3.2309 \(\int \frac{(a+b \sqrt [3]{x})^3}{x^2} \, dx\)

Optimal. Leaf size=39 \[ -\frac{9 a^2 b}{2 x^{2/3}}-\frac{a^3}{x}-\frac{9 a b^2}{\sqrt [3]{x}}+b^3 \log (x) \]

[Out]

-(a^3/x) - (9*a^2*b)/(2*x^(2/3)) - (9*a*b^2)/x^(1/3) + b^3*Log[x]

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Rubi [A]  time = 0.0190188, antiderivative size = 39, 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} \[ -\frac{9 a^2 b}{2 x^{2/3}}-\frac{a^3}{x}-\frac{9 a b^2}{\sqrt [3]{x}}+b^3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3/x^2,x]

[Out]

-(a^3/x) - (9*a^2*b)/(2*x^(2/3)) - (9*a*b^2)/x^(1/3) + b^3*Log[x]

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

Mathematica [A]  time = 0.0220569, size = 39, normalized size = 1. \[ -\frac{9 a^2 b}{2 x^{2/3}}-\frac{a^3}{x}-\frac{9 a b^2}{\sqrt [3]{x}}+b^3 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3/x^2,x]

[Out]

-(a^3/x) - (9*a^2*b)/(2*x^(2/3)) - (9*a*b^2)/x^(1/3) + b^3*Log[x]

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Maple [A]  time = 0.005, size = 34, normalized size = 0.9 \begin{align*} -{\frac{{a}^{3}}{x}}-{\frac{9\,b{a}^{2}}{2}{x}^{-{\frac{2}{3}}}}-9\,{\frac{{b}^{2}a}{\sqrt [3]{x}}}+{b}^{3}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3/x-9/2*a^2*b/x^(2/3)-9*a*b^2/x^(1/3)+b^3*ln(x)

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Maxima [A]  time = 0.971312, size = 49, normalized size = 1.26 \begin{align*} b^{3} \log \left (x\right ) - \frac{18 \, a b^{2} x^{\frac{2}{3}} + 9 \, a^{2} b x^{\frac{1}{3}} + 2 \, a^{3}}{2 \, x} \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]

b^3*log(x) - 1/2*(18*a*b^2*x^(2/3) + 9*a^2*b*x^(1/3) + 2*a^3)/x

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Fricas [A]  time = 1.53168, size = 101, normalized size = 2.59 \begin{align*} \frac{6 \, b^{3} x \log \left (x^{\frac{1}{3}}\right ) - 18 \, a b^{2} x^{\frac{2}{3}} - 9 \, a^{2} b x^{\frac{1}{3}} - 2 \, a^{3}}{2 \, x} \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]

1/2*(6*b^3*x*log(x^(1/3)) - 18*a*b^2*x^(2/3) - 9*a^2*b*x^(1/3) - 2*a^3)/x

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Sympy [A]  time = 0.836045, size = 36, normalized size = 0.92 \begin{align*} - \frac{a^{3}}{x} - \frac{9 a^{2} b}{2 x^{\frac{2}{3}}} - \frac{9 a b^{2}}{\sqrt [3]{x}} + b^{3} \log{\left (x \right )} \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 - 9*a**2*b/(2*x**(2/3)) - 9*a*b**2/x**(1/3) + b**3*log(x)

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Giac [A]  time = 1.12898, size = 50, normalized size = 1.28 \begin{align*} b^{3} \log \left ({\left | x \right |}\right ) - \frac{18 \, a b^{2} x^{\frac{2}{3}} + 9 \, a^{2} b x^{\frac{1}{3}} + 2 \, a^{3}}{2 \, x} \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]

b^3*log(abs(x)) - 1/2*(18*a*b^2*x^(2/3) + 9*a^2*b*x^(1/3) + 2*a^3)/x