\(\int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx\) [65]

Optimal result

Integrand size = 19, antiderivative size = 15 \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3}{2} \left (-x+x^3\right )^{2/3} \]

[Out]

3/2*(x^3-x)^(2/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1602} \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3}{2} \left (x^3-x\right )^{2/3} \]

[In]

Int[(-1 + 3*x^2)/(-x + x^3)^(1/3),x]

[Out]

(3*(-x + x^3)^(2/3))/2

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {3}{2} \left (-x+x^3\right )^{2/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3}{2} \left (x \left (-1+x^2\right )\right )^{2/3} \]

[In]

Integrate[(-1 + 3*x^2)/(-x + x^3)^(1/3),x]

[Out]

(3*(x*(-1 + x^2))^(2/3))/2

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {3 \left (x^{3}-x \right )^{\frac {2}{3}}}{2}\)	\(12\)
default	\(\frac {3 \left (x^{3}-x \right )^{\frac {2}{3}}}{2}\)	\(12\)
trager	\(\frac {3 \left (x^{3}-x \right )^{\frac {2}{3}}}{2}\)	\(12\)
pseudoelliptic	\(\frac {3 \left (x^{3}-x \right )^{\frac {2}{3}}}{2}\)	\(12\)
risch	\(\frac {3 x \left (x^{2}-1\right )}{2 {\left (x \left (x^{2}-1\right )\right )}^{\frac {1}{3}}}\)	\(18\)
gosper	\(\frac {3 \left (1+x \right ) x \left (x -1\right )}{2 \left (x^{3}-x \right )^{\frac {1}{3}}}\)	\(19\)
meijerg	\(-\frac {3 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{2}\right )}{2 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}}}+\frac {9 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {8}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], x^{2}\right )}{8 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}}}\)	\(66\)

[In]

int((3*x^2-1)/(x^3-x)^(1/3),x,method=_RETURNVERBOSE)

[Out]

3/2*(x^3-x)^(2/3)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3}{2} \, {\left (x^{3} - x\right )}^{\frac {2}{3}} \]

[In]

integrate((3*x^2-1)/(x^3-x)^(1/3),x, algorithm="fricas")

[Out]

3/2*(x^3 - x)^(2/3)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3 \left (x^{3} - x\right )^{\frac {2}{3}}}{2} \]

[In]

integrate((3*x**2-1)/(x**3-x)**(1/3),x)

[Out]

3*(x**3 - x)**(2/3)/2

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3}{2} \, {\left (x^{3} - x\right )}^{\frac {2}{3}} \]

[In]

integrate((3*x^2-1)/(x^3-x)^(1/3),x, algorithm="maxima")

[Out]

3/2*(x^3 - x)^(2/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3}{2} \, {\left (x^{3} - x\right )}^{\frac {2}{3}} \]

[In]

integrate((3*x^2-1)/(x^3-x)^(1/3),x, algorithm="giac")

[Out]

3/2*(x^3 - x)^(2/3)

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 x^2}{\sqrt [3]{-x+x^3}} \, dx=\frac {3\,{\left (x^3-x\right )}^{2/3}}{2} \]

[In]

int((3*x^2 - 1)/(x^3 - x)^(1/3),x)

[Out]

(3*(x^3 - x)^(2/3))/2