\(\int \frac {1}{\sqrt [3]{2+x^3}} \, dx\) [580]

Optimal result

Integrand size = 9, antiderivative size = 46 \[ \int \frac {1}{\sqrt [3]{2+x^3}} \, dx=\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{2+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (-x+\sqrt [3]{2+x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {245} \[ \int \frac {1}{\sqrt [3]{2+x^3}} \, dx=\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+2}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x^3+2}-x\right ) \]

[In]

Int[(2 + x^3)^(-1/3),x]

[Out]

ArcTan[(1 + (2*x)/(2 + x^3)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[-x + (2 + 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 {\tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{2+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (-x+\sqrt [3]{2+x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.70 \[ \int \frac {1}{\sqrt [3]{2+x^3}} \, dx=\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{2+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (1-\frac {x}{\sqrt [3]{2+x^3}}\right )+\frac {1}{6} \log \left (1+\frac {x^2}{\left (2+x^3\right )^{2/3}}+\frac {x}{\sqrt [3]{2+x^3}}\right ) \]

[In]

Integrate[(2 + x^3)^(-1/3),x]

[Out]

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

Maple [C] (verified)

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

Time = 4.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.39

method	result	size
meijerg	\(\frac {2^{\frac {2}{3}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};-\frac {x^{3}}{2}\right )}{2}\)	\(18\)
pseudoelliptic	\(-\frac {\ln \left (\frac {-x +\left (x^{3}+2\right )^{\frac {1}{3}}}{x}\right )}{3}+\frac {\ln \left (\frac {\left (x^{3}+2\right )^{\frac {2}{3}}+\left (x^{3}+2\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (x +2 \left (x^{3}+2\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 x}\right )}{3}\)	\(72\)
trager	\(-\frac {\ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {2}{3}} x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}+3 \left (x^{3}+2\right )^{\frac {2}{3}} x -3 \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}-2 x^{3}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-2\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {2}{3}} x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}+3 \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}-2 x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-4\right )}{3}\)	\(205\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.65 \[ \int \frac {1}{\sqrt [3]{2+x^3}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \frac {1}{3} \, \log \left (-\frac {x - {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{6} \, \log \left (\frac {x^{2} + {\left (x^{3} + 2\right )}^{\frac {1}{3}} x + {\left (x^{3} + 2\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt [3]{2+x^3}} \, dx=\frac {2^{\frac {2}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {x^{3} e^{i \pi }}{2}} \right )}}{6 \Gamma \left (\frac {4}{3}\right )} \]

[In]

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

[Out]

2**(2/3)*x*gamma(1/3)*hyper((1/3, 1/3), (4/3,), x**3*exp_polar(I*pi)/2)/(6*gamma(4/3))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt [3]{2+x^3}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) + \frac {1}{6} \, \log \left (\frac {{\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 2\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {1}{3} \, \log \left (\frac {{\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} - 1\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((x^3 + 2)^(-1/3), x)

Mupad [B] (verification not implemented)

Time = 5.80 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\sqrt [3]{2+x^3}} \, dx=\frac {2^{2/3}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -\frac {x^3}{2}\right )}{2} \]

[In]

int(1/(x^3 + 2)^(1/3),x)

[Out]

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