\(\int \frac {x}{(b x^2)^{5/2}} \, dx\) [55]

Optimal result

Integrand size = 11, antiderivative size = 19 \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=-\frac {1}{3 b^2 x^2 \sqrt {b x^2}} \]

[Out]

-1/3/b^2/x^2/(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 30} \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=-\frac {1}{3 b^2 x^2 \sqrt {b x^2}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=-\frac {x^2}{3 \left (b x^2\right )^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68

method	result	size
gosper	\(-\frac {x^{2}}{3 \left (b \,x^{2}\right )^{\frac {5}{2}}}\)	\(13\)
derivativedivides	\(-\frac {1}{3 \left (b \,x^{2}\right )^{\frac {3}{2}} b}\)	\(13\)
default	\(-\frac {x^{2}}{3 \left (b \,x^{2}\right )^{\frac {5}{2}}}\)	\(13\)
risch	\(-\frac {1}{3 b^{2} x^{2} \sqrt {b \,x^{2}}}\)	\(16\)
pseudoelliptic	\(-\frac {1}{3 b^{2} x^{2} \sqrt {b \,x^{2}}}\)	\(16\)
trager	\(\frac {\left (-1+x \right ) \left (x^{2}+x +1\right ) \sqrt {b \,x^{2}}}{3 b^{3} x^{4}}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=-\frac {\sqrt {b x^{2}}}{3 \, b^{3} x^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=- \frac {x^{2}}{3 \left (b x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

-x**2/(3*(b*x**2)**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=-\frac {1}{3 \, \left (b x^{2}\right )^{\frac {3}{2}} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=-\frac {1}{3 \, b^{\frac {5}{2}} x^{3} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.76 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {x}{\left (b x^2\right )^{5/2}} \, dx=-\frac {1}{3\,b^{5/2}\,{\left (x^2\right )}^{3/2}} \]

[In]

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

[Out]

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