3.2414 \(\int (a+\frac{b}{\sqrt [3]{x}})^3 x^2 \, dx\)

Optimal. Leaf size=47 \[ \frac{9}{8} a^2 b x^{8/3}+\frac{a^3 x^3}{3}+\frac{9}{7} a b^2 x^{7/3}+\frac{b^3 x^2}{2} \]

[Out]

(b^3*x^2)/2 + (9*a*b^2*x^(7/3))/7 + (9*a^2*b*x^(8/3))/8 + (a^3*x^3)/3

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Rubi [A]  time = 0.0300391, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {263, 266, 43} \[ \frac{9}{8} a^2 b x^{8/3}+\frac{a^3 x^3}{3}+\frac{9}{7} a b^2 x^{7/3}+\frac{b^3 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))^3*x^2,x]

[Out]

(b^3*x^2)/2 + (9*a*b^2*x^(7/3))/7 + (9*a^2*b*x^(8/3))/8 + (a^3*x^3)/3

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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 \left (a+\frac{b}{\sqrt [3]{x}}\right )^3 x^2 \, dx &=\int \left (b+a \sqrt [3]{x}\right )^3 x \, dx\\ &=3 \operatorname{Subst}\left (\int x^5 (b+a x)^3 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (b^3 x^5+3 a b^2 x^6+3 a^2 b x^7+a^3 x^8\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{b^3 x^2}{2}+\frac{9}{7} a b^2 x^{7/3}+\frac{9}{8} a^2 b x^{8/3}+\frac{a^3 x^3}{3}\\ \end{align*}

Mathematica [A]  time = 0.0184443, size = 47, normalized size = 1. \[ \frac{9}{8} a^2 b x^{8/3}+\frac{a^3 x^3}{3}+\frac{9}{7} a b^2 x^{7/3}+\frac{b^3 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))^3*x^2,x]

[Out]

(b^3*x^2)/2 + (9*a*b^2*x^(7/3))/7 + (9*a^2*b*x^(8/3))/8 + (a^3*x^3)/3

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Maple [A]  time = 0.002, size = 36, normalized size = 0.8 \begin{align*}{\frac{{b}^{3}{x}^{2}}{2}}+{\frac{9\,{b}^{2}a}{7}{x}^{{\frac{7}{3}}}}+{\frac{9\,b{a}^{2}}{8}{x}^{{\frac{8}{3}}}}+{\frac{{a}^{3}{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^(1/3))^3*x^2,x)

[Out]

1/2*b^3*x^2+9/7*a*b^2*x^(7/3)+9/8*a^2*b*x^(8/3)+1/3*a^3*x^3

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Maxima [A]  time = 0.956926, size = 50, normalized size = 1.06 \begin{align*} \frac{1}{168} \,{\left (56 \, a^{3} + \frac{189 \, a^{2} b}{x^{\frac{1}{3}}} + \frac{216 \, a b^{2}}{x^{\frac{2}{3}}} + \frac{84 \, b^{3}}{x}\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^(1/3))^3*x^2,x, algorithm="maxima")

[Out]

1/168*(56*a^3 + 189*a^2*b/x^(1/3) + 216*a*b^2/x^(2/3) + 84*b^3/x)*x^3

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Fricas [A]  time = 1.42199, size = 90, normalized size = 1.91 \begin{align*} \frac{1}{3} \, a^{3} x^{3} + \frac{9}{8} \, a^{2} b x^{\frac{8}{3}} + \frac{9}{7} \, a b^{2} x^{\frac{7}{3}} + \frac{1}{2} \, b^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^(1/3))^3*x^2,x, algorithm="fricas")

[Out]

1/3*a^3*x^3 + 9/8*a^2*b*x^(8/3) + 9/7*a*b^2*x^(7/3) + 1/2*b^3*x^2

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Sympy [A]  time = 1.42408, size = 42, normalized size = 0.89 \begin{align*} \frac{a^{3} x^{3}}{3} + \frac{9 a^{2} b x^{\frac{8}{3}}}{8} + \frac{9 a b^{2} x^{\frac{7}{3}}}{7} + \frac{b^{3} x^{2}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**(1/3))**3*x**2,x)

[Out]

a**3*x**3/3 + 9*a**2*b*x**(8/3)/8 + 9*a*b**2*x**(7/3)/7 + b**3*x**2/2

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Giac [A]  time = 1.15235, size = 47, normalized size = 1. \begin{align*} \frac{1}{3} \, a^{3} x^{3} + \frac{9}{8} \, a^{2} b x^{\frac{8}{3}} + \frac{9}{7} \, a b^{2} x^{\frac{7}{3}} + \frac{1}{2} \, b^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^(1/3))^3*x^2,x, algorithm="giac")

[Out]

1/3*a^3*x^3 + 9/8*a^2*b*x^(8/3) + 9/7*a*b^2*x^(7/3) + 1/2*b^3*x^2