\(\int x^{11} (a+b x^3)^{2/3} \, dx\) [527]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx=-\frac {a^3 \left (a+b x^3\right )^{5/3}}{5 b^4}+\frac {3 a^2 \left (a+b x^3\right )^{8/3}}{8 b^4}-\frac {3 a \left (a+b x^3\right )^{11/3}}{11 b^4}+\frac {\left (a+b x^3\right )^{14/3}}{14 b^4} \]

[Out]

-1/5*a^3*(b*x^3+a)^(5/3)/b^4+3/8*a^2*(b*x^3+a)^(8/3)/b^4-3/11*a*(b*x^3+a)^(11/3)/b^4+1/14*(b*x^3+a)^(14/3)/b^4

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx=-\frac {a^3 \left (a+b x^3\right )^{5/3}}{5 b^4}+\frac {3 a^2 \left (a+b x^3\right )^{8/3}}{8 b^4}+\frac {\left (a+b x^3\right )^{14/3}}{14 b^4}-\frac {3 a \left (a+b x^3\right )^{11/3}}{11 b^4} \]

[In]

Int[x^11*(a + b*x^3)^(2/3),x]

[Out]

-1/5*(a^3*(a + b*x^3)^(5/3))/b^4 + (3*a^2*(a + b*x^3)^(8/3))/(8*b^4) - (3*a*(a + b*x^3)^(11/3))/(11*b^4) + (a
+ b*x^3)^(14/3)/(14*b^4)

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 x^3 (a+b x)^{2/3} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (-\frac {a^3 (a+b x)^{2/3}}{b^3}+\frac {3 a^2 (a+b x)^{5/3}}{b^3}-\frac {3 a (a+b x)^{8/3}}{b^3}+\frac {(a+b x)^{11/3}}{b^3}\right ) \, dx,x,x^3\right ) \\ & = -\frac {a^3 \left (a+b x^3\right )^{5/3}}{5 b^4}+\frac {3 a^2 \left (a+b x^3\right )^{8/3}}{8 b^4}-\frac {3 a \left (a+b x^3\right )^{11/3}}{11 b^4}+\frac {\left (a+b x^3\right )^{14/3}}{14 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx=\frac {\left (a+b x^3\right )^{5/3} \left (-81 a^3+135 a^2 b x^3-180 a b^2 x^6+220 b^3 x^9\right )}{3080 b^4} \]

[In]

Integrate[x^11*(a + b*x^3)^(2/3),x]

[Out]

((a + b*x^3)^(5/3)*(-81*a^3 + 135*a^2*b*x^3 - 180*a*b^2*x^6 + 220*b^3*x^9))/(3080*b^4)

Maple [A] (verified)

Time = 3.85 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-220 b^{3} x^{9}+180 a \,b^{2} x^{6}-135 a^{2} b \,x^{3}+81 a^{3}\right )}{3080 b^{4}}\)	\(47\)
pseudoelliptic	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-220 b^{3} x^{9}+180 a \,b^{2} x^{6}-135 a^{2} b \,x^{3}+81 a^{3}\right )}{3080 b^{4}}\)	\(47\)
trager	\(-\frac {\left (-220 b^{4} x^{12}-40 a \,b^{3} x^{9}+45 a^{2} b^{2} x^{6}-54 a^{3} b \,x^{3}+81 a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{3080 b^{4}}\)	\(58\)
risch	\(-\frac {\left (-220 b^{4} x^{12}-40 a \,b^{3} x^{9}+45 a^{2} b^{2} x^{6}-54 a^{3} b \,x^{3}+81 a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{3080 b^{4}}\)	\(58\)

[In]

int(x^11*(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)

[Out]

-1/3080*(b*x^3+a)^(5/3)*(-220*b^3*x^9+180*a*b^2*x^6-135*a^2*b*x^3+81*a^3)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (220 \, b^{4} x^{12} + 40 \, a b^{3} x^{9} - 45 \, a^{2} b^{2} x^{6} + 54 \, a^{3} b x^{3} - 81 \, a^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{3080 \, b^{4}} \]

[In]

integrate(x^11*(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

1/3080*(220*b^4*x^12 + 40*a*b^3*x^9 - 45*a^2*b^2*x^6 + 54*a^3*b*x^3 - 81*a^4)*(b*x^3 + a)^(2/3)/b^4

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx=\begin {cases} - \frac {81 a^{4} \left (a + b x^{3}\right )^{\frac {2}{3}}}{3080 b^{4}} + \frac {27 a^{3} x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{1540 b^{3}} - \frac {9 a^{2} x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}{616 b^{2}} + \frac {a x^{9} \left (a + b x^{3}\right )^{\frac {2}{3}}}{77 b} + \frac {x^{12} \left (a + b x^{3}\right )^{\frac {2}{3}}}{14} & \text {for}\: b \neq 0 \\\frac {a^{\frac {2}{3}} x^{12}}{12} & \text {otherwise} \end {cases} \]

[In]

integrate(x**11*(b*x**3+a)**(2/3),x)

[Out]

Piecewise((-81*a**4*(a + b*x**3)**(2/3)/(3080*b**4) + 27*a**3*x**3*(a + b*x**3)**(2/3)/(1540*b**3) - 9*a**2*x*
*6*(a + b*x**3)**(2/3)/(616*b**2) + a*x**9*(a + b*x**3)**(2/3)/(77*b) + x**12*(a + b*x**3)**(2/3)/14, Ne(b, 0)
), (a**(2/3)*x**12/12, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {14}{3}}}{14 \, b^{4}} - \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} a}{11 \, b^{4}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a^{2}}{8 \, b^{4}} - \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{3}}{5 \, b^{4}} \]

[In]

integrate(x^11*(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

1/14*(b*x^3 + a)^(14/3)/b^4 - 3/11*(b*x^3 + a)^(11/3)*a/b^4 + 3/8*(b*x^3 + a)^(8/3)*a^2/b^4 - 1/5*(b*x^3 + a)^
(5/3)*a^3/b^4

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx=\frac {220 \, {\left (b x^{3} + a\right )}^{\frac {14}{3}} - 840 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} a + 1155 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a^{2} - 616 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{3}}{3080 \, b^{4}} \]

[In]

integrate(x^11*(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

1/3080*(220*(b*x^3 + a)^(14/3) - 840*(b*x^3 + a)^(11/3)*a + 1155*(b*x^3 + a)^(8/3)*a^2 - 616*(b*x^3 + a)^(5/3)
*a^3)/b^4

Mupad [B] (verification not implemented)

Time = 6.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.69 \[ \int x^{11} \left (a+b x^3\right )^{2/3} \, dx={\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {x^{12}}{14}-\frac {81\,a^4}{3080\,b^4}+\frac {a\,x^9}{77\,b}+\frac {27\,a^3\,x^3}{1540\,b^3}-\frac {9\,a^2\,x^6}{616\,b^2}\right ) \]

[In]

int(x^11*(a + b*x^3)^(2/3),x)

[Out]

(a + b*x^3)^(2/3)*(x^12/14 - (81*a^4)/(3080*b^4) + (a*x^9)/(77*b) + (27*a^3*x^3)/(1540*b^3) - (9*a^2*x^6)/(616
*b^2))