3.239 \(\int x^5 (a+b x^3)^3 \, dx\)

Optimal. Leaf size=34 \[ \frac{\left (a+b x^3\right )^5}{15 b^2}-\frac{a \left (a+b x^3\right )^4}{12 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0304055, antiderivative size = 34, 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} \[ \frac{\left (a+b x^3\right )^5}{15 b^2}-\frac{a \left (a+b x^3\right )^4}{12 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^3)^3,x]

[Out]

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

Mathematica [A]  time = 0.0016564, size = 43, normalized size = 1.26 \[ \frac{1}{3} a^2 b x^9+\frac{a^3 x^6}{6}+\frac{1}{4} a b^2 x^{12}+\frac{b^3 x^{15}}{15} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^3)^3,x]

[Out]

(a^3*x^6)/6 + (a^2*b*x^9)/3 + (a*b^2*x^12)/4 + (b^3*x^15)/15

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Maple [A]  time = 0.001, size = 36, normalized size = 1.1 \begin{align*}{\frac{{b}^{3}{x}^{15}}{15}}+{\frac{a{b}^{2}{x}^{12}}{4}}+{\frac{{a}^{2}b{x}^{9}}{3}}+{\frac{{a}^{3}{x}^{6}}{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^3+a)^3,x)

[Out]

1/15*b^3*x^15+1/4*a*b^2*x^12+1/3*a^2*b*x^9+1/6*a^3*x^6

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Maxima [A]  time = 0.949532, size = 47, normalized size = 1.38 \begin{align*} \frac{1}{15} \, b^{3} x^{15} + \frac{1}{4} \, a b^{2} x^{12} + \frac{1}{3} \, a^{2} b x^{9} + \frac{1}{6} \, a^{3} x^{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^3,x, algorithm="maxima")

[Out]

1/15*b^3*x^15 + 1/4*a*b^2*x^12 + 1/3*a^2*b*x^9 + 1/6*a^3*x^6

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Fricas [A]  time = 1.536, size = 84, normalized size = 2.47 \begin{align*} \frac{1}{15} x^{15} b^{3} + \frac{1}{4} x^{12} b^{2} a + \frac{1}{3} x^{9} b a^{2} + \frac{1}{6} x^{6} a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^3,x, algorithm="fricas")

[Out]

1/15*x^15*b^3 + 1/4*x^12*b^2*a + 1/3*x^9*b*a^2 + 1/6*x^6*a^3

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Sympy [A]  time = 0.073353, size = 36, normalized size = 1.06 \begin{align*} \frac{a^{3} x^{6}}{6} + \frac{a^{2} b x^{9}}{3} + \frac{a b^{2} x^{12}}{4} + \frac{b^{3} x^{15}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**3+a)**3,x)

[Out]

a**3*x**6/6 + a**2*b*x**9/3 + a*b**2*x**12/4 + b**3*x**15/15

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Giac [A]  time = 1.15908, size = 47, normalized size = 1.38 \begin{align*} \frac{1}{15} \, b^{3} x^{15} + \frac{1}{4} \, a b^{2} x^{12} + \frac{1}{3} \, a^{2} b x^{9} + \frac{1}{6} \, a^{3} x^{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^3,x, algorithm="giac")

[Out]

1/15*b^3*x^15 + 1/4*a*b^2*x^12 + 1/3*a^2*b*x^9 + 1/6*a^3*x^6