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

Optimal. Leaf size=80 \[ \frac{2 a^2 \left (a+b x^3\right )^{7/2}}{7 b^4}-\frac{2 a^3 \left (a+b x^3\right )^{5/2}}{15 b^4}+\frac{2 \left (a+b x^3\right )^{11/2}}{33 b^4}-\frac{2 a \left (a+b x^3\right )^{9/2}}{9 b^4} \]

[Out]

(-2*a^3*(a + b*x^3)^(5/2))/(15*b^4) + (2*a^2*(a + b*x^3)^(7/2))/(7*b^4) - (2*a*(a + b*x^3)^(9/2))/(9*b^4) + (2
*(a + b*x^3)^(11/2))/(33*b^4)

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Rubi [A]  time = 0.0469948, 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} \[ \frac{2 a^2 \left (a+b x^3\right )^{7/2}}{7 b^4}-\frac{2 a^3 \left (a+b x^3\right )^{5/2}}{15 b^4}+\frac{2 \left (a+b x^3\right )^{11/2}}{33 b^4}-\frac{2 a \left (a+b x^3\right )^{9/2}}{9 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a + b*x^3)^(3/2),x]

[Out]

(-2*a^3*(a + b*x^3)^(5/2))/(15*b^4) + (2*a^2*(a + b*x^3)^(7/2))/(7*b^4) - (2*a*(a + b*x^3)^(9/2))/(9*b^4) + (2
*(a + b*x^3)^(11/2))/(33*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 )^{3/2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x^3 (a+b x)^{3/2} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{a^3 (a+b x)^{3/2}}{b^3}+\frac{3 a^2 (a+b x)^{5/2}}{b^3}-\frac{3 a (a+b x)^{7/2}}{b^3}+\frac{(a+b x)^{9/2}}{b^3}\right ) \, dx,x,x^3\right )\\ &=-\frac{2 a^3 \left (a+b x^3\right )^{5/2}}{15 b^4}+\frac{2 a^2 \left (a+b x^3\right )^{7/2}}{7 b^4}-\frac{2 a \left (a+b x^3\right )^{9/2}}{9 b^4}+\frac{2 \left (a+b x^3\right )^{11/2}}{33 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0276563, size = 50, normalized size = 0.62 \[ \frac{2 \left (a+b x^3\right )^{5/2} \left (40 a^2 b x^3-16 a^3-70 a b^2 x^6+105 b^3 x^9\right )}{3465 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11*(a + b*x^3)^(3/2),x]

[Out]

(2*(a + b*x^3)^(5/2)*(-16*a^3 + 40*a^2*b*x^3 - 70*a*b^2*x^6 + 105*b^3*x^9))/(3465*b^4)

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Maple [A]  time = 0.007, size = 47, normalized size = 0.6 \begin{align*} -{\frac{-210\,{b}^{3}{x}^{9}+140\,a{b}^{2}{x}^{6}-80\,{a}^{2}b{x}^{3}+32\,{a}^{3}}{3465\,{b}^{4}} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(b*x^3+a)^(5/2)*(-105*b^3*x^9+70*a*b^2*x^6-40*a^2*b*x^3+16*a^3)/b^4

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Maxima [A]  time = 1.02037, size = 86, normalized size = 1.08 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{11}{2}}}{33 \, b^{4}} - \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{9}{2}} a}{9 \, b^{4}} + \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}} a^{2}}{7 \, b^{4}} - \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a^{3}}{15 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/33*(b*x^3 + a)^(11/2)/b^4 - 2/9*(b*x^3 + a)^(9/2)*a/b^4 + 2/7*(b*x^3 + a)^(7/2)*a^2/b^4 - 2/15*(b*x^3 + a)^(
5/2)*a^3/b^4

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Fricas [A]  time = 1.43779, size = 155, normalized size = 1.94 \begin{align*} \frac{2 \,{\left (105 \, b^{5} x^{15} + 140 \, a b^{4} x^{12} + 5 \, a^{2} b^{3} x^{9} - 6 \, a^{3} b^{2} x^{6} + 8 \, a^{4} b x^{3} - 16 \, a^{5}\right )} \sqrt{b x^{3} + a}}{3465 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(105*b^5*x^15 + 140*a*b^4*x^12 + 5*a^2*b^3*x^9 - 6*a^3*b^2*x^6 + 8*a^4*b*x^3 - 16*a^5)*sqrt(b*x^3 + a)/
b^4

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Sympy [A]  time = 10.0594, size = 136, normalized size = 1.7 \begin{align*} \begin{cases} - \frac{32 a^{5} \sqrt{a + b x^{3}}}{3465 b^{4}} + \frac{16 a^{4} x^{3} \sqrt{a + b x^{3}}}{3465 b^{3}} - \frac{4 a^{3} x^{6} \sqrt{a + b x^{3}}}{1155 b^{2}} + \frac{2 a^{2} x^{9} \sqrt{a + b x^{3}}}{693 b} + \frac{8 a x^{12} \sqrt{a + b x^{3}}}{99} + \frac{2 b x^{15} \sqrt{a + b x^{3}}}{33} & \text{for}\: b \neq 0 \\\frac{a^{\frac{3}{2}} 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)**(3/2),x)

[Out]

Piecewise((-32*a**5*sqrt(a + b*x**3)/(3465*b**4) + 16*a**4*x**3*sqrt(a + b*x**3)/(3465*b**3) - 4*a**3*x**6*sqr
t(a + b*x**3)/(1155*b**2) + 2*a**2*x**9*sqrt(a + b*x**3)/(693*b) + 8*a*x**12*sqrt(a + b*x**3)/99 + 2*b*x**15*s
qrt(a + b*x**3)/33, Ne(b, 0)), (a**(3/2)*x**12/12, True))

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Giac [B]  time = 1.12336, size = 181, normalized size = 2.26 \begin{align*} \frac{2 \,{\left (\frac{11 \,{\left (35 \,{\left (b x^{3} + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a^{3}\right )} a}{b^{3}} + \frac{315 \,{\left (b x^{3} + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x^{3} + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x^{3} + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a^{4}}{b^{3}}\right )}}{10395 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/10395*(11*(35*(b*x^3 + a)^(9/2) - 135*(b*x^3 + a)^(7/2)*a + 189*(b*x^3 + a)^(5/2)*a^2 - 105*(b*x^3 + a)^(3/2
)*a^3)*a/b^3 + (315*(b*x^3 + a)^(11/2) - 1540*(b*x^3 + a)^(9/2)*a + 2970*(b*x^3 + a)^(7/2)*a^2 - 2772*(b*x^3 +
a)^(5/2)*a^3 + 1155*(b*x^3 + a)^(3/2)*a^4)/b^3)/b