3.238 \(\int x^8 (a+b x^3)^3 \, dx\)

Optimal. Leaf size=43 \[ \frac{1}{4} a^2 b x^{12}+\frac{a^3 x^9}{9}+\frac{1}{5} a b^2 x^{15}+\frac{b^3 x^{18}}{18} \]

[Out]

(a^3*x^9)/9 + (a^2*b*x^12)/4 + (a*b^2*x^15)/5 + (b^3*x^18)/18

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Rubi [A]  time = 0.024537, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac{1}{4} a^2 b x^{12}+\frac{a^3 x^9}{9}+\frac{1}{5} a b^2 x^{15}+\frac{b^3 x^{18}}{18} \]

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*x^3)^3,x]

[Out]

(a^3*x^9)/9 + (a^2*b*x^12)/4 + (a*b^2*x^15)/5 + (b^3*x^18)/18

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 )^3 \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x^2 (a+b x)^3 \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (a^3 x^2+3 a^2 b x^3+3 a b^2 x^4+b^3 x^5\right ) \, dx,x,x^3\right )\\ &=\frac{a^3 x^9}{9}+\frac{1}{4} a^2 b x^{12}+\frac{1}{5} a b^2 x^{15}+\frac{b^3 x^{18}}{18}\\ \end{align*}

Mathematica [A]  time = 0.001715, size = 43, normalized size = 1. \[ \frac{1}{4} a^2 b x^{12}+\frac{a^3 x^9}{9}+\frac{1}{5} a b^2 x^{15}+\frac{b^3 x^{18}}{18} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*x^3)^3,x]

[Out]

(a^3*x^9)/9 + (a^2*b*x^12)/4 + (a*b^2*x^15)/5 + (b^3*x^18)/18

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Maple [A]  time = 0.001, size = 36, normalized size = 0.8 \begin{align*}{\frac{{a}^{3}{x}^{9}}{9}}+{\frac{{a}^{2}b{x}^{12}}{4}}+{\frac{a{b}^{2}{x}^{15}}{5}}+{\frac{{b}^{3}{x}^{18}}{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*a^3*x^9+1/4*a^2*b*x^12+1/5*a*b^2*x^15+1/18*b^3*x^18

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Maxima [A]  time = 0.953135, size = 47, normalized size = 1.09 \begin{align*} \frac{1}{18} \, b^{3} x^{18} + \frac{1}{5} \, a b^{2} x^{15} + \frac{1}{4} \, a^{2} b x^{12} + \frac{1}{9} \, a^{3} x^{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*b^3*x^18 + 1/5*a*b^2*x^15 + 1/4*a^2*b*x^12 + 1/9*a^3*x^9

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Fricas [A]  time = 1.44471, size = 85, normalized size = 1.98 \begin{align*} \frac{1}{18} x^{18} b^{3} + \frac{1}{5} x^{15} b^{2} a + \frac{1}{4} x^{12} b a^{2} + \frac{1}{9} x^{9} a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*x^18*b^3 + 1/5*x^15*b^2*a + 1/4*x^12*b*a^2 + 1/9*x^9*a^3

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Sympy [A]  time = 0.084648, size = 36, normalized size = 0.84 \begin{align*} \frac{a^{3} x^{9}}{9} + \frac{a^{2} b x^{12}}{4} + \frac{a b^{2} x^{15}}{5} + \frac{b^{3} x^{18}}{18} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*x**9/9 + a**2*b*x**12/4 + a*b**2*x**15/5 + b**3*x**18/18

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Giac [A]  time = 1.11864, size = 47, normalized size = 1.09 \begin{align*} \frac{1}{18} \, b^{3} x^{18} + \frac{1}{5} \, a b^{2} x^{15} + \frac{1}{4} \, a^{2} b x^{12} + \frac{1}{9} \, a^{3} x^{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*b^3*x^18 + 1/5*a*b^2*x^15 + 1/4*a^2*b*x^12 + 1/9*a^3*x^9