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

Optimal. Leaf size=63 \[ \frac {1}{3} \left (x^3+2\right )^{2/3} x+\frac {5}{6} \log \left (\sqrt [3]{x^3+2}-x\right )-\frac {5 \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+2}}+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.01, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {388, 239} \[ \frac {1}{3} \left (x^3+2\right )^{2/3} x+\frac {5}{6} \log \left (\sqrt [3]{x^3+2}-x\right )-\frac {5 \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+2}}+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x^3)/(2 + x^3)^(1/3),x]

[Out]

(x*(2 + x^3)^(2/3))/3 - (5*ArcTan[(1 + (2*x)/(2 + x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]) + (5*Log[-x + (2 + x^3)^(1
/3)])/6

Rule 239

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]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx &=\frac {1}{3} x \left (2+x^3\right )^{2/3}-\frac {5}{3} \int \frac {1}{\sqrt [3]{2+x^3}} \, dx\\ &=\frac {1}{3} x \left (2+x^3\right )^{2/3}-\frac {5 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{2+x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {5}{6} \log \left (-x+\sqrt [3]{2+x^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 91, normalized size = 1.44 \[ \frac {1}{18} \left (6 \left (x^3+2\right )^{2/3} x+10 \log \left (1-\frac {x}{\sqrt [3]{x^3+2}}\right )-10 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+2}}+1}{\sqrt {3}}\right )-5 \log \left (\frac {x}{\sqrt [3]{x^3+2}}+\frac {x^2}{\left (x^3+2\right )^{2/3}}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*x*(2 + x^3)^(2/3) - 10*Sqrt[3]*ArcTan[(1 + (2*x)/(2 + x^3)^(1/3))/Sqrt[3]] + 10*Log[1 - x/(2 + x^3)^(1/3)]
- 5*Log[1 + x^2/(2 + x^3)^(2/3) + x/(2 + x^3)^(1/3)])/18

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IntegrateAlgebraic [A]  time = 0.15, size = 94, normalized size = 1.49 \[ \frac {1}{3} \left (x^3+2\right )^{2/3} x+\frac {5}{9} \log \left (\sqrt [3]{x^3+2}-x\right )-\frac {5 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+2}+x}\right )}{3 \sqrt {3}}-\frac {5}{18} \log \left (\sqrt [3]{x^3+2} x+\left (x^3+2\right )^{2/3}+x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2 + x^3)^(2/3))/3 - (5*ArcTan[(Sqrt[3]*x)/(x + 2*(2 + x^3)^(1/3))])/(3*Sqrt[3]) + (5*Log[-x + (2 + x^3)^(1
/3)])/9 - (5*Log[x^2 + x*(2 + x^3)^(1/3) + (2 + x^3)^(2/3)])/18

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fricas [A]  time = 0.86, size = 86, normalized size = 1.37 \[ \frac {1}{3} \, {\left (x^{3} + 2\right )}^{\frac {2}{3}} x + \frac {5}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {5}{9} \, \log \left (-\frac {x - {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x}\right ) - \frac {5}{18} \, \log \left (\frac {x^{2} + {\left (x^{3} + 2\right )}^{\frac {1}{3}} x + {\left (x^{3} + 2\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^3 + 2)^(2/3)*x + 5/9*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 + 2)^(1/3))/x) + 5/9*log(-(x - (x^3
+ 2)^(1/3))/x) - 5/18*log((x^2 + (x^3 + 2)^(1/3)*x + (x^3 + 2)^(2/3))/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} - 1}{{\left (x^{3} + 2\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.63, size = 29, normalized size = 0.46

method	result	size
risch	\(\frac {x \left (x^{3}+2\right )^{\frac {2}{3}}}{3}-\frac {5 \,2^{\frac {2}{3}} x \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -\frac {x^{3}}{2}\right )}{6}\)	\(29\)
meijerg	\(-\frac {2^{\frac {2}{3}} x \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -\frac {x^{3}}{2}\right )}{2}+\frac {2^{\frac {2}{3}} x^{4} \hypergeom \left (\left [\frac {1}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -\frac {x^{3}}{2}\right )}{8}\)	\(38\)
trager	\(\frac {x \left (x^{3}+2\right )^{\frac {2}{3}}}{3}+\frac {5 \ln \left (-4 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}+6 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}-8 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}+2\right )^{\frac {2}{3}}-4 x^{3}-4 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-4\right )}{9}+\frac {10 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \ln \left (4 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}-6 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {2}{3}} x +6 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}+2 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{3}+3 \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}-2 x^{3}-4 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-4\right )}{9}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*(x^3+2)^(2/3)-5/6*2^(2/3)*x*hypergeom([1/3,1/3],[4/3],-1/2*x^3)

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maxima [A]  time = 1.21, size = 94, normalized size = 1.49 \[ \frac {5}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) + \frac {2 \, {\left (x^{3} + 2\right )}^{\frac {2}{3}}}{3 \, x^{2} {\left (\frac {x^{3} + 2}{x^{3}} - 1\right )}} - \frac {5}{18} \, \log \left (\frac {{\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 2\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + \frac {5}{9} \, \log \left (\frac {{\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 2)^(1/3)/x + 1)) + 2/3*(x^3 + 2)^(2/3)/(x^2*((x^3 + 2)/x^3 - 1)) - 5/
18*log((x^3 + 2)^(1/3)/x + (x^3 + 2)^(2/3)/x^2 + 1) + 5/9*log((x^3 + 2)^(1/3)/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^3-1}{{\left (x^3+2\right )}^{1/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - 1)/(x^3 + 2)^(1/3),x)

[Out]

int((x^3 - 1)/(x^3 + 2)^(1/3), x)

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sympy [C]  time = 2.00, size = 71, normalized size = 1.13 \[ \frac {2^{\frac {2}{3}} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {x^{3} e^{i \pi }}{2}} \right )}}{6 \Gamma \left (\frac {7}{3}\right )} - \frac {2^{\frac {2}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {x^{3} e^{i \pi }}{2}} \right )}}{6 \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(2/3)*x**4*gamma(4/3)*hyper((1/3, 4/3), (7/3,), x**3*exp_polar(I*pi)/2)/(6*gamma(7/3)) - 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))

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