3.2409 \(\int \frac{(a+\frac{b}{\sqrt [3]{x}})^2}{x^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac{a^2}{x}-\frac{3 a b}{2 x^{4/3}}-\frac{3 b^2}{5 x^{5/3}} \]

[Out]

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

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Rubi [A]  time = 0.0163252, antiderivative size = 32, 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} \[ -\frac{a^2}{x}-\frac{3 a b}{2 x^{4/3}}-\frac{3 b^2}{5 x^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))^2/x^2,x]

[Out]

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

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 \frac{\left (a+\frac{b}{\sqrt [3]{x}}\right )^2}{x^2} \, dx &=\int \frac{\left (b+a \sqrt [3]{x}\right )^2}{x^{8/3}} \, dx\\ &=3 \operatorname{Subst}\left (\int \frac{(b+a x)^2}{x^6} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (\frac{b^2}{x^6}+\frac{2 a b}{x^5}+\frac{a^2}{x^4}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{3 b^2}{5 x^{5/3}}-\frac{3 a b}{2 x^{4/3}}-\frac{a^2}{x}\\ \end{align*}

Mathematica [A]  time = 0.0167544, size = 32, normalized size = 1. \[ -\frac{a^2}{x}-\frac{3 a b}{2 x^{4/3}}-\frac{3 b^2}{5 x^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))^2/x^2,x]

[Out]

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

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Maple [A]  time = 0.006, size = 25, normalized size = 0.8 \begin{align*} -{\frac{3\,{b}^{2}}{5}{x}^{-{\frac{5}{3}}}}-{\frac{3\,ab}{2}{x}^{-{\frac{4}{3}}}}-{\frac{{a}^{2}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.989646, size = 63, normalized size = 1.97 \begin{align*} -\frac{3 \,{\left (a + \frac{b}{x^{\frac{1}{3}}}\right )}^{5}}{5 \, b^{3}} + \frac{3 \,{\left (a + \frac{b}{x^{\frac{1}{3}}}\right )}^{4} a}{2 \, b^{3}} - \frac{{\left (a + \frac{b}{x^{\frac{1}{3}}}\right )}^{3} a^{2}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.50915, size = 74, normalized size = 2.31 \begin{align*} -\frac{10 \, a^{2} x + 15 \, a b x^{\frac{2}{3}} + 6 \, b^{2} x^{\frac{1}{3}}}{10 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(10*a^2*x + 15*a*b*x^(2/3) + 6*b^2*x^(1/3))/x^2

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Sympy [A]  time = 0.825653, size = 29, normalized size = 0.91 \begin{align*} - \frac{a^{2}}{x} - \frac{3 a b}{2 x^{\frac{4}{3}}} - \frac{3 b^{2}}{5 x^{\frac{5}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2/x - 3*a*b/(2*x**(4/3)) - 3*b**2/(5*x**(5/3))

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Giac [A]  time = 1.16178, size = 35, normalized size = 1.09 \begin{align*} -\frac{10 \, a^{2} x^{\frac{2}{3}} + 15 \, a b x^{\frac{1}{3}} + 6 \, b^{2}}{10 \, x^{\frac{5}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(10*a^2*x^(2/3) + 15*a*b*x^(1/3) + 6*b^2)/x^(5/3)