\(\int \frac {1}{(\frac {b}{x^2})^{3/2}} \, dx\) [99]

Optimal result

Integrand size = 9, antiderivative size = 19 \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^3}{4 b \sqrt {\frac {b}{x^2}}} \]

[Out]

1/4*x^3/b/(b/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {15, 30} \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^3}{4 b \sqrt {\frac {b}{x^2}}} \]

[In]

Int[(b/x^2)^(-3/2),x]

[Out]

x^3/(4*b*Sqrt[b/x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int x^3 \, dx}{b \sqrt {\frac {b}{x^2}} x} \\ & = \frac {x^3}{4 b \sqrt {\frac {b}{x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x}{4 \left (\frac {b}{x^2}\right )^{3/2}} \]

[In]

Integrate[(b/x^2)^(-3/2),x]

[Out]

x/(4*(b/x^2)^(3/2))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58

method	result	size
gosper	\(\frac {x}{4 \left (\frac {b}{x^{2}}\right )^{\frac {3}{2}}}\)	\(11\)
default	\(\frac {x}{4 \left (\frac {b}{x^{2}}\right )^{\frac {3}{2}}}\)	\(11\)
risch	\(\frac {x^{3}}{4 b \sqrt {\frac {b}{x^{2}}}}\)	\(16\)
trager	\(\frac {x \left (x^{3}+x^{2}+x +1\right ) \left (-1+x \right ) \sqrt {\frac {b}{x^{2}}}}{4 b^{2}}\)	\(26\)

[In]

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

[Out]

1/4*x/(b/x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^{5} \sqrt {\frac {b}{x^{2}}}}{4 \, b^{2}} \]

[In]

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

[Out]

1/4*x^5*sqrt(b/x^2)/b^2

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x}{4 \left (\frac {b}{x^{2}}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

x/(4*(b/x**2)**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x}{4 \, \left (\frac {b}{x^{2}}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

1/4*x/(b/x^2)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^{4}}{4 \, b^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/4*x^4/(b^(3/2)*sgn(x))

Mupad [B] (verification not implemented)

Time = 5.46 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^5\,\sqrt {\frac {1}{x^2}}}{4\,b^{3/2}} \]

[In]

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

[Out]

(x^5*(1/x^2)^(1/2))/(4*b^(3/2))