\(\int (-1+3 x^2) \sqrt [3]{-x+x^3} \, dx\) [66]

Optimal result

Integrand size = 19, antiderivative size = 15 \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3}{4} \left (-x+x^3\right )^{4/3} \]

[Out]

3/4*(x^3-x)^(4/3)

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} \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3}{4} \left (x^3-x\right )^{4/3} \]

[In]

Int[(-1 + 3*x^2)*(-x + x^3)^(1/3),x]

[Out]

(3*(-x + x^3)^(4/3))/4

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}{4} \left (-x+x^3\right )^{4/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3}{4} \left (x \left (-1+x^2\right )\right )^{4/3} \]

[In]

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

[Out]

(3*(x*(-1 + x^2))^(4/3))/4

Maple [A] (verified)

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

method	result	size
derivativedivides	\(\frac {3 \left (x^{3}-x \right )^{\frac {4}{3}}}{4}\)	\(12\)
default	\(\frac {3 \left (x^{3}-x \right )^{\frac {4}{3}}}{4}\)	\(12\)
trager	\(\frac {3 x \left (x^{2}-1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{4}\)	\(18\)
risch	\(\frac {3 {\left (x \left (x^{2}-1\right )\right )}^{\frac {1}{3}} x \left (x^{2}-1\right )}{4}\)	\(18\)
gosper	\(\frac {3 \left (1+x \right ) x \left (x -1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{4}\)	\(19\)
pseudoelliptic	\(\frac {\left (3 x^{3}-3 x \right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{4}\)	\(21\)
meijerg	\(-\frac {3 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{2}\right )}{4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}}}+\frac {9 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} x^{\frac {10}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], x^{2}\right )}{10 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}}}\)	\(66\)

[In]

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

[Out]

3/4*(x^3-x)^(4/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3}{4} \, {\left (x^{3} - x\right )}^{\frac {4}{3}} \]

[In]

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

[Out]

3/4*(x^3 - x)^(4/3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (10) = 20\).

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.80 \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3 x^{3} \sqrt [3]{x^{3} - x}}{4} - \frac {3 x \sqrt [3]{x^{3} - x}}{4} \]

[In]

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

[Out]

3*x**3*(x**3 - x)**(1/3)/4 - 3*x*(x**3 - x)**(1/3)/4

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3}{4} \, {\left (x^{3} - x\right )}^{\frac {4}{3}} \]

[In]

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

[Out]

3/4*(x^3 - x)^(4/3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3}{4} \, {\left (x^{3} - x\right )}^{\frac {4}{3}} \]

[In]

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

[Out]

3/4*(x^3 - x)^(4/3)

Mupad [B] (verification not implemented)

Time = 5.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \left (-1+3 x^2\right ) \sqrt [3]{-x+x^3} \, dx=\frac {3\,{\left (x^3-x\right )}^{4/3}}{4} \]

[In]

int((x^3 - x)^(1/3)*(3*x^2 - 1),x)

[Out]

(3*(x^3 - x)^(4/3))/4