\(\int \frac {1}{(b x^2)^{3/2}} \, dx\) [50]

Optimal result

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

[Out]

-1/2/b/x/(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} \[ \int \frac {1}{\left (b x^2\right )^{3/2}} \, dx=-\frac {1}{2 b x \sqrt {b x^2}} \]

[In]

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

[Out]

-1/2*1/(b*x*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 {x \int \frac {1}{x^3} \, dx}{b \sqrt {b x^2}} \\ & = -\frac {1}{2 b x \sqrt {b x^2}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

-1/2*x/(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	\(-\frac {x}{2 \left (b \,x^{2}\right )^{\frac {3}{2}}}\)	\(11\)
default	\(-\frac {x}{2 \left (b \,x^{2}\right )^{\frac {3}{2}}}\)	\(11\)
risch	\(-\frac {1}{2 b x \sqrt {b \,x^{2}}}\)	\(16\)
trager	\(\frac {\left (-1+x \right ) \left (1+x \right ) \sqrt {b \,x^{2}}}{2 b^{2} x^{3}}\)	\(22\)

[In]

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

[Out]

-1/2*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 \[ \int \frac {1}{\left (b x^2\right )^{3/2}} \, dx=-\frac {\sqrt {b x^{2}}}{2 \, b^{2} x^{3}} \]

[In]

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

[Out]

-1/2*sqrt(b*x^2)/(b^2*x^3)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\left (b x^2\right )^{3/2}} \, dx=- \frac {x}{2 \left (b x^{2}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (b x^2\right )^{3/2}} \, dx=-\frac {1}{2 \, b^{\frac {3}{2}} x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\left (b x^2\right )^{3/2}} \, dx=-\frac {1}{2 \, b^{\frac {3}{2}} x^{2} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.71 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (b x^2\right )^{3/2}} \, dx=-\frac {1}{2\,b^{3/2}\,x\,\sqrt {x^2}} \]

[In]

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

[Out]

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