3.2406 \(\int (a+\frac{b}{\sqrt [3]{x}})^2 x \, dx\)

Optimal. Leaf size=34 \[ \frac{a^2 x^2}{2}+\frac{6}{5} a b x^{5/3}+\frac{3}{4} b^2 x^{4/3} \]

[Out]

(3*b^2*x^(4/3))/4 + (6*a*b*x^(5/3))/5 + (a^2*x^2)/2

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Rubi [A]  time = 0.0195002, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {263, 266, 43} \[ \frac{a^2 x^2}{2}+\frac{6}{5} a b x^{5/3}+\frac{3}{4} b^2 x^{4/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))^2*x,x]

[Out]

(3*b^2*x^(4/3))/4 + (6*a*b*x^(5/3))/5 + (a^2*x^2)/2

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 \, dx &=\int \left (b+a \sqrt [3]{x}\right )^2 \sqrt [3]{x} \, dx\\ &=3 \operatorname{Subst}\left (\int x^3 (b+a x)^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (b^2 x^3+2 a b x^4+a^2 x^5\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{3}{4} b^2 x^{4/3}+\frac{6}{5} a b x^{5/3}+\frac{a^2 x^2}{2}\\ \end{align*}

Mathematica [A]  time = 0.0098686, size = 34, normalized size = 1. \[ \frac{a^2 x^2}{2}+\frac{6}{5} a b x^{5/3}+\frac{3}{4} b^2 x^{4/3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))^2*x,x]

[Out]

(3*b^2*x^(4/3))/4 + (6*a*b*x^(5/3))/5 + (a^2*x^2)/2

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Maple [A]  time = 0.002, size = 25, normalized size = 0.7 \begin{align*}{\frac{3\,{b}^{2}}{4}{x}^{{\frac{4}{3}}}}+{\frac{6\,ab}{5}{x}^{{\frac{5}{3}}}}+{\frac{{a}^{2}{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*b^2*x^(4/3)+6/5*a*b*x^(5/3)+1/2*a^2*x^2

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Maxima [A]  time = 1.01205, size = 35, normalized size = 1.03 \begin{align*} \frac{1}{20} \,{\left (10 \, a^{2} + \frac{24 \, a b}{x^{\frac{1}{3}}} + \frac{15 \, b^{2}}{x^{\frac{2}{3}}}\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(10*a^2 + 24*a*b/x^(1/3) + 15*b^2/x^(2/3))*x^2

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Fricas [A]  time = 1.48472, size = 66, normalized size = 1.94 \begin{align*} \frac{1}{2} \, a^{2} x^{2} + \frac{6}{5} \, a b x^{\frac{5}{3}} + \frac{3}{4} \, b^{2} x^{\frac{4}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x^2 + 6/5*a*b*x^(5/3) + 3/4*b^2*x^(4/3)

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Sympy [A]  time = 0.326758, size = 31, normalized size = 0.91 \begin{align*} \frac{a^{2} x^{2}}{2} + \frac{6 a b x^{\frac{5}{3}}}{5} + \frac{3 b^{2} x^{\frac{4}{3}}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**2/2 + 6*a*b*x**(5/3)/5 + 3*b**2*x**(4/3)/4

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Giac [A]  time = 1.16425, size = 32, normalized size = 0.94 \begin{align*} \frac{1}{2} \, a^{2} x^{2} + \frac{6}{5} \, a b x^{\frac{5}{3}} + \frac{3}{4} \, b^{2} x^{\frac{4}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x^2 + 6/5*a*b*x^(5/3) + 3/4*b^2*x^(4/3)