\(\int \frac {1}{\sqrt [3]{1-x^3}} \, dx\) [35]

Optimal result

Integrand size = 11, antiderivative size = 49 \[ \int \frac {1}{\sqrt [3]{1-x^3}} \, dx=-\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (x+\sqrt [3]{1-x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {245} \[ \int \frac {1}{\sqrt [3]{1-x^3}} \, dx=\frac {1}{2} \log \left (\sqrt [3]{1-x^3}+x\right )-\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}} \]

[In]

Int[(1 - x^3)^(-1/3),x]

[Out]

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

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 {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (x+\sqrt [3]{1-x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.76 \[ \int \frac {1}{\sqrt [3]{1-x^3}} \, dx=\frac {\arctan \left (\frac {-1+\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{6} \log \left (1+\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}\right )+\frac {1}{3} \log \left (1+\frac {x}{\sqrt [3]{1-x^3}}\right ) \]

[In]

Integrate[(1 - x^3)^(-1/3),x]

[Out]

ArcTan[(-1 + (2*x)/(1 - x^3)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[1 + x^2/(1 - x^3)^(2/3) - x/(1 - x^3)^(1/3)]/6 + Lo
g[1 + x/(1 - x^3)^(1/3)]/3

Maple [C] (verified)

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

Time = 1.54 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.24

method	result	size
meijerg	\(x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};x^{3}\right )\)	\(12\)
pseudoelliptic	\(\frac {\ln \left (\frac {x +\left (-x^{3}+1\right )^{\frac {1}{3}}}{x}\right )}{3}-\frac {\ln \left (\frac {\left (-x^{3}+1\right )^{\frac {2}{3}}-x \left (-x^{3}+1\right )^{\frac {1}{3}}+x^{2}}{x^{2}}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-2 \left (-x^{3}+1\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )}{3}\)	\(79\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}} x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}} x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (-x^{3}+1\right )^{\frac {2}{3}}-3 x^{2} \left (-x^{3}+1\right )^{\frac {1}{3}}+4 x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2\right )}{3}-\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}} x +3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-1\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{3}-\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}} x +3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-1\right )}{3}\)	\(286\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (40) = 80\).

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.67 \[ \int \frac {1}{\sqrt [3]{1-x^3}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} x - 2 \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {1}{3} \, \log \left (\frac {x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{6} \, \log \left (\frac {x^{2} - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[In]

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

[Out]

-1/3*sqrt(3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(-x^3 + 1)^(1/3))/x) + 1/3*log((x + (-x^3 + 1)^(1/3))/x) - 1/6
*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 = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\sqrt [3]{1-x^3}} \, dx=\frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\sqrt [3]{1-x^3}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right )}\right ) + \frac {1}{3} \, \log \left (\frac {{\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right ) - \frac {1}{6} \, \log \left (-\frac {{\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) \]

[In]

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

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3)/x - 1)) + 1/3*log((-x^3 + 1)^(1/3)/x + 1) - 1/6*log(-(-x^3
+ 1)^(1/3)/x + (-x^3 + 1)^(2/3)/x^2 + 1)

Giac [F]

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

[In]

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

[Out]

integrate((-x^3 + 1)^(-1/3), x)

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.20 \[ \int \frac {1}{\sqrt [3]{1-x^3}} \, dx=x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x^3\right ) \]

[In]

int(1/(1 - x^3)^(1/3),x)

[Out]

x*hypergeom([1/3, 1/3], 4/3, x^3)