\(\int (1-x^3)^{2/3} \, dx\) [106]

Optimal result

Integrand size = 11, antiderivative size = 67 \[ \int \left (1-x^3\right )^{2/3} \, dx=\frac {1}{3} x \left (1-x^3\right )^{2/3}-\frac {2 \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{3} \log \left (x+\sqrt [3]{1-x^3}\right ) \]

[Out]

1/3*x*(-x^3+1)^(2/3)+1/3*ln(x+(-x^3+1)^(1/3))-2/9*arctan(1/3*(1-2*x/(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {201, 245} \[ \int \left (1-x^3\right )^{2/3} \, dx=-\frac {2 \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{3} \left (1-x^3\right )^{2/3} x+\frac {1}{3} \log \left (\sqrt [3]{1-x^3}+x\right ) \]

[In]

Int[(1 - x^3)^(2/3),x]

[Out]

(x*(1 - x^3)^(2/3))/3 - (2*ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]) + Log[x + (1 - x^3)^(1/3)]
/3

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x \left (1-x^3\right )^{2/3}+\frac {2}{3} \int \frac {1}{\sqrt [3]{1-x^3}} \, dx \\ & = \frac {1}{3} x \left (1-x^3\right )^{2/3}-\frac {2 \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{3} \log \left (x+\sqrt [3]{1-x^3}\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.51 \[ \int \left (1-x^3\right )^{2/3} \, dx=\frac {3 (-1+x) \left (1-x^3\right )^{2/3} \operatorname {AppellF1}\left (\frac {5}{3},-\frac {2}{3},-\frac {2}{3},\frac {8}{3},-\frac {-1+x}{1-(-1)^{2/3}},-\frac {-1+x}{1+\sqrt [3]{-1}}\right )}{5 \left (1+\frac {-1+x}{1+\sqrt [3]{-1}}\right )^{2/3} \left (1+\frac {-1+x}{1-(-1)^{2/3}}\right )^{2/3}} \]

[In]

Integrate[(1 - x^3)^(2/3),x]

[Out]

(3*(-1 + x)*(1 - x^3)^(2/3)*AppellF1[5/3, -2/3, -2/3, 8/3, -((-1 + x)/(1 - (-1)^(2/3))), -((-1 + x)/(1 + (-1)^
(1/3)))])/(5*(1 + (-1 + x)/(1 + (-1)^(1/3)))^(2/3)*(1 + (-1 + x)/(1 - (-1)^(2/3)))^(2/3))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.18

method	result	size
meijerg	\(x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {2}{3},\frac {1}{3};\frac {4}{3};x^{3}\right )\)	\(12\)
risch	\(-\frac {x \left (x^{3}-1\right )}{3 \left (-x^{3}+1\right )^{\frac {1}{3}}}+\frac {2 x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};x^{3}\right )}{3}\)	\(31\)
pseudoelliptic	\(\frac {3 x \left (-x^{3}+1\right )^{\frac {2}{3}}+2 \sqrt {3}\, \arctan \left (\frac {\left (-2 \left (-x^{3}+1\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )+2 \ln \left (\frac {x +\left (-x^{3}+1\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (-x^{3}+1\right )^{\frac {2}{3}}-x \left (-x^{3}+1\right )^{\frac {1}{3}}+x^{2}}{x^{2}}\right )}{9 \left (x +\left (-x^{3}+1\right )^{\frac {1}{3}}\right ) \left (\left (-x^{3}+1\right )^{\frac {2}{3}}-x \left (-x^{3}+1\right )^{\frac {1}{3}}+x^{2}\right )}\)	\(133\)
trager	\(\frac {x \left (-x^{3}+1\right )^{\frac {2}{3}}}{3}+\frac {2 \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}} x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (-x^{3}+1\right )^{\frac {2}{3}}+3 x^{2} \left (-x^{3}+1\right )^{\frac {1}{3}}-2 x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1\right )}{9}+\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}} x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}-3 x^{2} \left (-x^{3}+1\right )^{\frac {1}{3}}-2 x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+2\right )}{9}\)	\(206\)

[In]

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

[Out]

x*hypergeom([-2/3,1/3],[4/3],x^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int \left (1-x^3\right )^{2/3} \, dx=\frac {1}{3} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x - \frac {2}{9} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} x - 2 \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {2}{9} \, \log \left (\frac {x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{9} \, \log \left (\frac {x^{2} - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[In]

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

[Out]

1/3*(-x^3 + 1)^(2/3)*x - 2/9*sqrt(3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(-x^3 + 1)^(1/3))/x) + 2/9*log((x + (-
x^3 + 1)^(1/3))/x) - 1/9*log((x^2 - (-x^3 + 1)^(1/3)*x + (-x^3 + 1)^(2/3))/x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.46 \[ \int \left (1-x^3\right )^{2/3} \, dx=\frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

[In]

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

[Out]

x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), x**3*exp_polar(2*I*pi))/(3*gamma(4/3))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (52) = 104\).

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.57 \[ \int \left (1-x^3\right )^{2/3} \, dx=-\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right )}\right ) - \frac {{\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{3 \, x^{2} {\left (\frac {x^{3} - 1}{x^{3}} - 1\right )}} + \frac {2}{9} \, \log \left (\frac {{\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right ) - \frac {1}{9} \, \log \left (-\frac {{\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) \]

[In]

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

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3)/x - 1)) - 1/3*(-x^3 + 1)^(2/3)/(x^2*((x^3 - 1)/x^3 - 1)) +
2/9*log((-x^3 + 1)^(1/3)/x + 1) - 1/9*log(-(-x^3 + 1)^(1/3)/x + (-x^3 + 1)^(2/3)/x^2 + 1)

Giac [F]

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

[In]

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

[Out]

integrate((-x^3 + 1)^(2/3), x)

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.15 \[ \int \left (1-x^3\right )^{2/3} \, dx=x\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},\frac {1}{3};\ \frac {4}{3};\ x^3\right ) \]

[In]

int((1 - x^3)^(2/3),x)

[Out]

x*hypergeom([-2/3, 1/3], 4/3, x^3)