3.2301 \(\int \frac{(a+b \sqrt [3]{x})^2}{x^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.014822, antiderivative size = 34, 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} \[ -\frac{a^2}{2 x^2}-\frac{6 a b}{5 x^{5/3}}-\frac{3 b^2}{4 x^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^2/x^3,x]

[Out]

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

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 \frac{\left (a+b \sqrt [3]{x}\right )^2}{x^3} \, dx &=3 \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x^7} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (\frac{a^2}{x^7}+\frac{2 a b}{x^6}+\frac{b^2}{x^5}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{a^2}{2 x^2}-\frac{6 a b}{5 x^{5/3}}-\frac{3 b^2}{4 x^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.012257, size = 34, normalized size = 1. \[ -\frac{a^2}{2 x^2}-\frac{6 a b}{5 x^{5/3}}-\frac{3 b^2}{4 x^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^2/x^3,x]

[Out]

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

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Maple [A]  time = 0.006, size = 25, normalized size = 0.7 \begin{align*} -{\frac{{a}^{2}}{2\,{x}^{2}}}-{\frac{6\,ab}{5}{x}^{-{\frac{5}{3}}}}-{\frac{3\,{b}^{2}}{4}{x}^{-{\frac{4}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.968705, size = 35, normalized size = 1.03 \begin{align*} -\frac{15 \, b^{2} x^{\frac{2}{3}} + 24 \, a b x^{\frac{1}{3}} + 10 \, a^{2}}{20 \, x^{2}} \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]

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

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Fricas [A]  time = 1.44383, size = 73, normalized size = 2.15 \begin{align*} -\frac{15 \, b^{2} x^{\frac{2}{3}} + 24 \, a b x^{\frac{1}{3}} + 10 \, a^{2}}{20 \, x^{2}} \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]

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

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Sympy [A]  time = 1.37497, size = 32, normalized size = 0.94 \begin{align*} - \frac{a^{2}}{2 x^{2}} - \frac{6 a b}{5 x^{\frac{5}{3}}} - \frac{3 b^{2}}{4 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**3,x)

[Out]

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

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Giac [A]  time = 1.13355, size = 35, normalized size = 1.03 \begin{align*} -\frac{15 \, b^{2} x^{\frac{2}{3}} + 24 \, a b x^{\frac{1}{3}} + 10 \, a^{2}}{20 \, x^{2}} \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]

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