3.5.73 \(\int \frac {\sqrt {-9+4 x^2}}{x^5} \, dx\) [473]

Optimal. Leaf size=57 \[ -\frac {\sqrt {-9+4 x^2}}{4 x^4}+\frac {\sqrt {-9+4 x^2}}{18 x^2}+\frac {2}{27} \tan ^{-1}\left (\frac {1}{3} \sqrt {-9+4 x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 44, 65, 209} \begin {gather*} \frac {2}{27} \text {ArcTan}\left (\frac {1}{3} \sqrt {4 x^2-9}\right )+\frac {\sqrt {4 x^2-9}}{18 x^2}-\frac {\sqrt {4 x^2-9}}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-9 + 4*x^2]/x^5,x]

[Out]

-1/4*Sqrt[-9 + 4*x^2]/x^4 + Sqrt[-9 + 4*x^2]/(18*x^2) + (2*ArcTan[Sqrt[-9 + 4*x^2]/3])/27

Rule 43

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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*} \int \frac {\sqrt {-9+4 x^2}}{x^5} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {-9+4 x}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {-9+4 x^2}}{4 x^4}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {-9+4 x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {-9+4 x^2}}{4 x^4}+\frac {\sqrt {-9+4 x^2}}{18 x^2}+\frac {1}{9} \text {Subst}\left (\int \frac {1}{x \sqrt {-9+4 x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {-9+4 x^2}}{4 x^4}+\frac {\sqrt {-9+4 x^2}}{18 x^2}+\frac {1}{18} \text {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}}{4 x^4}+\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 = 0.04, size = 46, normalized size = 0.81 \begin {gather*} \frac {\left (-9+2 x^2\right ) \sqrt {-9+4 x^2}}{36 x^4}+\frac {2}{27} \tan ^{-1}\left (\frac {1}{3} \sqrt {-9+4 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-9 + 4*x^2]/x^5,x]

[Out]

((-9 + 2*x^2)*Sqrt[-9 + 4*x^2])/(36*x^4) + (2*ArcTan[Sqrt[-9 + 4*x^2]/3])/27

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Maple [A]
time = 0.13, size = 55, normalized size = 0.96

method	result	size
risch	\(\frac {8 x^{4}-54 x^{2}+81}{36 x^{4} \sqrt {4 x^{2}-9}}-\frac {2 \arctan \left (\frac {3}{\sqrt {4 x^{2}-9}}\right )}{27}\)	\(42\)
trager	\(\frac {\left (2 x^{2}-9\right ) \sqrt {4 x^{2}-9}}{36 x^{4}}-\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\sqrt {4 x^{2}-9}-3 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x}\right )}{27}\)	\(54\)
default	\(\frac {\left (4 x^{2}-9\right )^{\frac {3}{2}}}{36 x^{4}}+\frac {\left (4 x^{2}-9\right )^{\frac {3}{2}}}{162 x^{2}}-\frac {2 \sqrt {4 x^{2}-9}}{81}-\frac {2 \arctan \left (\frac {3}{\sqrt {4 x^{2}-9}}\right )}{27}\)	\(55\)
meijerg	\(-\frac {4 \sqrt {\mathrm {signum}\left (-1+\frac {4 x^{2}}{9}\right )}\, \left (-\frac {81 \sqrt {\pi }\, \left (\frac {16}{81} x^{4}-\frac {32}{9} x^{2}+8\right )}{128 x^{4}}+\frac {81 \sqrt {\pi }\, \left (-\frac {16 x^{2}}{9}+8\right ) \sqrt {1-\frac {4 x^{2}}{9}}}{128 x^{4}}-\frac {\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {4 x^{2}}{9}}}{2}\right )}{2}+\frac {\left (\frac {1}{2}+2 \ln \left (x \right )-2 \ln \left (3\right )+i \pi \right ) \sqrt {\pi }}{4}+\frac {81 \sqrt {\pi }}{16 x^{4}}-\frac {9 \sqrt {\pi }}{4 x^{2}}\right )}{27 \sqrt {\pi }\, \sqrt {-\mathrm {signum}\left (-1+\frac {4 x^{2}}{9}\right )}}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36/x^4*(4*x^2-9)^(3/2)+1/162/x^2*(4*x^2-9)^(3/2)-2/81*(4*x^2-9)^(1/2)-2/27*arctan(3/(4*x^2-9)^(1/2))

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Maxima [A]
time = 0.61, size = 49, normalized size = 0.86 \begin {gather*} -\frac {2}{81} \, \sqrt {4 \, x^{2} - 9} + \frac {{\left (4 \, x^{2} - 9\right )}^{\frac {3}{2}}}{162 \, x^{2}} + \frac {{\left (4 \, x^{2} - 9\right )}^{\frac {3}{2}}}{36 \, x^{4}} - \frac {2}{27} \, \arcsin \left (\frac {3}{2 \, {\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/81*sqrt(4*x^2 - 9) + 1/162*(4*x^2 - 9)^(3/2)/x^2 + 1/36*(4*x^2 - 9)^(3/2)/x^4 - 2/27*arcsin(3/2/abs(x))

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Fricas [A]
time = 0.79, size = 45, normalized size = 0.79 \begin {gather*} \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} - 9\right )}}{108 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/108*(16*x^4*arctan(-2/3*x + 1/3*sqrt(4*x^2 - 9)) + 3*sqrt(4*x^2 - 9)*(2*x^2 - 9))/x^4

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Sympy [C] Result contains complex when optimal does not.
time = 1.90, size = 139, normalized size = 2.44 \begin {gather*} \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 {3 i}{4 x^{3} \sqrt {-1 + \frac {9}{4 x^{2}}}} - \frac {9 i}{8 x^{5} \sqrt {-1 + \frac {9}{4 x^{2}}}} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > \frac {4}{9} \\- \frac {2 \operatorname {asin}{\left (\frac {3}{2 x} \right )}}{27} + \frac {1}{9 x \sqrt {1 - \frac {9}{4 x^{2}}}} - \frac {3}{4 x^{3} \sqrt {1 - \frac {9}{4 x^{2}}}} + \frac {9}{8 x^{5} \sqrt {1 - \frac {9}{4 x^{2}}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.69, size = 41, normalized size = 0.72 \begin {gather*} \frac {{\left (4 \, x^{2} - 9\right )}^{\frac {3}{2}} - 9 \, \sqrt {4 \, x^{2} - 9}}{72 \, x^{4}} + \frac {2}{27} \, \arctan \left (\frac {1}{3} \, \sqrt {4 \, x^{2} - 9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*((4*x^2 - 9)^(3/2) - 9*sqrt(4*x^2 - 9))/x^4 + 2/27*arctan(1/3*sqrt(4*x^2 - 9))

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Mupad [B]
time = 5.19, size = 43, normalized size = 0.75 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\sqrt {4\,x^2-9}}{3}\right )}{27}-\frac {\frac {\sqrt {4\,x^2-9}}{8}-\frac {{\left (4\,x^2-9\right )}^{3/2}}{72}}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 - 9)^(1/2)/x^5,x)

[Out]

(2*atan((4*x^2 - 9)^(1/2)/3))/27 - ((4*x^2 - 9)^(1/2)/8 - (4*x^2 - 9)^(3/2)/72)/x^4

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