3.1950 \(\int \frac{1}{(a+\frac{b}{x^2})^{5/2} x^9} \, dx\)

Optimal. Leaf size=76 \[ -\frac{a^3}{3 b^4 \left (a+\frac{b}{x^2}\right )^{3/2}}+\frac{3 a^2}{b^4 \sqrt{a+\frac{b}{x^2}}}+\frac{3 a \sqrt{a+\frac{b}{x^2}}}{b^4}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{3 b^4} \]

[Out]

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

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Rubi [A]  time = 0.0404859, antiderivative size = 76, 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} \[ -\frac{a^3}{3 b^4 \left (a+\frac{b}{x^2}\right )^{3/2}}+\frac{3 a^2}{b^4 \sqrt{a+\frac{b}{x^2}}}+\frac{3 a \sqrt{a+\frac{b}{x^2}}}{b^4}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{3 b^4} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(5/2)*x^9),x]

[Out]

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

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 )^{5/2} x^9} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3}{(a+b x)^{5/2}} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{a^3}{b^3 (a+b x)^{5/2}}+\frac{3 a^2}{b^3 (a+b x)^{3/2}}-\frac{3 a}{b^3 \sqrt{a+b x}}+\frac{\sqrt{a+b x}}{b^3}\right ) \, dx,x,\frac{1}{x^2}\right )\right )\\ &=-\frac{a^3}{3 b^4 \left (a+\frac{b}{x^2}\right )^{3/2}}+\frac{3 a^2}{b^4 \sqrt{a+\frac{b}{x^2}}}+\frac{3 a \sqrt{a+\frac{b}{x^2}}}{b^4}-\frac{\left (a+\frac{b}{x^2}\right )^{3/2}}{3 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0127014, size = 62, normalized size = 0.82 \[ \frac{24 a^2 b x^4+16 a^3 x^6+6 a b^2 x^2-b^3}{3 b^4 x^4 \sqrt{a+\frac{b}{x^2}} \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(5/2)*x^9),x]

[Out]

(-b^3 + 6*a*b^2*x^2 + 24*a^2*b*x^4 + 16*a^3*x^6)/(3*b^4*Sqrt[a + b/x^2]*x^4*(b + a*x^2))

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Maple [A]  time = 0.003, size = 61, normalized size = 0.8 \begin{align*}{\frac{ \left ( a{x}^{2}+b \right ) \left ( 16\,{a}^{3}{x}^{6}+24\,{a}^{2}b{x}^{4}+6\,a{b}^{2}{x}^{2}-{b}^{3} \right ) }{3\,{x}^{8}{b}^{4}} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x^2*b)^(5/2)/x^9,x)

[Out]

1/3*(a*x^2+b)*(16*a^3*x^6+24*a^2*b*x^4+6*a*b^2*x^2-b^3)/x^8/b^4/((a*x^2+b)/x^2)^(5/2)

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Maxima [A]  time = 0.979099, size = 86, normalized size = 1.13 \begin{align*} -\frac{{\left (a + \frac{b}{x^{2}}\right )}^{\frac{3}{2}}}{3 \, b^{4}} + \frac{3 \, \sqrt{a + \frac{b}{x^{2}}} a}{b^{4}} + \frac{3 \, a^{2}}{\sqrt{a + \frac{b}{x^{2}}} b^{4}} - \frac{a^{3}}{3 \,{\left (a + \frac{b}{x^{2}}\right )}^{\frac{3}{2}} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^9,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59622, size = 153, normalized size = 2.01 \begin{align*} \frac{{\left (16 \, a^{3} x^{6} + 24 \, a^{2} b x^{4} + 6 \, a b^{2} x^{2} - b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{3 \,{\left (a^{2} b^{4} x^{6} + 2 \, a b^{5} x^{4} + b^{6} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^9,x, algorithm="fricas")

[Out]

1/3*(16*a^3*x^6 + 24*a^2*b*x^4 + 6*a*b^2*x^2 - b^3)*sqrt((a*x^2 + b)/x^2)/(a^2*b^4*x^6 + 2*a*b^5*x^4 + b^6*x^2
)

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Sympy [A]  time = 15.9793, size = 201, normalized size = 2.64 \begin{align*} \begin{cases} \frac{16 a^{3} x^{6}}{3 a b^{4} x^{6} \sqrt{a + \frac{b}{x^{2}}} + 3 b^{5} x^{4} \sqrt{a + \frac{b}{x^{2}}}} + \frac{24 a^{2} b x^{4}}{3 a b^{4} x^{6} \sqrt{a + \frac{b}{x^{2}}} + 3 b^{5} x^{4} \sqrt{a + \frac{b}{x^{2}}}} + \frac{6 a b^{2} x^{2}}{3 a b^{4} x^{6} \sqrt{a + \frac{b}{x^{2}}} + 3 b^{5} x^{4} \sqrt{a + \frac{b}{x^{2}}}} - \frac{b^{3}}{3 a b^{4} x^{6} \sqrt{a + \frac{b}{x^{2}}} + 3 b^{5} x^{4} \sqrt{a + \frac{b}{x^{2}}}} & \text{for}\: b \neq 0 \\- \frac{1}{8 a^{\frac{5}{2}} x^{8}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(5/2)/x**9,x)

[Out]

Piecewise((16*a**3*x**6/(3*a*b**4*x**6*sqrt(a + b/x**2) + 3*b**5*x**4*sqrt(a + b/x**2)) + 24*a**2*b*x**4/(3*a*
b**4*x**6*sqrt(a + b/x**2) + 3*b**5*x**4*sqrt(a + b/x**2)) + 6*a*b**2*x**2/(3*a*b**4*x**6*sqrt(a + b/x**2) + 3
*b**5*x**4*sqrt(a + b/x**2)) - b**3/(3*a*b**4*x**6*sqrt(a + b/x**2) + 3*b**5*x**4*sqrt(a + b/x**2)), Ne(b, 0))
, (-1/(8*a**(5/2)*x**8), 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{5}{2}} x^{9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^9,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^2)^(5/2)*x^9), x)