3.1934 \(\int \frac{1}{(a+\frac{b}{x^2})^{3/2} x^5} \, dx\)

Optimal. Leaf size=34 \[ -\frac{a}{b^2 \sqrt{a+\frac{b}{x^2}}}-\frac{\sqrt{a+\frac{b}{x^2}}}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0217856, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ -\frac{a}{b^2 \sqrt{a+\frac{b}{x^2}}}-\frac{\sqrt{a+\frac{b}{x^2}}}{b^2} \]

Antiderivative was successfully verified.

[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 266

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]]

Rule 43

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])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x^2}\right )^{3/2} x^5} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(a+b x)^{3/2}} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=-\left (\frac{1}{2} \operatorname{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]  time = 0.0086135, size = 28, normalized size = 0.82 \[ \frac{-2 a x^2-b}{b^2 x^2 \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.003, size = 37, normalized size = 1.1 \begin{align*} -{\frac{ \left ( a{x}^{2}+b \right ) \left ( 2\,a{x}^{2}+b \right ) }{{b}^{2}{x}^{4}} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04155, size = 41, normalized size = 1.21 \begin{align*} -\frac{\sqrt{a + \frac{b}{x^{2}}}}{b^{2}} - \frac{a}{\sqrt{a + \frac{b}{x^{2}}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.4559, size = 76, normalized size = 2.24 \begin{align*} -\frac{{\left (2 \, a x^{2} + b\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a b^{2} x^{2} + b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 2.90804, size = 48, normalized size = 1.41 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)
)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x^{2}}\right )}^{\frac{3}{2}} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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