3.2369 \(\int \frac {1}{(a+b \sqrt [3]{x})^2 x} \, dx\)

Optimal. Leaf size=38 \[ -\frac {3 \log \left (a+b \sqrt [3]{x}\right )}{a^2}+\frac {\log (x)}{a^2}+\frac {3}{a \left (a+b \sqrt [3]{x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, 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, 44} \[ -\frac {3 \log \left (a+b \sqrt [3]{x}\right )}{a^2}+\frac {\log (x)}{a^2}+\frac {3}{a \left (a+b \sqrt [3]{x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^(1/3))^2*x),x]

[Out]

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 {1}{\left (a+b \sqrt [3]{x}\right )^2 x} \, dx &=3 \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (\frac {1}{a^2 x}-\frac {b}{a (a+b x)^2}-\frac {b}{a^2 (a+b x)}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {3}{a \left (a+b \sqrt [3]{x}\right )}-\frac {3 \log \left (a+b \sqrt [3]{x}\right )}{a^2}+\frac {\log (x)}{a^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 33, normalized size = 0.87 \[ \frac {\frac {3 a}{a+b \sqrt [3]{x}}-3 \log \left (a+b \sqrt [3]{x}\right )+\log (x)}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^(1/3))^2*x),x]

[Out]

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

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fricas [B]  time = 0.63, size = 70, normalized size = 1.84 \[ \frac {3 \, {\left (a b^{2} x^{\frac {2}{3}} - a^{2} b x^{\frac {1}{3}} + a^{3} - {\left (b^{3} x + a^{3}\right )} \log \left (b x^{\frac {1}{3}} + a\right ) + {\left (b^{3} x + a^{3}\right )} \log \left (x^{\frac {1}{3}}\right )\right )}}{a^{2} b^{3} x + a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^(1/3))^2/x,x, algorithm="fricas")

[Out]

3*(a*b^2*x^(2/3) - a^2*b*x^(1/3) + a^3 - (b^3*x + a^3)*log(b*x^(1/3) + a) + (b^3*x + a^3)*log(x^(1/3)))/(a^2*b
^3*x + a^5)

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giac [A]  time = 0.16, size = 36, normalized size = 0.95 \[ -\frac {3 \, \log \left ({\left | b x^{\frac {1}{3}} + a \right |}\right )}{a^{2}} + \frac {\log \left ({\left | x \right |}\right )}{a^{2}} + \frac {3}{{\left (b x^{\frac {1}{3}} + a\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^(1/3))^2/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 35, normalized size = 0.92 \[ \frac {3}{\left (b \,x^{\frac {1}{3}}+a \right ) a}+\frac {\ln \relax (x )}{a^{2}}-\frac {3 \ln \left (b \,x^{\frac {1}{3}}+a \right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 34, normalized size = 0.89 \[ \frac {3}{a b x^{\frac {1}{3}} + a^{2}} - \frac {3 \, \log \left (b x^{\frac {1}{3}} + a\right )}{a^{2}} + \frac {\log \relax (x)}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^(1/3))^2/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 32, normalized size = 0.84 \[ \frac {3}{a\,\left (a+b\,x^{1/3}\right )}-\frac {6\,\mathrm {atanh}\left (\frac {2\,b\,x^{1/3}}{a}+1\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^(1/3))^2),x)

[Out]

3/(a*(a + b*x^(1/3))) - (6*atanh((2*b*x^(1/3))/a + 1))/a^2

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sympy [A]  time = 1.38, size = 160, normalized size = 4.21 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {2}{3}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {\log {\relax (x )}}{a^{2}} & \text {for}\: b = 0 \\- \frac {3}{2 b^{2} x^{\frac {2}{3}}} & \text {for}\: a = 0 \\\frac {a x^{\frac {2}{3}} \log {\relax (x )}}{a^{3} x^{\frac {2}{3}} + a^{2} b x} - \frac {3 a x^{\frac {2}{3}} \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{a^{3} x^{\frac {2}{3}} + a^{2} b x} + \frac {3 a x^{\frac {2}{3}}}{a^{3} x^{\frac {2}{3}} + a^{2} b x} + \frac {b x \log {\relax (x )}}{a^{3} x^{\frac {2}{3}} + a^{2} b x} - \frac {3 b x \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{a^{3} x^{\frac {2}{3}} + a^{2} b x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(2/3), Eq(a, 0) & Eq(b, 0)), (log(x)/a**2, Eq(b, 0)), (-3/(2*b**2*x**(2/3)), Eq(a, 0)), (a*x
**(2/3)*log(x)/(a**3*x**(2/3) + a**2*b*x) - 3*a*x**(2/3)*log(a/b + x**(1/3))/(a**3*x**(2/3) + a**2*b*x) + 3*a*
x**(2/3)/(a**3*x**(2/3) + a**2*b*x) + b*x*log(x)/(a**3*x**(2/3) + a**2*b*x) - 3*b*x*log(a/b + x**(1/3))/(a**3*
x**(2/3) + a**2*b*x), True))

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