3.562 \(\int \frac{x^8}{(a+b x^3)^{2/3}} \, dx\)

Optimal. Leaf size=56 \[ \frac{a^2 \sqrt [3]{a+b x^3}}{b^3}+\frac{\left (a+b x^3\right )^{7/3}}{7 b^3}-\frac{a \left (a+b x^3\right )^{4/3}}{2 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0330715, antiderivative size = 56, 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} \[ \frac{a^2 \sqrt [3]{a+b x^3}}{b^3}+\frac{\left (a+b x^3\right )^{7/3}}{7 b^3}-\frac{a \left (a+b x^3\right )^{4/3}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^3)^(2/3),x]

[Out]

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

Mathematica [A]  time = 0.016996, size = 39, normalized size = 0.7 \[ \frac{\sqrt [3]{a+b x^3} \left (9 a^2-3 a b x^3+2 b^2 x^6\right )}{14 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^3)^(2/3),x]

[Out]

((a + b*x^3)^(1/3)*(9*a^2 - 3*a*b*x^3 + 2*b^2*x^6))/(14*b^3)

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Maple [A]  time = 0.004, size = 36, normalized size = 0.6 \begin{align*}{\frac{2\,{b}^{2}{x}^{6}-3\,{x}^{3}ab+9\,{a}^{2}}{14\,{b}^{3}}\sqrt [3]{b{x}^{3}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)^(2/3),x)

[Out]

1/14*(b*x^3+a)^(1/3)*(2*b^2*x^6-3*a*b*x^3+9*a^2)/b^3

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Maxima [A]  time = 0.993589, size = 62, normalized size = 1.11 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{7}{3}}}{7 \, b^{3}} - \frac{{\left (b x^{3} + a\right )}^{\frac{4}{3}} a}{2 \, b^{3}} + \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}} a^{2}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.43133, size = 81, normalized size = 1.45 \begin{align*} \frac{{\left (2 \, b^{2} x^{6} - 3 \, a b x^{3} + 9 \, a^{2}\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{14 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

1/14*(2*b^2*x^6 - 3*a*b*x^3 + 9*a^2)*(b*x^3 + a)^(1/3)/b^3

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Sympy [A]  time = 2.09424, size = 68, normalized size = 1.21 \begin{align*} \begin{cases} \frac{9 a^{2} \sqrt [3]{a + b x^{3}}}{14 b^{3}} - \frac{3 a x^{3} \sqrt [3]{a + b x^{3}}}{14 b^{2}} + \frac{x^{6} \sqrt [3]{a + b x^{3}}}{7 b} & \text{for}\: b \neq 0 \\\frac{x^{9}}{9 a^{\frac{2}{3}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**3+a)**(2/3),x)

[Out]

Piecewise((9*a**2*(a + b*x**3)**(1/3)/(14*b**3) - 3*a*x**3*(a + b*x**3)**(1/3)/(14*b**2) + x**6*(a + b*x**3)**
(1/3)/(7*b), Ne(b, 0)), (x**9/(9*a**(2/3)), True))

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Giac [A]  time = 1.11753, size = 58, normalized size = 1.04 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{7}{3}} - 7 \,{\left (b x^{3} + a\right )}^{\frac{4}{3}} a + 14 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} a^{2}}{14 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

1/14*(2*(b*x^3 + a)^(7/3) - 7*(b*x^3 + a)^(4/3)*a + 14*(b*x^3 + a)^(1/3)*a^2)/b^3