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

Optimal. Leaf size=59 \[ \frac{a^2 \left (a+b x^3\right )^{5/3}}{5 b^3}+\frac{\left (a+b x^3\right )^{11/3}}{11 b^3}-\frac{a \left (a+b x^3\right )^{8/3}}{4 b^3} \]

[Out]

(a^2*(a + b*x^3)^(5/3))/(5*b^3) - (a*(a + b*x^3)^(8/3))/(4*b^3) + (a + b*x^3)^(11/3)/(11*b^3)

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Rubi [A]  time = 0.0343528, 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} \[ \frac{a^2 \left (a+b x^3\right )^{5/3}}{5 b^3}+\frac{\left (a+b x^3\right )^{11/3}}{11 b^3}-\frac{a \left (a+b x^3\right )^{8/3}}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*x^3)^(2/3),x]

[Out]

(a^2*(a + b*x^3)^(5/3))/(5*b^3) - (a*(a + b*x^3)^(8/3))/(4*b^3) + (a + b*x^3)^(11/3)/(11*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 x^8 \left (a+b x^3\right )^{2/3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x^2 (a+b x)^{2/3} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{a^2 (a+b x)^{2/3}}{b^2}-\frac{2 a (a+b x)^{5/3}}{b^2}+\frac{(a+b x)^{8/3}}{b^2}\right ) \, dx,x,x^3\right )\\ &=\frac{a^2 \left (a+b x^3\right )^{5/3}}{5 b^3}-\frac{a \left (a+b x^3\right )^{8/3}}{4 b^3}+\frac{\left (a+b x^3\right )^{11/3}}{11 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0182656, size = 39, normalized size = 0.66 \[ \frac{\left (a+b x^3\right )^{5/3} \left (9 a^2-15 a b x^3+20 b^2 x^6\right )}{220 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*x^3)^(2/3),x]

[Out]

((a + b*x^3)^(5/3)*(9*a^2 - 15*a*b*x^3 + 20*b^2*x^6))/(220*b^3)

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Maple [A]  time = 0.006, size = 36, normalized size = 0.6 \begin{align*}{\frac{20\,{b}^{2}{x}^{6}-15\,{x}^{3}ab+9\,{a}^{2}}{220\,{b}^{3}} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/220*(b*x^3+a)^(5/3)*(20*b^2*x^6-15*a*b*x^3+9*a^2)/b^3

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Maxima [A]  time = 1.03002, size = 63, normalized size = 1.07 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{11}{3}}}{11 \, b^{3}} - \frac{{\left (b x^{3} + a\right )}^{\frac{8}{3}} a}{4 \, b^{3}} + \frac{{\left (b x^{3} + a\right )}^{\frac{5}{3}} a^{2}}{5 \, 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]

1/11*(b*x^3 + a)^(11/3)/b^3 - 1/4*(b*x^3 + a)^(8/3)*a/b^3 + 1/5*(b*x^3 + a)^(5/3)*a^2/b^3

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Fricas [A]  time = 1.5583, size = 105, normalized size = 1.78 \begin{align*} \frac{{\left (20 \, b^{3} x^{9} + 5 \, a b^{2} x^{6} - 6 \, a^{2} b x^{3} + 9 \, a^{3}\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{220 \, 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]

1/220*(20*b^3*x^9 + 5*a*b^2*x^6 - 6*a^2*b*x^3 + 9*a^3)*(b*x^3 + a)^(2/3)/b^3

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Sympy [A]  time = 4.05843, size = 87, normalized size = 1.47 \begin{align*} \begin{cases} \frac{9 a^{3} \left (a + b x^{3}\right )^{\frac{2}{3}}}{220 b^{3}} - \frac{3 a^{2} x^{3} \left (a + b x^{3}\right )^{\frac{2}{3}}}{110 b^{2}} + \frac{a x^{6} \left (a + b x^{3}\right )^{\frac{2}{3}}}{44 b} + \frac{x^{9} \left (a + b x^{3}\right )^{\frac{2}{3}}}{11} & \text{for}\: b \neq 0 \\\frac{a^{\frac{2}{3}} x^{9}}{9} & \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**3*(a + b*x**3)**(2/3)/(220*b**3) - 3*a**2*x**3*(a + b*x**3)**(2/3)/(110*b**2) + a*x**6*(a + b*
x**3)**(2/3)/(44*b) + x**9*(a + b*x**3)**(2/3)/11, Ne(b, 0)), (a**(2/3)*x**9/9, True))

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Giac [A]  time = 1.10149, size = 58, normalized size = 0.98 \begin{align*} \frac{20 \,{\left (b x^{3} + a\right )}^{\frac{11}{3}} - 55 \,{\left (b x^{3} + a\right )}^{\frac{8}{3}} a + 44 \,{\left (b x^{3} + a\right )}^{\frac{5}{3}} a^{2}}{220 \, 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]

1/220*(20*(b*x^3 + a)^(11/3) - 55*(b*x^3 + a)^(8/3)*a + 44*(b*x^3 + a)^(5/3)*a^2)/b^3