\(\int x^6 (-1+x^3)^{2/3} \, dx\) [1548]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 13, antiderivative size = 107 \[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=\frac {1}{81} \left (-1+x^3\right )^{2/3} \left (-4 x-3 x^4+9 x^7\right )-\frac {4 \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{81 \sqrt {3}}+\frac {4}{243} \log \left (-x+\sqrt [3]{-1+x^3}\right )-\frac {2}{243} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]

[Out]

1/81*(x^3-1)^(2/3)*(9*x^7-3*x^4-4*x)-4/243*arctan(3^(1/2)*x/(x+2*(x^3-1)^(1/3)))*3^(1/2)+4/243*ln(-x+(x^3-1)^(
1/3))-2/243*ln(x^2+x*(x^3-1)^(1/3)+(x^3-1)^(2/3))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {285, 327, 245} \[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=-\frac {4 \arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{81 \sqrt {3}}-\frac {4}{81} \left (x^3-1\right )^{2/3} x+\frac {2}{81} \log \left (\sqrt [3]{x^3-1}-x\right )+\frac {1}{9} \left (x^3-1\right )^{2/3} x^7-\frac {1}{27} \left (x^3-1\right )^{2/3} x^4 \]

[In]

Int[x^6*(-1 + x^3)^(2/3),x]

[Out]

(-4*x*(-1 + x^3)^(2/3))/81 - (x^4*(-1 + x^3)^(2/3))/27 + (x^7*(-1 + x^3)^(2/3))/9 - (4*ArcTan[(1 + (2*x)/(-1 +
 x^3)^(1/3))/Sqrt[3]])/(81*Sqrt[3]) + (2*Log[-x + (-1 + x^3)^(1/3)])/81

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]

Rule 285

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

Rule 327

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} x^7 \left (-1+x^3\right )^{2/3}-\frac {2}{9} \int \frac {x^6}{\sqrt [3]{-1+x^3}} \, dx \\ & = -\frac {1}{27} x^4 \left (-1+x^3\right )^{2/3}+\frac {1}{9} x^7 \left (-1+x^3\right )^{2/3}-\frac {4}{27} \int \frac {x^3}{\sqrt [3]{-1+x^3}} \, dx \\ & = -\frac {4}{81} x \left (-1+x^3\right )^{2/3}-\frac {1}{27} x^4 \left (-1+x^3\right )^{2/3}+\frac {1}{9} x^7 \left (-1+x^3\right )^{2/3}-\frac {4}{81} \int \frac {1}{\sqrt [3]{-1+x^3}} \, dx \\ & = -\frac {4}{81} x \left (-1+x^3\right )^{2/3}-\frac {1}{27} x^4 \left (-1+x^3\right )^{2/3}+\frac {1}{9} x^7 \left (-1+x^3\right )^{2/3}-\frac {4 \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{81 \sqrt {3}}+\frac {2}{81} \log \left (-x+\sqrt [3]{-1+x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.95 \[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=\frac {1}{243} \left (3 x \left (-1+x^3\right )^{2/3} \left (-4-3 x^3+9 x^6\right )-4 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )+4 \log \left (-x+\sqrt [3]{-1+x^3}\right )-2 \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )\right ) \]

[In]

Integrate[x^6*(-1 + x^3)^(2/3),x]

[Out]

(3*x*(-1 + x^3)^(2/3)*(-4 - 3*x^3 + 9*x^6) - 4*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*(-1 + x^3)^(1/3))] + 4*Log[-x
 + (-1 + x^3)^(1/3)] - 2*Log[x^2 + x*(-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)])/243

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.31

method result size
meijerg \(\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}} x^{7} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {7}{3}\right ], \left [\frac {10}{3}\right ], x^{3}\right )}{7 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}}}\) \(33\)
risch \(\frac {x \left (9 x^{6}-3 x^{3}-4\right ) \left (x^{3}-1\right )^{\frac {2}{3}}}{81}-\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{81 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}\) \(54\)
pseudoelliptic \(\frac {-2 \ln \left (\frac {x^{2}+x \left (x^{3}-1\right )^{\frac {1}{3}}+\left (x^{3}-1\right )^{\frac {2}{3}}}{x^{2}}\right )+4 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}-1\right )^{\frac {1}{3}}\right )}{3 x}\right )+4 \ln \left (\frac {-x +\left (x^{3}-1\right )^{\frac {1}{3}}}{x}\right )+\left (27 x^{7}-9 x^{4}-12 x \right ) \left (x^{3}-1\right )^{\frac {2}{3}}}{243 {\left (\left (x^{3}-1\right )^{\frac {2}{3}}+x \left (x +\left (x^{3}-1\right )^{\frac {1}{3}}\right )\right )}^{3} {\left (x -\left (x^{3}-1\right )^{\frac {1}{3}}\right )}^{3}}\) \(130\)
trager \(\frac {x \left (9 x^{6}-3 x^{3}-4\right ) \left (x^{3}-1\right )^{\frac {2}{3}}}{81}+\frac {4 \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}-1\right )^{\frac {2}{3}}-3 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-2 x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1\right )}{243}+\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}-1\right )^{\frac {2}{3}}-x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1\right )}{243}\) \(189\)

[In]

int(x^6*(x^3-1)^(2/3),x,method=_RETURNVERBOSE)

[Out]

1/7*signum(x^3-1)^(2/3)/(-signum(x^3-1))^(2/3)*x^7*hypergeom([-2/3,7/3],[10/3],x^3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=\frac {1}{81} \, {\left (9 \, x^{7} - 3 \, x^{4} - 4 \, x\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} + \frac {4}{243} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {4}{243} \, \log \left (-\frac {x - {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {2}{243} \, \log \left (\frac {x^{2} + {\left (x^{3} - 1\right )}^{\frac {1}{3}} x + {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[In]

integrate(x^6*(x^3-1)^(2/3),x, algorithm="fricas")

[Out]

1/81*(9*x^7 - 3*x^4 - 4*x)*(x^3 - 1)^(2/3) + 4/243*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 - 1)^(1/3))/
x) + 4/243*log(-(x - (x^3 - 1)^(1/3))/x) - 2/243*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 = 3.58 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.32 \[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=- \frac {x^{7} e^{- \frac {i \pi }{3}} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \]

[In]

integrate(x**6*(x**3-1)**(2/3),x)

[Out]

-x**7*exp(-I*pi/3)*gamma(7/3)*hyper((-2/3, 7/3), (10/3,), x**3)/(3*gamma(10/3))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.36 \[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=\frac {4}{243} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + \frac {11 \, {\left (x^{3} - 1\right )}^{\frac {5}{3}}}{x^{5}} - \frac {4 \, {\left (x^{3} - 1\right )}^{\frac {8}{3}}}{x^{8}}}{81 \, {\left (\frac {3 \, {\left (x^{3} - 1\right )}}{x^{3}} - \frac {3 \, {\left (x^{3} - 1\right )}^{2}}{x^{6}} + \frac {{\left (x^{3} - 1\right )}^{3}}{x^{9}} - 1\right )}} - \frac {2}{243} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + \frac {4}{243} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \]

[In]

integrate(x^6*(x^3-1)^(2/3),x, algorithm="maxima")

[Out]

4/243*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3)/x + 1)) - 1/81*(2*(x^3 - 1)^(2/3)/x^2 + 11*(x^3 - 1)^(5/3)
/x^5 - 4*(x^3 - 1)^(8/3)/x^8)/(3*(x^3 - 1)/x^3 - 3*(x^3 - 1)^2/x^6 + (x^3 - 1)^3/x^9 - 1) - 2/243*log((x^3 - 1
)^(1/3)/x + (x^3 - 1)^(2/3)/x^2 + 1) + 4/243*log((x^3 - 1)^(1/3)/x - 1)

Giac [F]

\[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=\int { {\left (x^{3} - 1\right )}^{\frac {2}{3}} x^{6} \,d x } \]

[In]

integrate(x^6*(x^3-1)^(2/3),x, algorithm="giac")

[Out]

integrate((x^3 - 1)^(2/3)*x^6, x)

Mupad [F(-1)]

Timed out. \[ \int x^6 \left (-1+x^3\right )^{2/3} \, dx=\int x^6\,{\left (x^3-1\right )}^{2/3} \,d x \]

[In]

int(x^6*(x^3 - 1)^(2/3),x)

[Out]

int(x^6*(x^3 - 1)^(2/3), x)