\(\int \frac {1}{x^2 (b x^2)^{3/2}} \, dx\) [51]

Optimal result

Integrand size = 13, antiderivative size = 19 \[ \int \frac {1}{x^2 \left (b x^2\right )^{3/2}} \, dx=-\frac {1}{4 b x^3 \sqrt {b x^2}} \]

[Out]

-1/4/b/x^3/(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {15, 30} \[ \int \frac {1}{x^2 \left (b x^2\right )^{3/2}} \, dx=-\frac {1}{4 b x^3 \sqrt {b x^2}} \]

[In]

Int[1/(x^2*(b*x^2)^(3/2)),x]

[Out]

-1/4*1/(b*x^3*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^5} \, dx}{b \sqrt {b x^2}} \\ & = -\frac {1}{4 b x^3 \sqrt {b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^2 \left (b x^2\right )^{3/2}} \, dx=-\frac {b x}{4 \left (b x^2\right )^{5/2}} \]

[In]

Integrate[1/(x^2*(b*x^2)^(3/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68

method	result	size
gosper	\(-\frac {1}{4 x \left (b \,x^{2}\right )^{\frac {3}{2}}}\)	\(13\)
default	\(-\frac {1}{4 x \left (b \,x^{2}\right )^{\frac {3}{2}}}\)	\(13\)
risch	\(-\frac {1}{4 b \,x^{3} \sqrt {b \,x^{2}}}\)	\(16\)
trager	\(\frac {\left (-1+x \right ) \left (x^{3}+x^{2}+x +1\right ) \sqrt {b \,x^{2}}}{4 b^{2} x^{5}}\)	\(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^2 \left (b x^2\right )^{3/2}} \, dx=-\frac {\sqrt {b x^{2}}}{4 \, b^{2} x^{5}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^2 \left (b x^2\right )^{3/2}} \, dx=- \frac {1}{4 x \left (b x^{2}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^2 \left (b x^2\right )^{3/2}} \, dx=-\frac {1}{4 \, b^{\frac {3}{2}} x^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^2 \left (b x^2\right )^{3/2}} \, dx=-\frac {1}{4\,b^{3/2}\,x\,{\left (x^2\right )}^{3/2}} \]

[In]

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

[Out]

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