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

Method (no overloads)

Summary

x/((a + b x^3) sqrt(c + d x^3)) at the three ratios where its integral is elementary,
Rubi's 1.1.3.4: 4 b c = a d, 8 b c + a d = 0 and
a^2 d^2 - 20 a b c d - 8 b^2 c^2 = 0, in closed form.

Remarks

With q = (d/c)^(1/3), the real root, and y = sqrt(c + d x^3), at
4 b c = a d, for r = sqrt(c) where c is positive and
r = sqrt(-c) where it is negative:

c > 0:   q/(3 2^(2/3) b r) (artanh(y/r)/3 + atan(y/(sqrt(3) r))/sqrt(3)
                     - atan(sqrt(3) r (1 + 2^(1/3) q x)/y)/sqrt(3) - artanh(r (1 - 2^(1/3) q x)/y))
c < 0:  -q/(3 2^(2/3) b r) (atan(y/r)/3 + artanh(y/(sqrt(3) r))/sqrt(3)
                     + artanh(sqrt(3) r (1 + 2^(1/3) q x)/y)/sqrt(3) + atan(r (1 - 2^(1/3) q x)/y))

and at 8 b c + a d = 0, where the integrand is -(d/b) x/((8 c - d x^3) y), that
times

c > 0:   (artanh(r (1 + q x)^2/(3 y))/18 - artanh(y/(3 r))/18 - atan(sqrt(3) r (1 + q x)/y)/(6 sqrt(3)))/(r^3 q^2)
c < 0:  -(atan(r (1 + q x)^2/(3 y))/18 + atan(y/(3 r))/18 - artanh(sqrt(3) r (1 + q x)/y)/(6 sqrt(3)))/(r^3 q^2)

Rubi splits the second into three integrals, a substitution's in x^3, a linear
over a linear and a quadratic over a quadratic, each elementary, and their sum is the
form above; asked as three, two come back piecewise in the symbols and their sum holds
every combination of the arms.
At the third, r = (a d - 10 b c)/(6 b c) is sqrt(3) or -sqrt(3), the
condition's two roots, and with s = sqrt(c r) Rubi's four terms are

-q (2 - r)/(sqrt(2) b) (atan(s (1 - r) y/(sqrt(2) c r^2))/(3 r s) + atan(s (1 + r) (1 + q x)/(sqrt(2) y))/(2 r s)
                       + artanh(s (1 - r) (1 + q x)/(sqrt(2) y))/(6 s) + artanh(s (1 + r - 2 q x)/(sqrt(2) y))/(3 s))

Each is atan(s v)/s or artanh(s v)/s, which depends on s^2 alone, so
where c r is negative the same terms are written with t = sqrt(-c r), each
atan becoming an artanh and each artanh an atan: one real form
for each sign, where Rubi has one for each sign of c and a power of r that is
complex at r = -sqrt(3).
Every condition is decided as a value, and the answer is differentiated back at sampled
points before it is returned.
Welz's x/((4 - x^3) sqrt(1 - x^3)) and x/((8 - d x^3) sqrt(1 + d x^3)) were
declined, with the rest of 1.1.3.4 at those ratios, and
x/((b x^3 + 2 a (5 + 3 sqrt(3))) sqrt(a + b x^3)) with its seven variants after them.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

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