3.545 \(\int \frac{x^8}{\sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=59 \[ \frac{a^2 \left (a+b x^3\right )^{2/3}}{2 b^3}+\frac{\left (a+b x^3\right )^{8/3}}{8 b^3}-\frac{2 a \left (a+b x^3\right )^{5/3}}{5 b^3} \]

[Out]

(a^2*(a + b*x^3)^(2/3))/(2*b^3) - (2*a*(a + b*x^3)^(5/3))/(5*b^3) + (a + b*x^3)^(8/3)/(8*b^3)

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Rubi [A]  time = 0.0349255, antiderivative size = 59, 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} \[ \frac{a^2 \left (a+b x^3\right )^{2/3}}{2 b^3}+\frac{\left (a+b x^3\right )^{8/3}}{8 b^3}-\frac{2 a \left (a+b x^3\right )^{5/3}}{5 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^3)^(1/3),x]

[Out]

(a^2*(a + b*x^3)^(2/3))/(2*b^3) - (2*a*(a + b*x^3)^(5/3))/(5*b^3) + (a + b*x^3)^(8/3)/(8*b^3)

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^8}{\sqrt [3]{a+b x^3}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt [3]{a+b x}} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{a^2}{b^2 \sqrt [3]{a+b x}}-\frac{2 a (a+b x)^{2/3}}{b^2}+\frac{(a+b x)^{5/3}}{b^2}\right ) \, dx,x,x^3\right )\\ &=\frac{a^2 \left (a+b x^3\right )^{2/3}}{2 b^3}-\frac{2 a \left (a+b x^3\right )^{5/3}}{5 b^3}+\frac{\left (a+b x^3\right )^{8/3}}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0179953, size = 39, normalized size = 0.66 \[ \frac{\left (a+b x^3\right )^{2/3} \left (9 a^2-6 a b x^3+5 b^2 x^6\right )}{40 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^3)^(1/3),x]

[Out]

((a + b*x^3)^(2/3)*(9*a^2 - 6*a*b*x^3 + 5*b^2*x^6))/(40*b^3)

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Maple [A]  time = 0.006, size = 36, normalized size = 0.6 \begin{align*}{\frac{5\,{b}^{2}{x}^{6}-6\,{x}^{3}ab+9\,{a}^{2}}{40\,{b}^{3}} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)^(1/3),x)

[Out]

1/40*(b*x^3+a)^(2/3)*(5*b^2*x^6-6*a*b*x^3+9*a^2)/b^3

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Maxima [A]  time = 1.0767, size = 63, normalized size = 1.07 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{8}{3}}}{8 \, b^{3}} - \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{5}{3}} a}{5 \, b^{3}} + \frac{{\left (b x^{3} + a\right )}^{\frac{2}{3}} a^{2}}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

1/8*(b*x^3 + a)^(8/3)/b^3 - 2/5*(b*x^3 + a)^(5/3)*a/b^3 + 1/2*(b*x^3 + a)^(2/3)*a^2/b^3

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Fricas [A]  time = 1.60175, size = 81, normalized size = 1.37 \begin{align*} \frac{{\left (5 \, b^{2} x^{6} - 6 \, a b x^{3} + 9 \, a^{2}\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{40 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

1/40*(5*b^2*x^6 - 6*a*b*x^3 + 9*a^2)*(b*x^3 + a)^(2/3)/b^3

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Sympy [A]  time = 1.89038, size = 68, normalized size = 1.15 \begin{align*} \begin{cases} \frac{9 a^{2} \left (a + b x^{3}\right )^{\frac{2}{3}}}{40 b^{3}} - \frac{3 a x^{3} \left (a + b x^{3}\right )^{\frac{2}{3}}}{20 b^{2}} + \frac{x^{6} \left (a + b x^{3}\right )^{\frac{2}{3}}}{8 b} & \text{for}\: b \neq 0 \\\frac{x^{9}}{9 \sqrt [3]{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**3+a)**(1/3),x)

[Out]

Piecewise((9*a**2*(a + b*x**3)**(2/3)/(40*b**3) - 3*a*x**3*(a + b*x**3)**(2/3)/(20*b**2) + x**6*(a + b*x**3)**
(2/3)/(8*b), Ne(b, 0)), (x**9/(9*a**(1/3)), True))

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Giac [A]  time = 1.12202, size = 58, normalized size = 0.98 \begin{align*} \frac{5 \,{\left (b x^{3} + a\right )}^{\frac{8}{3}} - 16 \,{\left (b x^{3} + a\right )}^{\frac{5}{3}} a + 20 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} a^{2}}{40 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

1/40*(5*(b*x^3 + a)^(8/3) - 16*(b*x^3 + a)^(5/3)*a + 20*(b*x^3 + a)^(2/3)*a^2)/b^3