\(\int \frac {\log (-1-x^2)}{\sqrt {1+x^2}} \, dx\) [629]

Optimal result

Integrand size = 18, antiderivative size = 75 \[ \int \frac {\log \left (-1-x^2\right )}{\sqrt {1+x^2}} \, dx=\text {arcsinh}(x) \log \left (-1-x^2\right )+\log ^2\left (x+\sqrt {1+x^2}\right )-2 \log \left (x+\sqrt {1+x^2}\right ) \log \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )-\operatorname {PolyLog}\left (2,-\left (x+\sqrt {1+x^2}\right )^2\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.80 \[ \int \frac {\log \left (-1-x^2\right )}{\sqrt {1+x^2}} \, dx=-\frac {\sqrt {-x^2} \sqrt {-1-x^2} \left (-\text {arcsinh}\left (\sqrt {-1-x^2}\right )^2-2 \text {arcsinh}\left (\sqrt {-1-x^2}\right ) \log \left (1-e^{-2 \text {arcsinh}\left (\sqrt {-1-x^2}\right )}\right )+\log \left (-1-x^2\right ) \log \left (\sqrt {-x^2}+\sqrt {-1-x^2}\right )+\operatorname {PolyLog}\left (2,e^{-2 \text {arcsinh}\left (\sqrt {-1-x^2}\right )}\right )\right )}{x \sqrt {1+x^2}} \] 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[Sqr 
t[-x^2] + Sqrt[-1 - x^2]] + PolyLog[2, E^(-2*ArcSinh[Sqrt[-1 - x^2]])]))/( 
x*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.

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

\(\Big \downarrow \) 2923

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

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

Sympy [F]

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

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

Output:

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

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

Output:

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