3.546 \(\int \frac{x^5}{\sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=38 \[ \frac{\left (a+b x^3\right )^{5/3}}{5 b^2}-\frac{a \left (a+b x^3\right )^{2/3}}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0222965, antiderivative size = 38, 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} \[ \frac{\left (a+b x^3\right )^{5/3}}{5 b^2}-\frac{a \left (a+b x^3\right )^{2/3}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^3)^(1/3),x]

[Out]

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

Mathematica [A]  time = 0.0124022, size = 28, normalized size = 0.74 \[ \frac{\left (a+b x^3\right )^{2/3} \left (2 b x^3-3 a\right )}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^3)^(1/3),x]

[Out]

((a + b*x^3)^(2/3)*(-3*a + 2*b*x^3))/(10*b^2)

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Maple [A]  time = 0.004, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-2\,b{x}^{3}+3\,a}{10\,{b}^{2}} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^3+a)^(1/3),x)

[Out]

-1/10*(b*x^3+a)^(2/3)*(-2*b*x^3+3*a)/b^2

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Maxima [A]  time = 0.97447, size = 41, normalized size = 1.08 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{5}{3}}}{5 \, b^{2}} - \frac{{\left (b x^{3} + a\right )}^{\frac{2}{3}} a}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

1/5*(b*x^3 + a)^(5/3)/b^2 - 1/2*(b*x^3 + a)^(2/3)*a/b^2

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Fricas [A]  time = 1.38708, size = 59, normalized size = 1.55 \begin{align*} \frac{{\left (2 \, b x^{3} - 3 \, a\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{10 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

1/10*(2*b*x^3 - 3*a)*(b*x^3 + a)^(2/3)/b^2

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Sympy [A]  time = 0.919107, size = 44, normalized size = 1.16 \begin{align*} \begin{cases} - \frac{3 a \left (a + b x^{3}\right )^{\frac{2}{3}}}{10 b^{2}} + \frac{x^{3} \left (a + b x^{3}\right )^{\frac{2}{3}}}{5 b} & \text{for}\: b \neq 0 \\\frac{x^{6}}{6 \sqrt [3]{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**3+a)**(1/3),x)

[Out]

Piecewise((-3*a*(a + b*x**3)**(2/3)/(10*b**2) + x**3*(a + b*x**3)**(2/3)/(5*b), Ne(b, 0)), (x**6/(6*a**(1/3)),
True))

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Giac [A]  time = 1.09071, size = 39, normalized size = 1.03 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{5}{3}} - 5 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} a}{10 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

1/10*(2*(b*x^3 + a)^(5/3) - 5*(b*x^3 + a)^(2/3)*a)/b^2