3.563 \(\int \frac{x^5}{(a+b x^3)^{2/3}} \, dx\)

Optimal. Leaf size=36 \[ \frac{\left (a+b x^3\right )^{4/3}}{4 b^2}-\frac{a \sqrt [3]{a+b x^3}}{b^2} \]

[Out]

-((a*(a + b*x^3)^(1/3))/b^2) + (a + b*x^3)^(4/3)/(4*b^2)

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Rubi [A]  time = 0.0219188, antiderivative size = 36, 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} \[ \frac{\left (a+b x^3\right )^{4/3}}{4 b^2}-\frac{a \sqrt [3]{a+b x^3}}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^3)^(2/3),x]

[Out]

-((a*(a + b*x^3)^(1/3))/b^2) + (a + b*x^3)^(4/3)/(4*b^2)

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

Mathematica [A]  time = 0.0124883, size = 27, normalized size = 0.75 \[ \frac{\left (b x^3-3 a\right ) \sqrt [3]{a+b x^3}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^3)^(2/3),x]

[Out]

((-3*a + b*x^3)*(a + b*x^3)^(1/3))/(4*b^2)

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Maple [A]  time = 0.003, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-b{x}^{3}+3\,a}{4\,{b}^{2}}\sqrt [3]{b{x}^{3}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^3+a)^(2/3),x)

[Out]

-1/4*(b*x^3+a)^(1/3)*(-b*x^3+3*a)/b^2

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Maxima [A]  time = 1.02863, size = 41, normalized size = 1.14 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{4}{3}}}{4 \, b^{2}} - \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}} a}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

1/4*(b*x^3 + a)^(4/3)/b^2 - (b*x^3 + a)^(1/3)*a/b^2

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Fricas [A]  time = 1.42485, size = 55, normalized size = 1.53 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b x^{3} - 3 \, a\right )}}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

1/4*(b*x^3 + a)^(1/3)*(b*x^3 - 3*a)/b^2

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Sympy [A]  time = 0.906806, size = 44, normalized size = 1.22 \begin{align*} \begin{cases} - \frac{3 a \sqrt [3]{a + b x^{3}}}{4 b^{2}} + \frac{x^{3} \sqrt [3]{a + b x^{3}}}{4 b} & \text{for}\: b \neq 0 \\\frac{x^{6}}{6 a^{\frac{2}{3}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**3+a)**(2/3),x)

[Out]

Piecewise((-3*a*(a + b*x**3)**(1/3)/(4*b**2) + x**3*(a + b*x**3)**(1/3)/(4*b), Ne(b, 0)), (x**6/(6*a**(2/3)),
True))

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Giac [A]  time = 1.07938, size = 36, normalized size = 1. \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{4}{3}} - 4 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} a}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

1/4*((b*x^3 + a)^(4/3) - 4*(b*x^3 + a)^(1/3)*a)/b^2