\(\int \frac {1}{(a+\frac {b}{\sqrt [3]{x}})^2 x^2} \, dx\) [2433]

Optimal result

Integrand size = 15, antiderivative size = 52 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=-\frac {3 a}{b^2 \left (b+a \sqrt [3]{x}\right )}-\frac {3}{b^2 \sqrt [3]{x}}+\frac {6 a \log \left (b+a \sqrt [3]{x}\right )}{b^3}-\frac {2 a \log (x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {269, 272, 46} \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=\frac {6 a \log \left (a \sqrt [3]{x}+b\right )}{b^3}-\frac {2 a \log (x)}{b^3}-\frac {3 a}{b^2 \left (a \sqrt [3]{x}+b\right )}-\frac {3}{b^2 \sqrt [3]{x}} \]

[In]

Int[1/((a + b/x^(1/3))^2*x^2),x]

[Out]

(-3*a)/(b^2*(b + a*x^(1/3))) - 3/(b^2*x^(1/3)) + (6*a*Log[b + a*x^(1/3)])/b^3 - (2*a*Log[x])/b^3

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 269

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 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*} \text {integral}& = \int \frac {1}{\left (b+a \sqrt [3]{x}\right )^2 x^{4/3}} \, dx \\ & = 3 \text {Subst}\left (\int \frac {1}{x^2 (b+a x)^2} \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (\frac {1}{b^2 x^2}-\frac {2 a}{b^3 x}+\frac {a^2}{b^2 (b+a x)^2}+\frac {2 a^2}{b^3 (b+a x)}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = -\frac {3 a}{b^2 \left (b+a \sqrt [3]{x}\right )}-\frac {3}{b^2 \sqrt [3]{x}}+\frac {6 a \log \left (b+a \sqrt [3]{x}\right )}{b^3}-\frac {2 a \log (x)}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=\frac {3 \left (-\frac {b \left (2 a+\frac {b}{\sqrt [3]{x}}\right )}{b+a \sqrt [3]{x}}+2 a \log \left (b+a \sqrt [3]{x}\right )-\frac {2}{3} a \log (x)\right )}{b^3} \]

[In]

Integrate[1/((a + b/x^(1/3))^2*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 12.91 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(-\frac {3 a}{b^{2} \left (b +a \,x^{\frac {1}{3}}\right )}-\frac {3}{x^{\frac {1}{3}} b^{2}}+\frac {6 a \ln \left (b +a \,x^{\frac {1}{3}}\right )}{b^{3}}-\frac {2 a \ln \left (x \right )}{b^{3}}\)	\(47\)
default	\(-\frac {3 a}{b^{2} \left (b +a \,x^{\frac {1}{3}}\right )}-\frac {3}{x^{\frac {1}{3}} b^{2}}+\frac {6 a \ln \left (b +a \,x^{\frac {1}{3}}\right )}{b^{3}}-\frac {2 a \ln \left (x \right )}{b^{3}}\)	\(47\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (46) = 92\).

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=\frac {3 \, {\left (a^{2} b^{2} x^{\frac {4}{3}} - a b^{3} x + 2 \, {\left (a^{4} x^{2} + a b^{3} x\right )} \log \left (a x^{\frac {1}{3}} + b\right ) - 2 \, {\left (a^{4} x^{2} + a b^{3} x\right )} \log \left (x^{\frac {1}{3}}\right ) - {\left (2 \, a^{3} b x + b^{4}\right )} x^{\frac {2}{3}}\right )}}{a^{3} b^{3} x^{2} + b^{6} x} \]

[In]

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

[Out]

3*(a^2*b^2*x^(4/3) - a*b^3*x + 2*(a^4*x^2 + a*b^3*x)*log(a*x^(1/3) + b) - 2*(a^4*x^2 + a*b^3*x)*log(x^(1/3)) -
(2*a^3*b*x + b^4)*x^(2/3))/(a^3*b^3*x^2 + b^6*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (51) = 102\).

Time = 0.79 (sec) , antiderivative size = 211, normalized size of antiderivative = 4.06 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=\begin {cases} \frac {\tilde {\infty }}{\sqrt [3]{x}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {3}{b^{2} \sqrt [3]{x}} & \text {for}\: a = 0 \\- \frac {1}{a^{2} x} & \text {for}\: b = 0 \\- \frac {2 a^{2} x^{2} \log {\left (x \right )}}{a b^{3} x^{2} + b^{4} x^{\frac {5}{3}}} + \frac {6 a^{2} x^{2} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{a b^{3} x^{2} + b^{4} x^{\frac {5}{3}}} - \frac {2 a b x^{\frac {5}{3}} \log {\left (x \right )}}{a b^{3} x^{2} + b^{4} x^{\frac {5}{3}}} + \frac {6 a b x^{\frac {5}{3}} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{a b^{3} x^{2} + b^{4} x^{\frac {5}{3}}} - \frac {6 a b x^{\frac {5}{3}}}{a b^{3} x^{2} + b^{4} x^{\frac {5}{3}}} - \frac {3 b^{2} x^{\frac {4}{3}}}{a b^{3} x^{2} + b^{4} x^{\frac {5}{3}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo/x**(1/3), Eq(a, 0) & Eq(b, 0)), (-3/(b**2*x**(1/3)), Eq(a, 0)), (-1/(a**2*x), Eq(b, 0)), (-2*a*
*2*x**2*log(x)/(a*b**3*x**2 + b**4*x**(5/3)) + 6*a**2*x**2*log(x**(1/3) + b/a)/(a*b**3*x**2 + b**4*x**(5/3)) -
2*a*b*x**(5/3)*log(x)/(a*b**3*x**2 + b**4*x**(5/3)) + 6*a*b*x**(5/3)*log(x**(1/3) + b/a)/(a*b**3*x**2 + b**4*
x**(5/3)) - 6*a*b*x**(5/3)/(a*b**3*x**2 + b**4*x**(5/3)) - 3*b**2*x**(4/3)/(a*b**3*x**2 + b**4*x**(5/3)), True
))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=\frac {6 \, a \log \left (a + \frac {b}{x^{\frac {1}{3}}}\right )}{b^{3}} - \frac {3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}}{b^{3}} + \frac {3 \, a^{2}}{{\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=\frac {6 \, a \log \left ({\left | a x^{\frac {1}{3}} + b \right |}\right )}{b^{3}} - \frac {2 \, a \log \left ({\left | x \right |}\right )}{b^{3}} - \frac {3 \, {\left (2 \, a x^{\frac {1}{3}} + b\right )}}{{\left (a x^{\frac {2}{3}} + b x^{\frac {1}{3}}\right )} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^2} \, dx=\frac {12\,a\,\mathrm {atanh}\left (\frac {2\,a\,x^{1/3}}{b}+1\right )}{b^3}-\frac {\frac {3}{b}+\frac {6\,a\,x^{1/3}}{b^2}}{a\,x^{2/3}+b\,x^{1/3}} \]

[In]

int(1/(x^2*(a + b/x^(1/3))^2),x)

[Out]

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