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

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

[Out]

2/27*arctanh(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 = 0.02, 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 = {266, 47, 51, 63, 207} \[ -\frac {\sqrt {4 x^2+9}}{18 x^2}-\frac {\sqrt {4 x^2+9}}{4 x^4}+\frac {2}{27} \tanh ^{-1}\left (\frac {1}{3} \sqrt {4 x^2+9}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[9 + 4*x^2]/x^5,x]

[Out]

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

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[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 {\sqrt {9+4 x^2}}{x^5} \, dx &=\frac {1}{2} \operatorname {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} \operatorname {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} \operatorname {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} \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}}{4 x^4}-\frac {\sqrt {9+4 x^2}}{18 x^2}+\frac {2}{27} \tanh ^{-1}\left (\frac {1}{3} \sqrt {9+4 x^2}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.56 \[ -\frac {16 \left (4 x^2+9\right )^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {4 x^2}{9}+1\right )}{2187} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[9 + 4*x^2]/x^5,x]

[Out]

(-16*(9 + 4*x^2)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (4*x^2)/9])/2187

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fricas [A]  time = 1.10, size = 64, normalized size = 1.12 \[ \frac {8 \, x^{4} \log \left (-2 \, x + \sqrt {4 \, x^{2} + 9} + 3\right ) - 8 \, x^{4} \log \left (-2 \, x + \sqrt {4 \, x^{2} + 9} - 3\right ) - 3 \, \sqrt {4 \, x^{2} + 9} {\left (2 \, x^{2} + 9\right )}}{108 \, x^{4}} \]

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*(8*x^4*log(-2*x + sqrt(4*x^2 + 9) + 3) - 8*x^4*log(-2*x + sqrt(4*x^2 + 9) - 3) - 3*sqrt(4*x^2 + 9)*(2*x^
2 + 9))/x^4

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giac [A]  time = 1.05, size = 55, normalized size = 0.96 \[ -\frac {{\left (4 \, x^{2} + 9\right )}^{\frac {3}{2}} + 9 \, \sqrt {4 \, x^{2} + 9}}{72 \, x^{4}} + \frac {1}{27} \, \log \left (\sqrt {4 \, x^{2} + 9} + 3\right ) - \frac {1}{27} \, \log \left (\sqrt {4 \, x^{2} + 9} - 3\right ) \]

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

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maple [A]  time = 0.01, size = 55, normalized size = 0.96 \[ \frac {2 \arctanh \left (\frac {3}{\sqrt {4 x^{2}+9}}\right )}{27}+\frac {\left (4 x^{2}+9\right )^{\frac {3}{2}}}{162 x^{2}}-\frac {\left (4 x^{2}+9\right )^{\frac {3}{2}}}{36 x^{4}}-\frac {2 \sqrt {4 x^{2}+9}}{81} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.01, size = 49, normalized size = 0.86 \[ -\frac {2}{81} \, \sqrt {4 \, x^{2} + 9} + \frac {{\left (4 \, x^{2} + 9\right )}^{\frac {3}{2}}}{162 \, x^{2}} - \frac {{\left (4 \, x^{2} + 9\right )}^{\frac {3}{2}}}{36 \, x^{4}} + \frac {2}{27} \, \operatorname {arsinh}\left (\frac {3}{2 \, {\left | x \right |}}\right ) \]

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*arcsinh(3/2/abs(x))

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mupad [B]  time = 0.03, size = 45, normalized size = 0.79 \[ \frac {2\,\mathrm {atanh}\left (\frac {2\,\sqrt {x^2+\frac {9}{4}}}{3}\right )}{27}+\frac {\sqrt {x^2+\frac {9}{4}}\,\left (\frac {2}{3\,x^2}-\frac {1}{x^4}\right )}{2}-\frac {4\,\sqrt {x^2+\frac {9}{4}}}{9\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 + 9)^(1/2)/x^5,x)

[Out]

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

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sympy [A]  time = 3.20, size = 63, normalized size = 1.11 \[ \frac {2 \operatorname {asinh}{\left (\frac {3}{2 x} \right )}}{27} - \frac {1}{9 x \sqrt {1 + \frac {9}{4 x^{2}}}} - \frac {3}{4 x^{3} \sqrt {1 + \frac {9}{4 x^{2}}}} - \frac {9}{8 x^{5} \sqrt {1 + \frac {9}{4 x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*asinh(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)))

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