\(\int \frac {\log (-1+x^2)}{\sqrt {1-x^2}} \, dx\) [631]

Optimal result

Integrand size = 18, antiderivative size = 105 \[ \int \frac {\log \left (-1+x^2\right )}{\sqrt {1-x^2}} \, dx=\arcsin (x) \log \left (-1+x^2\right )-i \log ^2\left (i x+\sqrt {1-x^2}\right )+2 i \log \left (i x+\sqrt {1-x^2}\right ) \log \left (1+\left (i x+\sqrt {1-x^2}\right )^2\right )+i \operatorname {PolyLog}\left (2,-\left (i x+\sqrt {1-x^2}\right )^2\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.12 \[ \int \frac {\log \left (-1+x^2\right )}{\sqrt {1-x^2}} \, dx=\frac {\sqrt {x^2} \sqrt {-1+x^2} \left (-\text {arcsinh}\left (\sqrt {-1+x^2}\right )^2-2 \text {arcsinh}\left (\sqrt {-1+x^2}\right ) \log \left (1-e^{-2 \text {arcsinh}\left (\sqrt {-1+x^2}\right )}\right )+\log \left (-1+x^2\right ) \log \left (\sqrt {x^2}+\sqrt {-1+x^2}\right )+\operatorname {PolyLog}\left (2,e^{-2 \text {arcsinh}\left (\sqrt {-1+x^2}\right )}\right )\right )}{x \sqrt {1-x^2}} \] Input:

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

Output:

(Sqrt[x^2]*Sqrt[-1 + x^2]*(-ArcSinh[Sqrt[-1 + x^2]]^2 - 2*ArcSinh[Sqrt[-1 
+ x^2]]*Log[1 - E^(-2*ArcSinh[Sqrt[-1 + x^2]])] + Log[-1 + x^2]*Log[Sqrt[x 
^2] + Sqrt[-1 + x^2]] + PolyLog[2, E^(-2*ArcSinh[Sqrt[-1 + x^2]])]))/(x*Sq 
rt[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.

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

\(\Big \downarrow \) 2923

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 2923

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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