3.2311 \(\int \frac {(a+b \sqrt [3]{x})^3}{x^4} \, dx\)

Optimal. Leaf size=47 \[ -\frac {a^3}{3 x^3}-\frac {9 a^2 b}{8 x^{8/3}}-\frac {9 a b^2}{7 x^{7/3}}-\frac {b^3}{2 x^2} \]

[Out]

-1/3*a^3/x^3-9/8*a^2*b/x^(8/3)-9/7*a*b^2/x^(7/3)-1/2*b^3/x^2

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, 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} \[ -\frac {9 a^2 b}{8 x^{8/3}}-\frac {a^3}{3 x^3}-\frac {9 a b^2}{7 x^{7/3}}-\frac {b^3}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3/x^4,x]

[Out]

-a^3/(3*x^3) - (9*a^2*b)/(8*x^(8/3)) - (9*a*b^2)/(7*x^(7/3)) - b^3/(2*x^2)

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])

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]]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sqrt [3]{x}\right )^3}{x^4} \, dx &=3 \operatorname {Subst}\left (\int \frac {(a+b x)^3}{x^{10}} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (\frac {a^3}{x^{10}}+\frac {3 a^2 b}{x^9}+\frac {3 a b^2}{x^8}+\frac {b^3}{x^7}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {a^3}{3 x^3}-\frac {9 a^2 b}{8 x^{8/3}}-\frac {9 a b^2}{7 x^{7/3}}-\frac {b^3}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.87 \[ -\frac {56 a^3+189 a^2 b \sqrt [3]{x}+216 a b^2 x^{2/3}+84 b^3 x}{168 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3/x^4,x]

[Out]

-1/168*(56*a^3 + 189*a^2*b*x^(1/3) + 216*a*b^2*x^(2/3) + 84*b^3*x)/x^3

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fricas [A]  time = 0.71, size = 35, normalized size = 0.74 \[ -\frac {84 \, b^{3} x + 216 \, a b^{2} x^{\frac {2}{3}} + 189 \, a^{2} b x^{\frac {1}{3}} + 56 \, a^{3}}{168 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3/x^4,x, algorithm="fricas")

[Out]

-1/168*(84*b^3*x + 216*a*b^2*x^(2/3) + 189*a^2*b*x^(1/3) + 56*a^3)/x^3

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giac [A]  time = 0.16, size = 35, normalized size = 0.74 \[ -\frac {84 \, b^{3} x + 216 \, a b^{2} x^{\frac {2}{3}} + 189 \, a^{2} b x^{\frac {1}{3}} + 56 \, a^{3}}{168 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3/x^4,x, algorithm="giac")

[Out]

-1/168*(84*b^3*x + 216*a*b^2*x^(2/3) + 189*a^2*b*x^(1/3) + 56*a^3)/x^3

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maple [A]  time = 0.01, size = 36, normalized size = 0.77 \[ -\frac {b^{3}}{2 x^{2}}-\frac {9 a \,b^{2}}{7 x^{\frac {7}{3}}}-\frac {9 a^{2} b}{8 x^{\frac {8}{3}}}-\frac {a^{3}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/3))^3/x^4,x)

[Out]

-1/3*a^3/x^3-9/8*a^2*b/x^(8/3)-9/7*a*b^2/x^(7/3)-1/2*b^3/x^2

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maxima [A]  time = 0.81, size = 35, normalized size = 0.74 \[ -\frac {84 \, b^{3} x + 216 \, a b^{2} x^{\frac {2}{3}} + 189 \, a^{2} b x^{\frac {1}{3}} + 56 \, a^{3}}{168 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3/x^4,x, algorithm="maxima")

[Out]

-1/168*(84*b^3*x + 216*a*b^2*x^(2/3) + 189*a^2*b*x^(1/3) + 56*a^3)/x^3

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mupad [B]  time = 1.11, size = 35, normalized size = 0.74 \[ -\frac {84\,b^3\,x+56\,a^3+189\,a^2\,b\,x^{1/3}+216\,a\,b^2\,x^{2/3}}{168\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/3))^3/x^4,x)

[Out]

-(84*b^3*x + 56*a^3 + 189*a^2*b*x^(1/3) + 216*a*b^2*x^(2/3))/(168*x^3)

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sympy [A]  time = 1.95, size = 44, normalized size = 0.94 \[ - \frac {a^{3}}{3 x^{3}} - \frac {9 a^{2} b}{8 x^{\frac {8}{3}}} - \frac {9 a b^{2}}{7 x^{\frac {7}{3}}} - \frac {b^{3}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/3))**3/x**4,x)

[Out]

-a**3/(3*x**3) - 9*a**2*b/(8*x**(8/3)) - 9*a*b**2/(7*x**(7/3)) - b**3/(2*x**2)

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