3.564 \(\int \frac{1}{x^5 \sqrt{-9+4 x^2}} \, dx\)

Optimal. Leaf size=57 \[ \frac{\sqrt{4 x^2-9}}{54 x^2}+\frac{\sqrt{4 x^2-9}}{36 x^4}+\frac{2}{81} \tan ^{-1}\left (\frac{1}{3} \sqrt{4 x^2-9}\right ) \]

[Out]

Sqrt[-9 + 4*x^2]/(36*x^4) + Sqrt[-9 + 4*x^2]/(54*x^2) + (2*ArcTan[Sqrt[-9 + 4*x^2]/3])/81

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Rubi [A]  time = 0.0232267, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 203} \[ \frac{\sqrt{4 x^2-9}}{54 x^2}+\frac{\sqrt{4 x^2-9}}{36 x^4}+\frac{2}{81} \tan ^{-1}\left (\frac{1}{3} \sqrt{4 x^2-9}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[-9 + 4*x^2]),x]

[Out]

Sqrt[-9 + 4*x^2]/(36*x^4) + Sqrt[-9 + 4*x^2]/(54*x^2) + (2*ArcTan[Sqrt[-9 + 4*x^2]/3])/81

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^5 \sqrt{-9+4 x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{-9+4 x}} \, dx,x,x^2\right )\\ &=\frac{\sqrt{-9+4 x^2}}{36 x^4}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{-9+4 x}} \, dx,x,x^2\right )\\ &=\frac{\sqrt{-9+4 x^2}}{36 x^4}+\frac{\sqrt{-9+4 x^2}}{54 x^2}+\frac{1}{27} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{-9+4 x}} \, dx,x,x^2\right )\\ &=\frac{\sqrt{-9+4 x^2}}{36 x^4}+\frac{\sqrt{-9+4 x^2}}{54 x^2}+\frac{1}{54} \operatorname{Subst}\left (\int \frac{1}{\frac{9}{4}+\frac{x^2}{4}} \, dx,x,\sqrt{-9+4 x^2}\right )\\ &=\frac{\sqrt{-9+4 x^2}}{36 x^4}+\frac{\sqrt{-9+4 x^2}}{54 x^2}+\frac{2}{81} \tan ^{-1}\left (\frac{1}{3} \sqrt{-9+4 x^2}\right )\\ \end{align*}

Mathematica [C]  time = 0.0047386, size = 32, normalized size = 0.56 \[ \frac{16}{729} \sqrt{4 x^2-9} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};1-\frac{4 x^2}{9}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[-9 + 4*x^2]),x]

[Out]

(16*Sqrt[-9 + 4*x^2]*Hypergeometric2F1[1/2, 3, 3/2, 1 - (4*x^2)/9])/729

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Maple [A]  time = 0.004, size = 44, normalized size = 0.8 \begin{align*}{\frac{1}{36\,{x}^{4}}\sqrt{4\,{x}^{2}-9}}+{\frac{1}{54\,{x}^{2}}\sqrt{4\,{x}^{2}-9}}-{\frac{2}{81}\arctan \left ( 3\,{\frac{1}{\sqrt{4\,{x}^{2}-9}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(4*x^2-9)^(1/2),x)

[Out]

1/36*(4*x^2-9)^(1/2)/x^4+1/54*(4*x^2-9)^(1/2)/x^2-2/81*arctan(3/(4*x^2-9)^(1/2))

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Maxima [A]  time = 3.16137, size = 51, normalized size = 0.89 \begin{align*} \frac{\sqrt{4 \, x^{2} - 9}}{54 \, x^{2}} + \frac{\sqrt{4 \, x^{2} - 9}}{36 \, x^{4}} - \frac{2}{81} \, \arcsin \left (\frac{3}{2 \,{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(4*x^2-9)^(1/2),x, algorithm="maxima")

[Out]

1/54*sqrt(4*x^2 - 9)/x^2 + 1/36*sqrt(4*x^2 - 9)/x^4 - 2/81*arcsin(3/2/abs(x))

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Fricas [A]  time = 1.25057, size = 120, normalized size = 2.11 \begin{align*} \frac{16 \, x^{4} \arctan \left (-\frac{2}{3} \, x + \frac{1}{3} \, \sqrt{4 \, x^{2} - 9}\right ) + 3 \, \sqrt{4 \, x^{2} - 9}{\left (2 \, x^{2} + 3\right )}}{324 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(4*x^2-9)^(1/2),x, algorithm="fricas")

[Out]

1/324*(16*x^4*arctan(-2/3*x + 1/3*sqrt(4*x^2 - 9)) + 3*sqrt(4*x^2 - 9)*(2*x^2 + 3))/x^4

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Sympy [A]  time = 4.39706, size = 136, normalized size = 2.39 \begin{align*} \begin{cases} \frac{2 i \operatorname{acosh}{\left (\frac{3}{2 x} \right )}}{81} - \frac{i}{27 x \sqrt{-1 + \frac{9}{4 x^{2}}}} + \frac{i}{36 x^{3} \sqrt{-1 + \frac{9}{4 x^{2}}}} + \frac{i}{8 x^{5} \sqrt{-1 + \frac{9}{4 x^{2}}}} & \text{for}\: \frac{9}{4 \left |{x^{2}}\right |} > 1 \\- \frac{2 \operatorname{asin}{\left (\frac{3}{2 x} \right )}}{81} + \frac{1}{27 x \sqrt{1 - \frac{9}{4 x^{2}}}} - \frac{1}{36 x^{3} \sqrt{1 - \frac{9}{4 x^{2}}}} - \frac{1}{8 x^{5} \sqrt{1 - \frac{9}{4 x^{2}}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(4*x**2-9)**(1/2),x)

[Out]

Piecewise((2*I*acosh(3/(2*x))/81 - I/(27*x*sqrt(-1 + 9/(4*x**2))) + I/(36*x**3*sqrt(-1 + 9/(4*x**2))) + I/(8*x
**5*sqrt(-1 + 9/(4*x**2))), 9/(4*Abs(x**2)) > 1), (-2*asin(3/(2*x))/81 + 1/(27*x*sqrt(1 - 9/(4*x**2))) - 1/(36
*x**3*sqrt(1 - 9/(4*x**2))) - 1/(8*x**5*sqrt(1 - 9/(4*x**2))), True))

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Giac [A]  time = 2.49245, size = 55, normalized size = 0.96 \begin{align*} \frac{{\left (4 \, x^{2} - 9\right )}^{\frac{3}{2}} + 15 \, \sqrt{4 \, x^{2} - 9}}{216 \, x^{4}} + \frac{2}{81} \, \arctan \left (\frac{1}{3} \, \sqrt{4 \, x^{2} - 9}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(4*x^2-9)^(1/2),x, algorithm="giac")

[Out]

1/216*((4*x^2 - 9)^(3/2) + 15*sqrt(4*x^2 - 9))/x^4 + 2/81*arctan(1/3*sqrt(4*x^2 - 9))