3.562 \(\int \frac {1}{x^3 \sqrt {-9+4 x^2}} \, dx\)

Optimal. Leaf size=39 \[ \frac {\sqrt {4 x^2-9}}{18 x^2}+\frac {2}{27} \tan ^{-1}\left (\frac {1}{3} \sqrt {4 x^2-9}\right ) \]

[Out]

2/27*arctan(1/3*(4*x^2-9)^(1/2))+1/18*(4*x^2-9)^(1/2)/x^2

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 51, 63, 203} \[ \frac {\sqrt {4 x^2-9}}{18 x^2}+\frac {2}{27} \tan ^{-1}\left (\frac {1}{3} \sqrt {4 x^2-9}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[-9 + 4*x^2]),x]

[Out]

Sqrt[-9 + 4*x^2]/(18*x^2) + (2*ArcTan[Sqrt[-9 + 4*x^2]/3])/27

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])

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]]

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt {-9+4 x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {-9+4 x}} \, dx,x,x^2\right )\\ &=\frac {\sqrt {-9+4 x^2}}{18 x^2}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-9+4 x}} \, dx,x,x^2\right )\\ &=\frac {\sqrt {-9+4 x^2}}{18 x^2}+\frac {1}{18} \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}}{18 x^2}+\frac {2}{27} \tan ^{-1}\left (\frac {1}{3} \sqrt {-9+4 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 54, normalized size = 1.38 \[ \frac {4}{81} \sqrt {4 x^2-9} \left (\frac {9}{8 x^2}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {4 x^2}{9}}\right )}{2 \sqrt {1-\frac {4 x^2}{9}}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[-9 + 4*x^2]),x]

[Out]

(4*Sqrt[-9 + 4*x^2]*(9/(8*x^2) + ArcTanh[Sqrt[1 - (4*x^2)/9]]/(2*Sqrt[1 - (4*x^2)/9])))/81

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fricas [A]  time = 0.92, size = 38, normalized size = 0.97 \[ \frac {8 \, x^{2} \arctan \left (-\frac {2}{3} \, x + \frac {1}{3} \, \sqrt {4 \, x^{2} - 9}\right ) + 3 \, \sqrt {4 \, x^{2} - 9}}{54 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(4*x^2-9)^(1/2),x, algorithm="fricas")

[Out]

1/54*(8*x^2*arctan(-2/3*x + 1/3*sqrt(4*x^2 - 9)) + 3*sqrt(4*x^2 - 9))/x^2

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giac [A]  time = 1.10, size = 29, normalized size = 0.74 \[ \frac {\sqrt {4 \, x^{2} - 9}}{18 \, x^{2}} + \frac {2}{27} \, \arctan \left (\frac {1}{3} \, \sqrt {4 \, x^{2} - 9}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(4*x^2-9)^(1/2),x, algorithm="giac")

[Out]

1/18*sqrt(4*x^2 - 9)/x^2 + 2/27*arctan(1/3*sqrt(4*x^2 - 9))

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maple [A]  time = 0.01, size = 30, normalized size = 0.77 \[ -\frac {2 \arctan \left (\frac {3}{\sqrt {4 x^{2}-9}}\right )}{27}+\frac {\sqrt {4 x^{2}-9}}{18 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(4*x^2-9)^(1/2),x)

[Out]

1/18*(4*x^2-9)^(1/2)/x^2-2/27*arctan(3/(4*x^2-9)^(1/2))

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maxima [A]  time = 2.93, size = 24, normalized size = 0.62 \[ \frac {\sqrt {4 \, x^{2} - 9}}{18 \, x^{2}} - \frac {2}{27} \, \arcsin \left (\frac {3}{2 \, {\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(4*x^2-9)^(1/2),x, algorithm="maxima")

[Out]

1/18*sqrt(4*x^2 - 9)/x^2 - 2/27*arcsin(3/2/abs(x))

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mupad [B]  time = 4.88, size = 29, normalized size = 0.74 \[ \frac {2\,\mathrm {atan}\left (\frac {\sqrt {4\,x^2-9}}{3}\right )}{27}+\frac {\sqrt {4\,x^2-9}}{18\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(4*x^2 - 9)^(1/2)),x)

[Out]

(2*atan((4*x^2 - 9)^(1/2)/3))/27 + (4*x^2 - 9)^(1/2)/(18*x^2)

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sympy [A]  time = 2.13, size = 99, normalized size = 2.54 \[ \begin {cases} \frac {2 i \operatorname {acosh}{\left (\frac {3}{2 x} \right )}}{27} - \frac {i}{9 x \sqrt {-1 + \frac {9}{4 x^{2}}}} + \frac {i}{4 x^{3} \sqrt {-1 + \frac {9}{4 x^{2}}}} & \text {for}\: \frac {9}{4 \left |{x^{2}}\right |} > 1 \\- \frac {2 \operatorname {asin}{\left (\frac {3}{2 x} \right )}}{27} + \frac {1}{9 x \sqrt {1 - \frac {9}{4 x^{2}}}} - \frac {1}{4 x^{3} \sqrt {1 - \frac {9}{4 x^{2}}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(4*x**2-9)**(1/2),x)

[Out]

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

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