\(\int \frac {1}{x (b x^2)^{5/2}} \, dx\) [56]

Optimal result

Integrand size = 13, antiderivative size = 19 \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=-\frac {1}{5 b^2 x^4 \sqrt {b x^2}} \]

[Out]

-1/5/b^2/x^4/(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} \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=-\frac {1}{5 b^2 x^4 \sqrt {b x^2}} \]

[In]

Int[1/(x*(b*x^2)^(5/2)),x]

[Out]

-1/5*1/(b^2*x^4*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^6} \, dx}{b^2 \sqrt {b x^2}} \\ & = -\frac {1}{5 b^2 x^4 \sqrt {b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=-\frac {b x^2}{5 \left (b x^2\right )^{7/2}} \]

[In]

Integrate[1/(x*(b*x^2)^(5/2)),x]

[Out]

-1/5*(b*x^2)/(b*x^2)^(7/2)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53

method	result	size
gosper	\(-\frac {1}{5 \left (b \,x^{2}\right )^{\frac {5}{2}}}\)	\(10\)
derivativedivides	\(-\frac {1}{5 \left (b \,x^{2}\right )^{\frac {5}{2}}}\)	\(10\)
default	\(-\frac {1}{5 \left (b \,x^{2}\right )^{\frac {5}{2}}}\)	\(10\)
risch	\(-\frac {1}{5 b^{2} x^{4} \sqrt {b \,x^{2}}}\)	\(16\)
pseudoelliptic	\(-\frac {1}{5 b^{2} x^{4} \sqrt {b \,x^{2}}}\)	\(16\)
trager	\(\frac {\left (-1+x \right ) \left (x^{4}+x^{3}+x^{2}+x +1\right ) \sqrt {b \,x^{2}}}{5 b^{3} x^{6}}\)	\(31\)

[In]

int(1/x/(b*x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=-\frac {\sqrt {b x^{2}}}{5 \, b^{3} x^{6}} \]

[In]

integrate(1/x/(b*x^2)^(5/2),x, algorithm="fricas")

[Out]

-1/5*sqrt(b*x^2)/(b^3*x^6)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=- \frac {1}{5 \left (b x^{2}\right )^{\frac {5}{2}}} \]

[In]

integrate(1/x/(b*x**2)**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=-\frac {1}{5 \, b^{\frac {5}{2}} x^{5}} \]

[In]

integrate(1/x/(b*x^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=-\frac {1}{5 \, b^{\frac {5}{2}} x^{5} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/x/(b*x^2)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x \left (b x^2\right )^{5/2}} \, dx=-\frac {1}{5\,b^{5/2}\,{\left (x^2\right )}^{5/2}} \]

[In]

int(1/(x*(b*x^2)^(5/2)),x)

[Out]

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