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

Optimal. Leaf size=78 \[ \frac{2 a^3}{3 b^4 \sqrt{a+b x^3}}+\frac{2 a^2 \sqrt{a+b x^3}}{b^4}-\frac{2 a \left (a+b x^3\right )^{3/2}}{3 b^4}+\frac{2 \left (a+b x^3\right )^{5/2}}{15 b^4} \]

[Out]

(2*a^3)/(3*b^4*Sqrt[a + b*x^3]) + (2*a^2*Sqrt[a + b*x^3])/b^4 - (2*a*(a + b*x^3)^(3/2))/(3*b^4) + (2*(a + b*x^
3)^(5/2))/(15*b^4)

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Rubi [A]  time = 0.0437219, antiderivative size = 78, 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} \[ \frac{2 a^3}{3 b^4 \sqrt{a+b x^3}}+\frac{2 a^2 \sqrt{a+b x^3}}{b^4}-\frac{2 a \left (a+b x^3\right )^{3/2}}{3 b^4}+\frac{2 \left (a+b x^3\right )^{5/2}}{15 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^11/(a + b*x^3)^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.0222087, size = 49, normalized size = 0.63 \[ \frac{2 \left (8 a^2 b x^3+16 a^3-2 a b^2 x^6+b^3 x^9\right )}{15 b^4 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a + b*x^3)^(3/2),x]

[Out]

(2*(16*a^3 + 8*a^2*b*x^3 - 2*a*b^2*x^6 + b^3*x^9))/(15*b^4*Sqrt[a + b*x^3])

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Maple [A]  time = 0.005, size = 46, normalized size = 0.6 \begin{align*}{\frac{2\,{b}^{3}{x}^{9}-4\,a{b}^{2}{x}^{6}+16\,{a}^{2}b{x}^{3}+32\,{a}^{3}}{15\,{b}^{4}}{\frac{1}{\sqrt{b{x}^{3}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/(b*x^3+a)^(1/2)*(b^3*x^9-2*a*b^2*x^6+8*a^2*b*x^3+16*a^3)/b^4

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Maxima [A]  time = 0.953148, size = 86, normalized size = 1.1 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}}}{15 \, b^{4}} - \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a}{3 \, b^{4}} + \frac{2 \, \sqrt{b x^{3} + a} a^{2}}{b^{4}} + \frac{2 \, a^{3}}{3 \, \sqrt{b x^{3} + a} 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]

2/15*(b*x^3 + a)^(5/2)/b^4 - 2/3*(b*x^3 + a)^(3/2)*a/b^4 + 2*sqrt(b*x^3 + a)*a^2/b^4 + 2/3*a^3/(sqrt(b*x^3 + a
)*b^4)

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Fricas [A]  time = 1.47078, size = 117, normalized size = 1.5 \begin{align*} \frac{2 \,{\left (b^{3} x^{9} - 2 \, a b^{2} x^{6} + 8 \, a^{2} b x^{3} + 16 \, a^{3}\right )} \sqrt{b x^{3} + a}}{15 \,{\left (b^{5} x^{3} + a b^{4}\right )}} \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]

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

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Sympy [A]  time = 3.85958, size = 94, normalized size = 1.21 \begin{align*} \begin{cases} \frac{32 a^{3}}{15 b^{4} \sqrt{a + b x^{3}}} + \frac{16 a^{2} x^{3}}{15 b^{3} \sqrt{a + b x^{3}}} - \frac{4 a x^{6}}{15 b^{2} \sqrt{a + b x^{3}}} + \frac{2 x^{9}}{15 b \sqrt{a + b x^{3}}} & \text{for}\: b \neq 0 \\\frac{x^{12}}{12 a^{\frac{3}{2}}} & \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**3/(15*b**4*sqrt(a + b*x**3)) + 16*a**2*x**3/(15*b**3*sqrt(a + b*x**3)) - 4*a*x**6/(15*b**2*sq
rt(a + b*x**3)) + 2*x**9/(15*b*sqrt(a + b*x**3)), Ne(b, 0)), (x**12/(12*a**(3/2)), True))

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Giac [A]  time = 1.11318, size = 74, normalized size = 0.95 \begin{align*} \frac{2 \,{\left ({\left (b x^{3} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{b x^{3} + a} a^{2} + \frac{5 \, a^{3}}{\sqrt{b x^{3} + a}}\right )}}{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="giac")

[Out]

2/15*((b*x^3 + a)^(5/2) - 5*(b*x^3 + a)^(3/2)*a + 15*sqrt(b*x^3 + a)*a^2 + 5*a^3/sqrt(b*x^3 + a))/b^4