3.24.9 \(\int \frac {(a+b \sqrt [3]{x})^3}{x^2} \, dx\) [2309]

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

[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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Rubi [A]
time = 0.01, antiderivative size = 39, 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 = {272, 45} \begin {gather*} -\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 {gather*}

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 45

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 272

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^2} \, dx &=3 \text {Subst}\left (\int \frac {(a+b x)^3}{x^4} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \text {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*}

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Mathematica [A]
time = 0.03, size = 40, normalized size = 1.03 \begin {gather*} -\frac {a \left (2 a^2+9 a b \sqrt [3]{x}+18 b^2 x^{2/3}\right )}{2 x}+b^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3/x^2,x]

[Out]

-1/2*(a*(2*a^2 + 9*a*b*x^(1/3) + 18*b^2*x^(2/3)))/x + b^3*Log[x]

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Maple [A]
time = 0.21, size = 34, normalized size = 0.87

method	result	size
derivativedivides	\(-\frac {a^{3}}{x}-\frac {9 a^{2} b}{2 x^{\frac {2}{3}}}-\frac {9 a \,b^{2}}{x^{\frac {1}{3}}}+b^{3} \ln \left (x \right )\)	\(34\)
default	\(-\frac {a^{3}}{x}-\frac {9 a^{2} b}{2 x^{\frac {2}{3}}}-\frac {9 a \,b^{2}}{x^{\frac {1}{3}}}+b^{3} \ln \left (x \right )\)	\(34\)
trager	\(\frac {a^{3} \left (x -1\right )}{x}-\frac {9 a^{2} b}{2 x^{\frac {2}{3}}}-\frac {9 a \,b^{2}}{x^{\frac {1}{3}}}-b^{3} \ln \left (\frac {1}{x}\right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.30, size = 36, normalized size = 0.92 \begin {gather*} 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 {gather*}

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 = 0.38, size = 39, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 = 0.20, size = 36, normalized size = 0.92 \begin {gather*} - \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 {gather*}

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 = 1.91, size = 37, normalized size = 0.95 \begin {gather*} 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 {gather*}

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

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Mupad [B]
time = 0.04, size = 37, normalized size = 0.95 \begin {gather*} 3\,b^3\,\ln \left (x^{1/3}\right )-\frac {a^3+\frac {9\,a^2\,b\,x^{1/3}}{2}+9\,a\,b^2\,x^{2/3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/3))^3/x^2,x)

[Out]

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

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