\(\int x^8 \sqrt [3]{-1+x^3} \, dx\) [356]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=\frac {1}{140} \sqrt [3]{-1+x^3} \left (-9-3 x^3-2 x^6+14 x^9\right ) \]

[Out]

1/140*(x^3-1)^(1/3)*(14*x^9-2*x^6-3*x^3-9)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=\frac {1}{10} \left (x^3-1\right )^{10/3}+\frac {2}{7} \left (x^3-1\right )^{7/3}+\frac {1}{4} \left (x^3-1\right )^{4/3} \]

[In]

Int[x^8*(-1 + x^3)^(1/3),x]

[Out]

(-1 + x^3)^(4/3)/4 + (2*(-1 + x^3)^(7/3))/7 + (-1 + x^3)^(10/3)/10

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^2 \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\sqrt [3]{-1+x}+2 (-1+x)^{4/3}+(-1+x)^{7/3}\right ) \, dx,x,x^3\right ) \\ & = \frac {1}{4} \left (-1+x^3\right )^{4/3}+\frac {2}{7} \left (-1+x^3\right )^{7/3}+\frac {1}{10} \left (-1+x^3\right )^{10/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=\frac {1}{140} \left (-1+x^3\right )^{4/3} \left (9+12 x^3+14 x^6\right ) \]

[In]

Integrate[x^8*(-1 + x^3)^(1/3),x]

[Out]

((-1 + x^3)^(4/3)*(9 + 12*x^3 + 14*x^6))/140

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(\frac {\left (x^{3}-1\right )^{\frac {4}{3}} \left (14 x^{6}+12 x^{3}+9\right )}{140}\)	\(22\)
trager	\(\left (\frac {1}{10} x^{9}-\frac {1}{70} x^{6}-\frac {3}{140} x^{3}-\frac {9}{140}\right ) \left (x^{3}-1\right )^{\frac {1}{3}}\)	\(26\)
risch	\(\frac {\left (x^{3}-1\right )^{\frac {1}{3}} \left (14 x^{9}-2 x^{6}-3 x^{3}-9\right )}{140}\)	\(27\)
gosper	\(\frac {\left (x -1\right ) \left (x^{2}+x +1\right ) \left (14 x^{6}+12 x^{3}+9\right ) \left (x^{3}-1\right )^{\frac {1}{3}}}{140}\)	\(31\)
meijerg	\(\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{9} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, 3\right ], \left [4\right ], x^{3}\right )}{9 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}\)	\(33\)

[In]

int(x^8*(x^3-1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/140*(x^3-1)^(4/3)*(14*x^6+12*x^3+9)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=\frac {1}{140} \, {\left (14 \, x^{9} - 2 \, x^{6} - 3 \, x^{3} - 9\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} \]

[In]

integrate(x^8*(x^3-1)^(1/3),x, algorithm="fricas")

[Out]

1/140*(14*x^9 - 2*x^6 - 3*x^3 - 9)*(x^3 - 1)^(1/3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=\frac {x^{9} \sqrt [3]{x^{3} - 1}}{10} - \frac {x^{6} \sqrt [3]{x^{3} - 1}}{70} - \frac {3 x^{3} \sqrt [3]{x^{3} - 1}}{140} - \frac {9 \sqrt [3]{x^{3} - 1}}{140} \]

[In]

integrate(x**8*(x**3-1)**(1/3),x)

[Out]

x**9*(x**3 - 1)**(1/3)/10 - x**6*(x**3 - 1)**(1/3)/70 - 3*x**3*(x**3 - 1)**(1/3)/140 - 9*(x**3 - 1)**(1/3)/140

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=\frac {1}{10} \, {\left (x^{3} - 1\right )}^{\frac {10}{3}} + \frac {2}{7} \, {\left (x^{3} - 1\right )}^{\frac {7}{3}} + \frac {1}{4} \, {\left (x^{3} - 1\right )}^{\frac {4}{3}} \]

[In]

integrate(x^8*(x^3-1)^(1/3),x, algorithm="maxima")

[Out]

1/10*(x^3 - 1)^(10/3) + 2/7*(x^3 - 1)^(7/3) + 1/4*(x^3 - 1)^(4/3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=\frac {1}{10} \, {\left (x^{3} - 1\right )}^{\frac {10}{3}} + \frac {2}{7} \, {\left (x^{3} - 1\right )}^{\frac {7}{3}} + \frac {1}{4} \, {\left (x^{3} - 1\right )}^{\frac {4}{3}} \]

[In]

integrate(x^8*(x^3-1)^(1/3),x, algorithm="giac")

[Out]

1/10*(x^3 - 1)^(10/3) + 2/7*(x^3 - 1)^(7/3) + 1/4*(x^3 - 1)^(4/3)

Mupad [B] (verification not implemented)

Time = 5.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int x^8 \sqrt [3]{-1+x^3} \, dx=-{\left (x^3-1\right )}^{1/3}\,\left (-\frac {x^9}{10}+\frac {x^6}{70}+\frac {3\,x^3}{140}+\frac {9}{140}\right ) \]

[In]

int(x^8*(x^3 - 1)^(1/3),x)

[Out]

-(x^3 - 1)^(1/3)*((3*x^3)/140 + x^6/70 - x^9/10 + 9/140)