\(\int \frac {\log (1-x^2)}{x \sqrt {-1+x^2}} \, dx\) [653]

Optimal result

Integrand size = 21, antiderivative size = 58 \[ \int \frac {\log \left (1-x^2\right )}{x \sqrt {-1+x^2}} \, dx=\arctan \left (\sqrt {-1+x^2}\right ) \log \left (1-x^2\right )-i \operatorname {PolyLog}\left (2,-i \sqrt {-1+x^2}\right )+i \operatorname {PolyLog}\left (2,i \sqrt {-1+x^2}\right ) \] Output:

arctan((x^2-1)^(1/2))*ln(-x^2+1)-I*polylog(2,-I*(x^2-1)^(1/2))+I*polylog(2 
,I*(x^2-1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.81 \[ \int \frac {\log \left (1-x^2\right )}{x \sqrt {-1+x^2}} \, dx=-\frac {1}{2} i \log \left (1-x^2\right ) \log \left (-i \left (i-\sqrt {-1+x^2}\right )\right )+\frac {1}{2} i \log \left (1-x^2\right ) \log \left (-i \left (i+\sqrt {-1+x^2}\right )\right )-i \operatorname {PolyLog}\left (2,-i \sqrt {-1+x^2}\right )+i \operatorname {PolyLog}\left (2,i \sqrt {-1+x^2}\right ) \] Input:

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

Output:

(-1/2*I)*Log[1 - x^2]*Log[(-I)*(I - Sqrt[-1 + x^2])] + (I/2)*Log[1 - x^2]* 
Log[(-I)*(I + Sqrt[-1 + x^2])] - I*PolyLog[2, (-I)*Sqrt[-1 + x^2]] + I*Pol 
yLog[2, I*Sqrt[-1 + x^2]]
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\log \left (1-x^2\right )}{x \sqrt {x^2-1}} \, dx\)

\(\Big \downarrow \) 2929

\(\displaystyle \int \frac {\log \left (1-x^2\right )}{x \sqrt {x^2-1}}dx\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 2929

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((h_.)* 
(x_))^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Unintegrable[(h*x) 
^m*(f + g*x^s)^r*(a + b*Log[c*(d + e*x^n)^p])^q, x] /; FreeQ[{a, b, c, d, e 
, f, g, h, m, n, p, q, r, s}, x]
 

Maple [F]

\[\int \frac {\ln \left (-x^{2}+1\right )}{x \sqrt {x^{2}-1}}d x\]

Input:

int(ln(-x^2+1)/x/(x^2-1)^(1/2),x)
 

Output:

int(ln(-x^2+1)/x/(x^2-1)^(1/2),x)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(log(-x^2 + 1)/(sqrt(x^2 - 1)*x), x)
 

Giac [F]

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

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

Output:

integrate(log(-x^2 + 1)/(sqrt(x^2 - 1)*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (1-x^2\right )}{x \sqrt {-1+x^2}} \, dx=\int \frac {\ln \left (1-x^2\right )}{x\,\sqrt {x^2-1}} \,d x \] Input:

int(log(1 - x^2)/(x*(x^2 - 1)^(1/2)),x)
 

Output:

int(log(1 - x^2)/(x*(x^2 - 1)^(1/2)), x)
 

Reduce [F]

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

int(log(-x^2+1)/x/(x^2-1)^(1/2),x)
 

Output:

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