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

Optimal. Leaf size=80 \[ \frac{3 a^2 \left (a+b x^3\right )^{8/3}}{8 b^4}-\frac{a^3 \left (a+b x^3\right )^{5/3}}{5 b^4}+\frac{\left (a+b x^3\right )^{14/3}}{14 b^4}-\frac{3 a \left (a+b x^3\right )^{11/3}}{11 b^4} \]

[Out]

-(a^3*(a + b*x^3)^(5/3))/(5*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)

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Rubi [A]  time = 0.0489658, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{3 a^2 \left (a+b x^3\right )^{8/3}}{8 b^4}-\frac{a^3 \left (a+b x^3\right )^{5/3}}{5 b^4}+\frac{\left (a+b x^3\right )^{14/3}}{14 b^4}-\frac{3 a \left (a+b x^3\right )^{11/3}}{11 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a + b*x^3)^(2/3),x]

[Out]

-(a^3*(a + b*x^3)^(5/3))/(5*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 266

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]]

Rule 43

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])

Rubi steps

\begin{align*} \int x^{11} \left (a+b x^3\right )^{2/3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x^3 (a+b x)^{2/3} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{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]  time = 0.0254731, size = 50, normalized size = 0.62 \[ \frac{\left (a+b x^3\right )^{5/3} \left (135 a^2 b x^3-81 a^3-180 a b^2 x^6+220 b^3 x^9\right )}{3080 b^4} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.005, size = 47, normalized size = 0.6 \begin{align*} -{\frac{-220\,{b}^{3}{x}^{9}+180\,a{b}^{2}{x}^{6}-135\,{a}^{2}b{x}^{3}+81\,{a}^{3}}{3080\,{b}^{4}} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [A]  time = 1.06835, size = 86, normalized size = 1.08 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.72476, size = 136, normalized size = 1.7 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 8.96267, size = 110, normalized size = 1.38 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.1111, size = 77, normalized size = 0.96 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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