3.2404 \(\int (a+\frac{b}{\sqrt [3]{x}})^2 x^3 \, dx\)

Optimal. Leaf size=34 \[ \frac{a^2 x^4}{4}+\frac{6}{11} a b x^{11/3}+\frac{3}{10} b^2 x^{10/3} \]

[Out]

(3*b^2*x^(10/3))/10 + (6*a*b*x^(11/3))/11 + (a^2*x^4)/4

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Rubi [A]  time = 0.027699, antiderivative size = 34, 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} \[ \frac{a^2 x^4}{4}+\frac{6}{11} a b x^{11/3}+\frac{3}{10} b^2 x^{10/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))^2*x^3,x]

[Out]

(3*b^2*x^(10/3))/10 + (6*a*b*x^(11/3))/11 + (a^2*x^4)/4

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 )^2 x^3 \, dx &=\int \left (b+a \sqrt [3]{x}\right )^2 x^{7/3} \, dx\\ &=3 \operatorname{Subst}\left (\int x^9 (b+a x)^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (b^2 x^9+2 a b x^{10}+a^2 x^{11}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{3}{10} b^2 x^{10/3}+\frac{6}{11} a b x^{11/3}+\frac{a^2 x^4}{4}\\ \end{align*}

Mathematica [A]  time = 0.0102911, size = 34, normalized size = 1. \[ \frac{a^2 x^4}{4}+\frac{6}{11} a b x^{11/3}+\frac{3}{10} b^2 x^{10/3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))^2*x^3,x]

[Out]

(3*b^2*x^(10/3))/10 + (6*a*b*x^(11/3))/11 + (a^2*x^4)/4

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Maple [A]  time = 0.001, size = 25, normalized size = 0.7 \begin{align*}{\frac{3\,{b}^{2}}{10}{x}^{{\frac{10}{3}}}}+{\frac{6\,ab}{11}{x}^{{\frac{11}{3}}}}+{\frac{{a}^{2}{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*b^2*x^(10/3)+6/11*a*b*x^(11/3)+1/4*a^2*x^4

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Maxima [A]  time = 0.999271, size = 35, normalized size = 1.03 \begin{align*} \frac{1}{220} \,{\left (55 \, a^{2} + \frac{120 \, a b}{x^{\frac{1}{3}}} + \frac{66 \, b^{2}}{x^{\frac{2}{3}}}\right )} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/220*(55*a^2 + 120*a*b/x^(1/3) + 66*b^2/x^(2/3))*x^4

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Fricas [A]  time = 1.43084, size = 72, normalized size = 2.12 \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{6}{11} \, a b x^{\frac{11}{3}} + \frac{3}{10} \, b^{2} x^{\frac{10}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 + 6/11*a*b*x^(11/3) + 3/10*b^2*x^(10/3)

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Sympy [A]  time = 1.49883, size = 31, normalized size = 0.91 \begin{align*} \frac{a^{2} x^{4}}{4} + \frac{6 a b x^{\frac{11}{3}}}{11} + \frac{3 b^{2} x^{\frac{10}{3}}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**4/4 + 6*a*b*x**(11/3)/11 + 3*b**2*x**(10/3)/10

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Giac [A]  time = 1.19087, size = 32, normalized size = 0.94 \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{6}{11} \, a b x^{\frac{11}{3}} + \frac{3}{10} \, b^{2} x^{\frac{10}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 + 6/11*a*b*x^(11/3) + 3/10*b^2*x^(10/3)