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

Optimal. Leaf size=38 \[ \frac{\left (a+b x^3\right )^{7/3}}{7 b^2}-\frac{a \left (a+b x^3\right )^{4/3}}{4 b^2} \]

[Out]

-(a*(a + b*x^3)^(4/3))/(4*b^2) + (a + b*x^3)^(7/3)/(7*b^2)

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Rubi [A]  time = 0.0222906, 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} \[ \frac{\left (a+b x^3\right )^{7/3}}{7 b^2}-\frac{a \left (a+b x^3\right )^{4/3}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^3)^(1/3),x]

[Out]

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

Mathematica [A]  time = 0.0128967, size = 28, normalized size = 0.74 \[ \frac{\left (a+b x^3\right )^{4/3} \left (4 b x^3-3 a\right )}{28 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^3)^(1/3),x]

[Out]

((a + b*x^3)^(4/3)*(-3*a + 4*b*x^3))/(28*b^2)

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Maple [A]  time = 0.003, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-4\,b{x}^{3}+3\,a}{28\,{b}^{2}} \left ( b{x}^{3}+a \right ) ^{{\frac{4}{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]

-1/28*(b*x^3+a)^(4/3)*(-4*b*x^3+3*a)/b^2

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Maxima [A]  time = 1.03449, size = 41, normalized size = 1.08 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{7}{3}}}{7 \, b^{2}} - \frac{{\left (b x^{3} + a\right )}^{\frac{4}{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)^(1/3),x, algorithm="maxima")

[Out]

1/7*(b*x^3 + a)^(7/3)/b^2 - 1/4*(b*x^3 + a)^(4/3)*a/b^2

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Fricas [A]  time = 1.66456, size = 78, normalized size = 2.05 \begin{align*} \frac{{\left (4 \, b^{2} x^{6} + a b x^{3} - 3 \, a^{2}\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{28 \, 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]

1/28*(4*b^2*x^6 + a*b*x^3 - 3*a^2)*(b*x^3 + a)^(1/3)/b^2

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Sympy [A]  time = 1.01663, size = 63, normalized size = 1.66 \begin{align*} \begin{cases} - \frac{3 a^{2} \sqrt [3]{a + b x^{3}}}{28 b^{2}} + \frac{a x^{3} \sqrt [3]{a + b x^{3}}}{28 b} + \frac{x^{6} \sqrt [3]{a + b x^{3}}}{7} & \text{for}\: b \neq 0 \\\frac{\sqrt [3]{a} x^{6}}{6} & \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**2*(a + b*x**3)**(1/3)/(28*b**2) + a*x**3*(a + b*x**3)**(1/3)/(28*b) + x**6*(a + b*x**3)**(1/3
)/7, Ne(b, 0)), (a**(1/3)*x**6/6, True))

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Giac [A]  time = 1.11871, size = 39, normalized size = 1.03 \begin{align*} \frac{4 \,{\left (b x^{3} + a\right )}^{\frac{7}{3}} - 7 \,{\left (b x^{3} + a\right )}^{\frac{4}{3}} a}{28 \, 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]

1/28*(4*(b*x^3 + a)^(7/3) - 7*(b*x^3 + a)^(4/3)*a)/b^2