\(\int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx\) [623]

Optimal result

Integrand size = 18, antiderivative size = 18 \[ \int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx=\text {arctanh}\left (\frac {\log (x)}{\sqrt {-a^2+\log ^2(x)}}\right ) \]

[Out]

arctanh(ln(x)/(-a^2+ln(x)^2)^(1/2))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {223, 212} \[ \int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx=\text {arctanh}\left (\frac {\log (x)}{\sqrt {\log ^2(x)-a^2}}\right ) \]

[In]

Int[1/(x*Sqrt[-a^2 + Log[x]^2]),x]

[Out]

ArcTanh[Log[x]/Sqrt[-a^2 + Log[x]^2]]

Rule 212

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

Rule 223

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(50\) vs. \(2(18)=36\).

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.78 \[ \int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx=-\frac {1}{2} \log \left (1-\frac {\log (x)}{\sqrt {-a^2+\log ^2(x)}}\right )+\frac {1}{2} \log \left (1+\frac {\log (x)}{\sqrt {-a^2+\log ^2(x)}}\right ) \]

[In]

Integrate[1/(x*Sqrt[-a^2 + Log[x]^2]),x]

[Out]

-1/2*Log[1 - Log[x]/Sqrt[-a^2 + Log[x]^2]] + Log[1 + Log[x]/Sqrt[-a^2 + Log[x]^2]]/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\ln \left (\ln \left (x \right )+\sqrt {-a^{2}+\ln \left (x \right )^{2}}\right )\)	\(17\)
default	\(\ln \left (\ln \left (x \right )+\sqrt {-a^{2}+\ln \left (x \right )^{2}}\right )\)	\(17\)

[In]

int(1/x/(-a^2+ln(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x)+(-a^2+ln(x)^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx=-\log \left (\sqrt {-a^{2} + \log \left (x\right )^{2}} - \log \left (x\right )\right ) \]

[In]

integrate(1/x/(-a^2+log(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-log(sqrt(-a^2 + log(x)^2) - log(x))

Sympy [F]

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

[In]

integrate(1/x/(-a**2+ln(x)**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx=\log \left (2 \, \sqrt {-a^{2} + \log \left (x\right )^{2}} + 2 \, \log \left (x\right )\right ) \]

[In]

integrate(1/x/(-a^2+log(x)^2)^(1/2),x, algorithm="maxima")

[Out]

log(2*sqrt(-a^2 + log(x)^2) + 2*log(x))

Giac [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx=\text {Timed out} \]

[In]

integrate(1/x/(-a^2+log(x)^2)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \sqrt {-a^2+\log ^2(x)}} \, dx=\ln \left (\ln \left (x\right )+\sqrt {{\ln \left (x\right )}^2-a^2}\right ) \]

[In]

int(1/(x*(log(x)^2 - a^2)^(1/2)),x)

[Out]

log(log(x) + (log(x)^2 - a^2)^(1/2))