\(\int \frac {(a+\frac {b}{x})^{3/2}}{x^2} \, dx\) [1706]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=-\frac {2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {267} \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=-\frac {2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b} \]

[In]

Int[(a + b/x)^(3/2)/x^2,x]

[Out]

(-2*(a + b/x)^(5/2))/(5*b)

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 {2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=-\frac {2 \left (\frac {b+a x}{x}\right )^{5/2}}{5 b} \]

[In]

Integrate[(a + b/x)^(3/2)/x^2,x]

[Out]

(-2*((b + a*x)/x)^(5/2))/(5*b)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method result size
derivativedivides \(-\frac {2 \left (a +\frac {b}{x}\right )^{\frac {5}{2}}}{5 b}\) \(15\)
gosper \(-\frac {2 \left (a x +b \right ) \left (\frac {a x +b}{x}\right )^{\frac {3}{2}}}{5 x b}\) \(25\)
risch \(-\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (a^{2} x^{2}+2 a b x +b^{2}\right )}{5 x^{2} b}\) \(36\)
trager \(-\frac {2 \left (a^{2} x^{2}+2 a b x +b^{2}\right ) \sqrt {-\frac {-a x -b}{x}}}{5 x^{2} b}\) \(40\)
default \(-\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \left (a x +b \right )}{5 x^{3} b \sqrt {x \left (a x +b \right )}}\) \(45\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=-\frac {2 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {\frac {a x + b}{x}}}{5 \, b x^{2}} \]

[In]

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

[Out]

-2/5*(a^2*x^2 + 2*a*b*x + b^2)*sqrt((a*x + b)/x)/(b*x^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (14) = 28\).

Time = 0.57 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.61 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=- \frac {2 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}}{5 b} - \frac {4 a^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}}{5 x} - \frac {2 \sqrt {a} b \sqrt {1 + \frac {b}{a x}}}{5 x^{2}} \]

[In]

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

[Out]

-2*a**(5/2)*sqrt(1 + b/(a*x))/(5*b) - 4*a**(3/2)*sqrt(1 + b/(a*x))/(5*x) - 2*sqrt(a)*b*sqrt(1 + b/(a*x))/(5*x*
*2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=-\frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}}}{5 \, b} \]

[In]

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

[Out]

-2/5*(a + b/x)^(5/2)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (14) = 28\).

Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 8.06 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=\frac {2 \, {\left (5 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{4} a^{2} \mathrm {sgn}\left (x\right ) + 10 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{3} a^{\frac {3}{2}} b \mathrm {sgn}\left (x\right ) + 10 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{2} a b^{2} \mathrm {sgn}\left (x\right ) + 5 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} b^{3} \mathrm {sgn}\left (x\right ) + b^{4} \mathrm {sgn}\left (x\right )\right )}}{5 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{5}} \]

[In]

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

[Out]

2/5*(5*(sqrt(a)*x - sqrt(a*x^2 + b*x))^4*a^2*sgn(x) + 10*(sqrt(a)*x - sqrt(a*x^2 + b*x))^3*a^(3/2)*b*sgn(x) +
10*(sqrt(a)*x - sqrt(a*x^2 + b*x))^2*a*b^2*sgn(x) + 5*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a)*b^3*sgn(x) + b^4
*sgn(x))/(sqrt(a)*x - sqrt(a*x^2 + b*x))^5

Mupad [B] (verification not implemented)

Time = 6.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^2} \, dx=-\frac {2\,\sqrt {a+\frac {b}{x}}\,{\left (b+a\,x\right )}^2}{5\,b\,x^2} \]

[In]

int((a + b/x)^(3/2)/x^2,x)

[Out]

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