\(\int \frac {1}{(a+b \sqrt [3]{x}) x^2} \, dx\) [2361]

Optimal result

Integrand size = 15, antiderivative size = 63 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=-\frac {1}{a x}+\frac {3 b}{2 a^2 x^{2/3}}-\frac {3 b^2}{a^3 \sqrt [3]{x}}+\frac {3 b^3 \log \left (a+b \sqrt [3]{x}\right )}{a^4}-\frac {b^3 \log (x)}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 46} \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=\frac {3 b^3 \log \left (a+b \sqrt [3]{x}\right )}{a^4}-\frac {b^3 \log (x)}{a^4}-\frac {3 b^2}{a^3 \sqrt [3]{x}}+\frac {3 b}{2 a^2 x^{2/3}}-\frac {1}{a x} \]

[In]

Int[1/((a + b*x^(1/3))*x^2),x]

[Out]

-(1/(a*x)) + (3*b)/(2*a^2*x^(2/3)) - (3*b^2)/(a^3*x^(1/3)) + (3*b^3*Log[a + b*x^(1/3)])/a^4 - (b^3*Log[x])/a^4

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=-\frac {2 a^3-3 a^2 b \sqrt [3]{x}+6 a b^2 x^{2/3}-6 b^3 x \log \left (a+b \sqrt [3]{x}\right )+2 b^3 x \log (x)}{2 a^4 x} \]

[In]

Integrate[1/((a + b*x^(1/3))*x^2),x]

[Out]

-1/2*(2*a^3 - 3*a^2*b*x^(1/3) + 6*a*b^2*x^(2/3) - 6*b^3*x*Log[a + b*x^(1/3)] + 2*b^3*x*Log[x])/(a^4*x)

Maple [A] (verified)

Time = 12.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(-\frac {1}{a x}+\frac {3 b}{2 a^{2} x^{\frac {2}{3}}}-\frac {3 b^{2}}{a^{3} x^{\frac {1}{3}}}+\frac {3 b^{3} \ln \left (a +b \,x^{\frac {1}{3}}\right )}{a^{4}}-\frac {b^{3} \ln \left (x \right )}{a^{4}}\)	\(56\)
default	\(-\frac {1}{a x}+\frac {3 b}{2 a^{2} x^{\frac {2}{3}}}-\frac {3 b^{2}}{a^{3} x^{\frac {1}{3}}}+\frac {3 b^{3} \ln \left (a +b \,x^{\frac {1}{3}}\right )}{a^{4}}-\frac {b^{3} \ln \left (x \right )}{a^{4}}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=\frac {6 \, b^{3} x \log \left (b x^{\frac {1}{3}} + a\right ) - 6 \, b^{3} x \log \left (x^{\frac {1}{3}}\right ) - 6 \, a b^{2} x^{\frac {2}{3}} + 3 \, a^{2} b x^{\frac {1}{3}} - 2 \, a^{3}}{2 \, a^{4} x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {4}{3}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {3}{4 b x^{\frac {4}{3}}} & \text {for}\: a = 0 \\- \frac {1}{a x} & \text {for}\: b = 0 \\- \frac {1}{a x} + \frac {3 b}{2 a^{2} x^{\frac {2}{3}}} - \frac {3 b^{2}}{a^{3} \sqrt [3]{x}} - \frac {b^{3} \log {\left (x \right )}}{a^{4}} + \frac {3 b^{3} \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{a^{4}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=\frac {3 \, b^{3} \log \left (b x^{\frac {1}{3}} + a\right )}{a^{4}} - \frac {b^{3} \log \left (x\right )}{a^{4}} - \frac {6 \, b^{2} x^{\frac {2}{3}} - 3 \, a b x^{\frac {1}{3}} + 2 \, a^{2}}{2 \, a^{3} x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=\frac {3 \, b^{3} \log \left ({\left | b x^{\frac {1}{3}} + a \right |}\right )}{a^{4}} - \frac {b^{3} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {6 \, a b^{2} x^{\frac {2}{3}} - 3 \, a^{2} b x^{\frac {1}{3}} + 2 \, a^{3}}{2 \, a^{4} x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right ) x^2} \, dx=\frac {6\,b^3\,\mathrm {atanh}\left (\frac {2\,b\,x^{1/3}}{a}+1\right )}{a^4}-\frac {\frac {1}{a}-\frac {3\,b\,x^{1/3}}{2\,a^2}+\frac {3\,b^2\,x^{2/3}}{a^3}}{x} \]

[In]

int(1/(x^2*(a + b*x^(1/3))),x)

[Out]

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