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

Optimal. Leaf size=59 \[ \frac{a^2 \left (a+b x^3\right )^{4/3}}{4 b^3}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^3}-\frac{2 a \left (a+b x^3\right )^{7/3}}{7 b^3} \]

[Out]

(a^2*(a + b*x^3)^(4/3))/(4*b^3) - (2*a*(a + b*x^3)^(7/3))/(7*b^3) + (a + b*x^3)^(10/3)/(10*b^3)

________________________________________________________________________________________

Rubi [A]  time = 0.0321434, 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} \[ \frac{a^2 \left (a+b x^3\right )^{4/3}}{4 b^3}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^3}-\frac{2 a \left (a+b x^3\right )^{7/3}}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*x^3)^(1/3),x]

[Out]

(a^2*(a + b*x^3)^(4/3))/(4*b^3) - (2*a*(a + b*x^3)^(7/3))/(7*b^3) + (a + b*x^3)^(10/3)/(10*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 \sqrt [3]{a+b x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x^2 \sqrt [3]{a+b x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{a^2 \sqrt [3]{a+b x}}{b^2}-\frac{2 a (a+b x)^{4/3}}{b^2}+\frac{(a+b x)^{7/3}}{b^2}\right ) \, dx,x,x^3\right )\\ &=\frac{a^2 \left (a+b x^3\right )^{4/3}}{4 b^3}-\frac{2 a \left (a+b x^3\right )^{7/3}}{7 b^3}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0168128, size = 39, normalized size = 0.66 \[ \frac{\left (a+b x^3\right )^{4/3} \left (9 a^2-12 a b x^3+14 b^2 x^6\right )}{140 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*x^3)^(1/3),x]

[Out]

((a + b*x^3)^(4/3)*(9*a^2 - 12*a*b*x^3 + 14*b^2*x^6))/(140*b^3)

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 36, normalized size = 0.6 \begin{align*}{\frac{14\,{b}^{2}{x}^{6}-12\,{x}^{3}ab+9\,{a}^{2}}{140\,{b}^{3}} \left ( b{x}^{3}+a \right ) ^{{\frac{4}{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]

1/140*(b*x^3+a)^(4/3)*(14*b^2*x^6-12*a*b*x^3+9*a^2)/b^3

________________________________________________________________________________________

Maxima [A]  time = 1.03995, size = 63, normalized size = 1.07 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{\frac{10}{3}}}{10 \, b^{3}} - \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{7}{3}} a}{7 \, b^{3}} + \frac{{\left (b x^{3} + a\right )}^{\frac{4}{3}} a^{2}}{4 \, 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]

1/10*(b*x^3 + a)^(10/3)/b^3 - 2/7*(b*x^3 + a)^(7/3)*a/b^3 + 1/4*(b*x^3 + a)^(4/3)*a^2/b^3

________________________________________________________________________________________

Fricas [A]  time = 1.65881, size = 105, normalized size = 1.78 \begin{align*} \frac{{\left (14 \, b^{3} x^{9} + 2 \, a b^{2} x^{6} - 3 \, a^{2} b x^{3} + 9 \, a^{3}\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{140 \, 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]

1/140*(14*b^3*x^9 + 2*a*b^2*x^6 - 3*a^2*b*x^3 + 9*a^3)*(b*x^3 + a)^(1/3)/b^3

________________________________________________________________________________________

Sympy [A]  time = 2.44823, size = 87, normalized size = 1.47 \begin{align*} \begin{cases} \frac{9 a^{3} \sqrt [3]{a + b x^{3}}}{140 b^{3}} - \frac{3 a^{2} x^{3} \sqrt [3]{a + b x^{3}}}{140 b^{2}} + \frac{a x^{6} \sqrt [3]{a + b x^{3}}}{70 b} + \frac{x^{9} \sqrt [3]{a + b x^{3}}}{10} & \text{for}\: b \neq 0 \\\frac{\sqrt [3]{a} 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)**(1/3),x)

[Out]

Piecewise((9*a**3*(a + b*x**3)**(1/3)/(140*b**3) - 3*a**2*x**3*(a + b*x**3)**(1/3)/(140*b**2) + a*x**6*(a + b*
x**3)**(1/3)/(70*b) + x**9*(a + b*x**3)**(1/3)/10, Ne(b, 0)), (a**(1/3)*x**9/9, True))

________________________________________________________________________________________

Giac [A]  time = 1.11542, size = 58, normalized size = 0.98 \begin{align*} \frac{14 \,{\left (b x^{3} + a\right )}^{\frac{10}{3}} - 40 \,{\left (b x^{3} + a\right )}^{\frac{7}{3}} a + 35 \,{\left (b x^{3} + a\right )}^{\frac{4}{3}} a^{2}}{140 \, 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]

1/140*(14*(b*x^3 + a)^(10/3) - 40*(b*x^3 + a)^(7/3)*a + 35*(b*x^3 + a)^(4/3)*a^2)/b^3