\(\int x^{11} \sqrt [3]{-1+x^3} \, dx\) [429]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=\frac {\sqrt [3]{-1+x^3} \left (-81-27 x^3-18 x^6-14 x^9+140 x^{12}\right )}{1820} \]

[Out]

1/1820*(x^3-1)^(1/3)*(140*x^12-14*x^9-18*x^6-27*x^3-81)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.51, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=\frac {1}{13} \left (x^3-1\right )^{13/3}+\frac {3}{10} \left (x^3-1\right )^{10/3}+\frac {3}{7} \left (x^3-1\right )^{7/3}+\frac {1}{4} \left (x^3-1\right )^{4/3} \]

[In]

Int[x^11*(-1 + x^3)^(1/3),x]

[Out]

(-1 + x^3)^(4/3)/4 + (3*(-1 + x^3)^(7/3))/7 + (3*(-1 + x^3)^(10/3))/10 + (-1 + x^3)^(13/3)/13

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \sqrt [3]{-1+x} x^3 \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\sqrt [3]{-1+x}+3 (-1+x)^{4/3}+3 (-1+x)^{7/3}+(-1+x)^{10/3}\right ) \, dx,x,x^3\right ) \\ & = \frac {1}{4} \left (-1+x^3\right )^{4/3}+\frac {3}{7} \left (-1+x^3\right )^{7/3}+\frac {3}{10} \left (-1+x^3\right )^{10/3}+\frac {1}{13} \left (-1+x^3\right )^{13/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=\frac {\sqrt [3]{-1+x^3} \left (-81-27 x^3-18 x^6-14 x^9+140 x^{12}\right )}{1820} \]

[In]

Integrate[x^11*(-1 + x^3)^(1/3),x]

[Out]

((-1 + x^3)^(1/3)*(-81 - 27*x^3 - 18*x^6 - 14*x^9 + 140*x^12))/1820

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(\frac {\left (x^{3}-1\right )^{\frac {4}{3}} \left (140 x^{9}+126 x^{6}+108 x^{3}+81\right )}{1820}\)	\(27\)
trager	\(\left (\frac {1}{13} x^{12}-\frac {1}{130} x^{9}-\frac {9}{910} x^{6}-\frac {27}{1820} x^{3}-\frac {81}{1820}\right ) \left (x^{3}-1\right )^{\frac {1}{3}}\)	\(31\)
risch	\(\frac {\left (x^{3}-1\right )^{\frac {1}{3}} \left (140 x^{12}-14 x^{9}-18 x^{6}-27 x^{3}-81\right )}{1820}\)	\(32\)
meijerg	\(\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{12} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, 4\right ], \left [5\right ], x^{3}\right )}{12 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}\)	\(33\)
gosper	\(\frac {\left (x -1\right ) \left (x^{2}+x +1\right ) \left (140 x^{9}+126 x^{6}+108 x^{3}+81\right ) \left (x^{3}-1\right )^{\frac {1}{3}}}{1820}\)	\(36\)

[In]

int(x^11*(x^3-1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/1820*(x^3-1)^(4/3)*(140*x^9+126*x^6+108*x^3+81)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=\frac {1}{1820} \, {\left (140 \, x^{12} - 14 \, x^{9} - 18 \, x^{6} - 27 \, x^{3} - 81\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} \]

[In]

integrate(x^11*(x^3-1)^(1/3),x, algorithm="fricas")

[Out]

1/1820*(140*x^12 - 14*x^9 - 18*x^6 - 27*x^3 - 81)*(x^3 - 1)^(1/3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (31) = 62\).

Time = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.94 \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=\frac {x^{12} \sqrt [3]{x^{3} - 1}}{13} - \frac {x^{9} \sqrt [3]{x^{3} - 1}}{130} - \frac {9 x^{6} \sqrt [3]{x^{3} - 1}}{910} - \frac {27 x^{3} \sqrt [3]{x^{3} - 1}}{1820} - \frac {81 \sqrt [3]{x^{3} - 1}}{1820} \]

[In]

integrate(x**11*(x**3-1)**(1/3),x)

[Out]

x**12*(x**3 - 1)**(1/3)/13 - x**9*(x**3 - 1)**(1/3)/130 - 9*x**6*(x**3 - 1)**(1/3)/910 - 27*x**3*(x**3 - 1)**(
1/3)/1820 - 81*(x**3 - 1)**(1/3)/1820

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=\frac {1}{13} \, {\left (x^{3} - 1\right )}^{\frac {13}{3}} + \frac {3}{10} \, {\left (x^{3} - 1\right )}^{\frac {10}{3}} + \frac {3}{7} \, {\left (x^{3} - 1\right )}^{\frac {7}{3}} + \frac {1}{4} \, {\left (x^{3} - 1\right )}^{\frac {4}{3}} \]

[In]

integrate(x^11*(x^3-1)^(1/3),x, algorithm="maxima")

[Out]

1/13*(x^3 - 1)^(13/3) + 3/10*(x^3 - 1)^(10/3) + 3/7*(x^3 - 1)^(7/3) + 1/4*(x^3 - 1)^(4/3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=\frac {1}{13} \, {\left (x^{3} - 1\right )}^{\frac {13}{3}} + \frac {3}{10} \, {\left (x^{3} - 1\right )}^{\frac {10}{3}} + \frac {3}{7} \, {\left (x^{3} - 1\right )}^{\frac {7}{3}} + \frac {1}{4} \, {\left (x^{3} - 1\right )}^{\frac {4}{3}} \]

[In]

integrate(x^11*(x^3-1)^(1/3),x, algorithm="giac")

[Out]

1/13*(x^3 - 1)^(13/3) + 3/10*(x^3 - 1)^(10/3) + 3/7*(x^3 - 1)^(7/3) + 1/4*(x^3 - 1)^(4/3)

Mupad [B] (verification not implemented)

Time = 5.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int x^{11} \sqrt [3]{-1+x^3} \, dx=-{\left (x^3-1\right )}^{1/3}\,\left (-\frac {x^{12}}{13}+\frac {x^9}{130}+\frac {9\,x^6}{910}+\frac {27\,x^3}{1820}+\frac {81}{1820}\right ) \]

[In]

int(x^11*(x^3 - 1)^(1/3),x)

[Out]

-(x^3 - 1)^(1/3)*((27*x^3)/1820 + (9*x^6)/910 + x^9/130 - x^12/13 + 81/1820)