\(\int \frac {1}{(a+\frac {b}{x^2})^{3/2} x^5} \, dx\) [1934]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=-\frac {a}{b^2 \sqrt {a+\frac {b}{x^2}}}-\frac {\sqrt {a+\frac {b}{x^2}}}{b^2} \]

[Out]

-a/b^2/(a+b/x^2)^(1/2)-(a+b/x^2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=-\frac {a}{b^2 \sqrt {a+\frac {b}{x^2}}}-\frac {\sqrt {a+\frac {b}{x^2}}}{b^2} \]

[In]

Int[1/((a + b/x^2)^(3/2)*x^5),x]

[Out]

-(a/(b^2*Sqrt[a + b/x^2])) - Sqrt[a + b/x^2]/b^2

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {x}{(a+b x)^{3/2}} \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^{3/2}}+\frac {1}{b \sqrt {a+b x}}\right ) \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = -\frac {a}{b^2 \sqrt {a+\frac {b}{x^2}}}-\frac {\sqrt {a+\frac {b}{x^2}}}{b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=\frac {-b-2 a x^2}{b^2 \sqrt {a+\frac {b}{x^2}} x^2} \]

[In]

Integrate[1/((a + b/x^2)^(3/2)*x^5),x]

[Out]

(-b - 2*a*x^2)/(b^2*Sqrt[a + b/x^2]*x^2)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09

method	result	size
gosper	\(-\frac {\left (a \,x^{2}+b \right ) \left (2 a \,x^{2}+b \right )}{x^{4} b^{2} \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}}}\)	\(37\)
default	\(-\frac {\left (a \,x^{2}+b \right ) \left (2 a \,x^{2}+b \right )}{x^{4} b^{2} \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}}}\)	\(37\)
trager	\(-\frac {\left (2 a \,x^{2}+b \right ) \sqrt {-\frac {-a \,x^{2}-b}{x^{2}}}}{b^{2} \left (a \,x^{2}+b \right )}\)	\(40\)
risch	\(-\frac {a \,x^{2}+b}{b^{2} x^{2} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}-\frac {a}{b^{2} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}\)	\(49\)

[In]

int(1/(a+b/x^2)^(3/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

-(a*x^2+b)*(2*a*x^2+b)/x^4/b^2/((a*x^2+b)/x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=-\frac {{\left (2 \, a x^{2} + b\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a b^{2} x^{2} + b^{3}} \]

[In]

integrate(1/(a+b/x^2)^(3/2)/x^5,x, algorithm="fricas")

[Out]

-(2*a*x^2 + b)*sqrt((a*x^2 + b)/x^2)/(a*b^2*x^2 + b^3)

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=\begin {cases} - \frac {2 a}{b^{2} \sqrt {a + \frac {b}{x^{2}}}} - \frac {1}{b x^{2} \sqrt {a + \frac {b}{x^{2}}}} & \text {for}\: b \neq 0 \\- \frac {1}{4 a^{\frac {3}{2}} x^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a+b/x**2)**(3/2)/x**5,x)

[Out]

Piecewise((-2*a/(b**2*sqrt(a + b/x**2)) - 1/(b*x**2*sqrt(a + b/x**2)), Ne(b, 0)), (-1/(4*a**(3/2)*x**4), True)
)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=-\frac {\sqrt {a + \frac {b}{x^{2}}}}{b^{2}} - \frac {a}{\sqrt {a + \frac {b}{x^{2}}} b^{2}} \]

[In]

integrate(1/(a+b/x^2)^(3/2)/x^5,x, algorithm="maxima")

[Out]

-sqrt(a + b/x^2)/b^2 - a/(sqrt(a + b/x^2)*b^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=-\frac {a x}{\sqrt {a x^{2} + b} b^{2} \mathrm {sgn}\left (x\right )} + \frac {2 \, \sqrt {a}}{{\left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} - b\right )} b \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/(a+b/x^2)^(3/2)/x^5,x, algorithm="giac")

[Out]

-a*x/(sqrt(a*x^2 + b)*b^2*sgn(x)) + 2*sqrt(a)/(((sqrt(a)*x - sqrt(a*x^2 + b))^2 - b)*b*sgn(x))

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^5} \, dx=-\frac {x\,\sqrt {a+\frac {b}{x^2}}\,\left (\frac {1}{b}+\frac {2\,a\,x^2}{b^2}\right )}{a\,x^3+b\,x} \]

[In]

int(1/(x^5*(a + b/x^2)^(3/2)),x)

[Out]

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