3.2432 \(\int \frac {1}{(a+\frac {b}{\sqrt [3]{x}})^2 x} \, dx\)

Optimal. Leaf size=33 \[ \frac {3 b}{a^2 \left (a \sqrt [3]{x}+b\right )}+\frac {3 \log \left (a \sqrt [3]{x}+b\right )}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {263, 266, 43} \[ \frac {3 b}{a^2 \left (a \sqrt [3]{x}+b\right )}+\frac {3 \log \left (a \sqrt [3]{x}+b\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^(1/3))^2*x),x]

[Out]

(3*b)/(a^2*(b + a*x^(1/3))) + (3*Log[b + a*x^(1/3)])/a^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 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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+\frac {b}{\sqrt [3]{x}}\right )^2 x} \, dx &=\int \frac {1}{\left (b+a \sqrt [3]{x}\right )^2 \sqrt [3]{x}} \, dx\\ &=3 \operatorname {Subst}\left (\int \frac {x}{(b+a x)^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (-\frac {b}{a (b+a x)^2}+\frac {1}{a (b+a x)}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {3 b}{a^2 \left (b+a \sqrt [3]{x}\right )}+\frac {3 \log \left (b+a \sqrt [3]{x}\right )}{a^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 33, normalized size = 1.00 \[ \frac {-\frac {3 a}{a+\frac {b}{\sqrt [3]{x}}}+3 \log \left (a+\frac {b}{\sqrt [3]{x}}\right )+\log (x)}{a^2} \]

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 [A]  time = 0.56, size = 56, normalized size = 1.70 \[ \frac {3 \, {\left (a^{2} b x^{\frac {2}{3}} - a b^{2} x^{\frac {1}{3}} + b^{3} + {\left (a^{3} x + b^{3}\right )} \log \left (a x^{\frac {1}{3}} + b\right )\right )}}{a^{5} x + a^{2} b^{3}} \]

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

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giac [A]  time = 0.16, size = 30, normalized size = 0.91 \[ \frac {3 \, \log \left ({\left | a x^{\frac {1}{3}} + b \right |}\right )}{a^{2}} + \frac {3 \, b}{{\left (a x^{\frac {1}{3}} + b\right )} a^{2}} \]

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

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maple [A]  time = 0.01, size = 30, normalized size = 0.91 \[ \frac {3 b}{\left (a \,x^{\frac {1}{3}}+b \right ) a^{2}}+\frac {3 \ln \left (a \,x^{\frac {1}{3}}+b \right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 34, normalized size = 1.03 \[ -\frac {3}{a^{2} + \frac {a b}{x^{\frac {1}{3}}}} + \frac {3 \, \log \left (a + \frac {b}{x^{\frac {1}{3}}}\right )}{a^{2}} + \frac {\log \relax (x)}{a^{2}} \]

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

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mupad [B]  time = 0.04, size = 29, normalized size = 0.88 \[ \frac {3\,\ln \left (b+a\,x^{1/3}\right )}{a^2}+\frac {3\,b}{a^2\,\left (b+a\,x^{1/3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b/x^(1/3))^2),x)

[Out]

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

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sympy [A]  time = 1.08, size = 99, normalized size = 3.00 \[ \begin {cases} \frac {3 a x \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{a^{3} x + a^{2} b x^{\frac {2}{3}}} + \frac {3 b x^{\frac {2}{3}} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{a^{3} x + a^{2} b x^{\frac {2}{3}}} + \frac {3 b x^{\frac {2}{3}}}{a^{3} x + a^{2} b x^{\frac {2}{3}}} & \text {for}\: a \neq 0 \\\frac {3 x^{\frac {2}{3}}}{2 b^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a*x*log(x**(1/3) + b/a)/(a**3*x + a**2*b*x**(2/3)) + 3*b*x**(2/3)*log(x**(1/3) + b/a)/(a**3*x + a
**2*b*x**(2/3)) + 3*b*x**(2/3)/(a**3*x + a**2*b*x**(2/3)), Ne(a, 0)), (3*x**(2/3)/(2*b**2), True))

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