3.5.69 \(\int \frac {\sqrt {-9+4 x^2}}{x} \, dx\) [469]

Optimal. Leaf size=30 \[ \sqrt {-9+4 x^2}-3 \tan ^{-1}\left (\frac {1}{3} \sqrt {-9+4 x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 30, 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 = {272, 52, 65, 209} \begin {gather*} \sqrt {4 x^2-9}-3 \text {ArcTan}\left (\frac {1}{3} \sqrt {4 x^2-9}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-9 + 4*x^2]/x,x]

[Out]

Sqrt[-9 + 4*x^2] - 3*ArcTan[Sqrt[-9 + 4*x^2]/3]

Rule 52

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

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

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Mathematica [A]
time = 0.02, size = 30, normalized size = 1.00 \begin {gather*} \sqrt {-9+4 x^2}-3 \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,x]

[Out]

Sqrt[-9 + 4*x^2] - 3*ArcTan[Sqrt[-9 + 4*x^2]/3]

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Maple [A]
time = 0.14, size = 25, normalized size = 0.83

method	result	size
default	\(\sqrt {4 x^{2}-9}+3 \arctan \left (\frac {3}{\sqrt {4 x^{2}-9}}\right )\)	\(25\)
trager	\(\sqrt {4 x^{2}-9}+3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\sqrt {4 x^{2}-9}-3 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x}\right )\)	\(42\)
meijerg	\(-\frac {3 \sqrt {\mathrm {signum}\left (-1+\frac {4 x^{2}}{9}\right )}\, \left (4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {1-\frac {4 x^{2}}{9}}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {4 x^{2}}{9}}}{2}\right )-2 \left (2+2 \ln \left (x \right )-2 \ln \left (3\right )+i \pi \right ) \sqrt {\pi }\right )}{4 \sqrt {\pi }\, \sqrt {-\mathrm {signum}\left (-1+\frac {4 x^{2}}{9}\right )}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.56, size = 19, normalized size = 0.63 \begin {gather*} \sqrt {4 \, x^{2} - 9} + 3 \, \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,x, algorithm="maxima")

[Out]

sqrt(4*x^2 - 9) + 3*arcsin(3/2/abs(x))

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Fricas [A]
time = 1.25, size = 28, normalized size = 0.93 \begin {gather*} \sqrt {4 \, x^{2} - 9} - 6 \, \arctan \left (-\frac {2}{3} \, x + \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,x, algorithm="fricas")

[Out]

sqrt(4*x^2 - 9) - 6*arctan(-2/3*x + 1/3*sqrt(4*x^2 - 9))

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Sympy [C] Result contains complex when optimal does not.
time = 0.67, size = 80, normalized size = 2.67 \begin {gather*} \begin {cases} \sqrt {4 x^{2} - 9} - 3 i \log {\left (x \right )} + \frac {3 i \log {\left (x^{2} \right )}}{2} + 3 \operatorname {asin}{\left (\frac {3}{2 x} \right )} & \text {for}\: \left |{x^{2}}\right | > \frac {9}{4} \\i \sqrt {9 - 4 x^{2}} + \frac {3 i \log {\left (x^{2} \right )}}{2} - 3 i \log {\left (\sqrt {1 - \frac {4 x^{2}}{9}} + 1 \right )} & \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,x)

[Out]

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

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Giac [A]
time = 0.75, size = 24, normalized size = 0.80 \begin {gather*} \sqrt {4 \, x^{2} - 9} - 3 \, \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,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.34, size = 24, normalized size = 0.80 \begin {gather*} \sqrt {4\,x^2-9}-3\,\mathrm {atan}\left (\frac {\sqrt {4\,x^2-9}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 - 9)^(1/2)/x,x)

[Out]

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

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