3.2310 \(\int \frac{(a+b \sqrt [3]{x})^3}{x^3} \, dx\)

Optimal. Leaf size=45 \[ -\frac{9 a^2 b}{5 x^{5/3}}-\frac{a^3}{2 x^2}-\frac{9 a b^2}{4 x^{4/3}}-\frac{b^3}{x} \]

[Out]

-a^3/(2*x^2) - (9*a^2*b)/(5*x^(5/3)) - (9*a*b^2)/(4*x^(4/3)) - b^3/x

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Rubi [A]  time = 0.0191243, antiderivative size = 45, 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} \[ -\frac{9 a^2 b}{5 x^{5/3}}-\frac{a^3}{2 x^2}-\frac{9 a b^2}{4 x^{4/3}}-\frac{b^3}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3/x^3,x]

[Out]

-a^3/(2*x^2) - (9*a^2*b)/(5*x^(5/3)) - (9*a*b^2)/(4*x^(4/3)) - b^3/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^3} \, dx &=3 \operatorname{Subst}\left (\int \frac{(a+b x)^3}{x^7} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (\frac{a^3}{x^7}+\frac{3 a^2 b}{x^6}+\frac{3 a b^2}{x^5}+\frac{b^3}{x^4}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{a^3}{2 x^2}-\frac{9 a^2 b}{5 x^{5/3}}-\frac{9 a b^2}{4 x^{4/3}}-\frac{b^3}{x}\\ \end{align*}

Mathematica [A]  time = 0.0157946, size = 41, normalized size = 0.91 \[ -\frac{36 a^2 b \sqrt [3]{x}+10 a^3+45 a b^2 x^{2/3}+20 b^3 x}{20 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3/x^3,x]

[Out]

-(10*a^3 + 36*a^2*b*x^(1/3) + 45*a*b^2*x^(2/3) + 20*b^3*x)/(20*x^2)

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Maple [A]  time = 0.005, size = 36, normalized size = 0.8 \begin{align*} -{\frac{{a}^{3}}{2\,{x}^{2}}}-{\frac{9\,b{a}^{2}}{5}{x}^{-{\frac{5}{3}}}}-{\frac{9\,{b}^{2}a}{4}{x}^{-{\frac{4}{3}}}}-{\frac{{b}^{3}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^3/x^2-9/5*a^2*b/x^(5/3)-9/4*a*b^2/x^(4/3)-b^3/x

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Maxima [A]  time = 0.963843, size = 47, normalized size = 1.04 \begin{align*} -\frac{20 \, b^{3} x + 45 \, a b^{2} x^{\frac{2}{3}} + 36 \, a^{2} b x^{\frac{1}{3}} + 10 \, a^{3}}{20 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(20*b^3*x + 45*a*b^2*x^(2/3) + 36*a^2*b*x^(1/3) + 10*a^3)/x^2

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Fricas [A]  time = 1.42144, size = 93, normalized size = 2.07 \begin{align*} -\frac{20 \, b^{3} x + 45 \, a b^{2} x^{\frac{2}{3}} + 36 \, a^{2} b x^{\frac{1}{3}} + 10 \, a^{3}}{20 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(20*b^3*x + 45*a*b^2*x^(2/3) + 36*a^2*b*x^(1/3) + 10*a^3)/x^2

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Sympy [A]  time = 1.40515, size = 41, normalized size = 0.91 \begin{align*} - \frac{a^{3}}{2 x^{2}} - \frac{9 a^{2} b}{5 x^{\frac{5}{3}}} - \frac{9 a b^{2}}{4 x^{\frac{4}{3}}} - \frac{b^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**3/(2*x**2) - 9*a**2*b/(5*x**(5/3)) - 9*a*b**2/(4*x**(4/3)) - b**3/x

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Giac [A]  time = 1.10696, size = 47, normalized size = 1.04 \begin{align*} -\frac{20 \, b^{3} x + 45 \, a b^{2} x^{\frac{2}{3}} + 36 \, a^{2} b x^{\frac{1}{3}} + 10 \, a^{3}}{20 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(20*b^3*x + 45*a*b^2*x^(2/3) + 36*a^2*b*x^(1/3) + 10*a^3)/x^2