\(\int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx\) [135]

Optimal result

Integrand size = 16, antiderivative size = 11 \[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=\frac {1}{3} \arcsin \left (\frac {3 \log (x)}{2}\right ) \]

[Out]

1/3*arcsin(3/2*ln(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {222} \[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=\frac {1}{3} \arcsin \left (\frac {3 \log (x)}{2}\right ) \]

[In]

Int[1/(x*Sqrt[4 - 9*Log[x]^2]),x]

[Out]

ArcSin[(3*Log[x])/2]/3

Rule 222

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\sqrt {4-9 x^2}} \, dx,x,\log (x)\right ) \\ & = \frac {1}{3} \sin ^{-1}\left (\frac {3 \log (x)}{2}\right ) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(25\) vs. \(2(11)=22\).

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 2.27 \[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=\frac {2}{3} \arctan \left (\frac {3 \log (x)}{-2+\sqrt {4-9 \log ^2(x)}}\right ) \]

[In]

Integrate[1/(x*Sqrt[4 - 9*Log[x]^2]),x]

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\arcsin \left (\frac {3 \ln \left (x \right )}{2}\right )}{3}\)	\(8\)
default	\(\frac {\arcsin \left (\frac {3 \ln \left (x \right )}{2}\right )}{3}\)	\(8\)

[In]

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

[Out]

1/3*arcsin(3/2*ln(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 21 vs. \(2 (7) = 14\).

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.91 \[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=-\frac {2}{3} \, \arctan \left (\frac {\sqrt {-9 \, \log \left (x\right )^{2} + 4} - 2}{3 \, \log \left (x\right )}\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=\int \frac {1}{x \sqrt {- \left (3 \log {\left (x \right )} - 2\right ) \left (3 \log {\left (x \right )} + 2\right )}}\, dx \]

[In]

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

[Out]

Integral(1/(x*sqrt(-(3*log(x) - 2)*(3*log(x) + 2))), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=\frac {1}{3} \, \arcsin \left (\frac {3}{2} \, \log \left (x\right )\right ) \]

[In]

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

[Out]

1/3*arcsin(3/2*log(x))

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=\frac {1}{3} \, \arcsin \left (\frac {3}{2} \, \log \left (x\right )\right ) \]

[In]

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

[Out]

1/3*arcsin(3/2*log(x))

Mupad [B] (verification not implemented)

Time = 1.53 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x \sqrt {4-9 \log ^2(x)}} \, dx=\frac {\mathrm {asin}\left (\frac {3\,\ln \left (x\right )}{2}\right )}{3} \]

[In]

int(1/(x*(4 - 9*log(x)^2)^(1/2)),x)

[Out]

asin((3*log(x))/2)/3