\(\int \frac {1}{(a+\frac {b}{x})^3 x^2} \, dx\) [1639]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=\frac {1}{2 b \left (a+\frac {b}{x}\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=\frac {1}{2 b \left (a+\frac {b}{x}\right )^2} \]

[In]

Int[1/((a + b/x)^3*x^2),x]

[Out]

1/(2*b*(a + b/x)^2)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 b \left (a+\frac {b}{x}\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=-\frac {b+2 a x}{2 a^2 (b+a x)^2} \]

[In]

Integrate[1/((a + b/x)^3*x^2),x]

[Out]

-1/2*(b + 2*a*x)/(a^2*(b + a*x)^2)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {1}{2 b \left (a +\frac {b}{x}\right )^{2}}\)	\(15\)
gosper	\(-\frac {2 a x +b}{2 \left (a x +b \right )^{2} a^{2}}\)	\(19\)
parallelrisch	\(\frac {-2 a x -b}{2 a^{2} \left (a x +b \right )^{2}}\)	\(21\)
risch	\(\frac {-\frac {x}{a}-\frac {b}{2 a^{2}}}{\left (a x +b \right )^{2}}\)	\(22\)
default	\(-\frac {1}{\left (a x +b \right ) a^{2}}+\frac {b}{2 a^{2} \left (a x +b \right )^{2}}\)	\(27\)
norman	\(\frac {-\frac {x^{2}}{a}-\frac {b x}{2 a^{2}}}{x \left (a x +b \right )^{2}}\)	\(28\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (14) = 28\).

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=-\frac {2 \, a x + b}{2 \, {\left (a^{4} x^{2} + 2 \, a^{3} b x + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (10) = 20\).

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=\frac {- 2 a x - b}{2 a^{4} x^{2} + 4 a^{3} b x + 2 a^{2} b^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=\frac {1}{2 \, {\left (a + \frac {b}{x}\right )}^{2} b} \]

[In]

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

[Out]

1/2/((a + b/x)^2*b)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=\frac {1}{2 \, {\left (a + \frac {b}{x}\right )}^{2} b} \]

[In]

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

[Out]

1/2/((a + b/x)^2*b)

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^2} \, dx=-\frac {\frac {b}{2\,a^2}+\frac {x}{a}}{a^2\,x^2+2\,a\,b\,x+b^2} \]

[In]

int(1/(x^2*(a + b/x)^3),x)

[Out]

-(b/(2*a^2) + x/a)/(b^2 + a^2*x^2 + 2*a*b*x)