2)^(3/2) under u = e^x is a rational function over (6u^2 - u^4 - 1)^(3/2).
 https://github.com/asc-community/AngouriMath/issues/718

























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