Integrand size = 18, antiderivative size = 154 \[ \int \frac {\log \left (1+x^2\right )}{\sqrt {-1-x^2}} \, dx=\frac {\sqrt {1+x^2} \text {arcsinh}(x) \log \left (1+x^2\right )}{\sqrt {-1-x^2}}+\frac {\sqrt {1+x^2} \log ^2\left (x+\sqrt {1+x^2}\right )}{\sqrt {-1-x^2}}-\frac {2 \sqrt {1+x^2} \log \left (x+\sqrt {1+x^2}\right ) \log \left (1+\left (x+\sqrt {1+x^2}\right )^2\right )}{\sqrt {-1-x^2}}-\frac {\sqrt {1+x^2} \operatorname {PolyLog}\left (2,-\left (x+\sqrt {1+x^2}\right )^2\right )}{\sqrt {-1-x^2}} \] Output:
(x^2+1)^(1/2)*arcsinh(x)*ln(x^2+1)/(-x^2-1)^(1/2)+(x^2+1)^(1/2)*ln(x+(x^2+ 1)^(1/2))^2/(-x^2-1)^(1/2)-2*(x^2+1)^(1/2)*ln(x+(x^2+1)^(1/2))*ln(1+(x+(x^ 2+1)^(1/2))^2)/(-x^2-1)^(1/2)-(x^2+1)^(1/2)*polylog(2,-(x+(x^2+1)^(1/2))^2 )/(-x^2-1)^(1/2)
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.78 \[ \int \frac {\log \left (1+x^2\right )}{\sqrt {-1-x^2}} \, dx=-\frac {i \sqrt {1+\frac {1}{x^2}} x \left (\arcsin \left (\sqrt {1+x^2}\right )^2-2 i \arcsin \left (\sqrt {1+x^2}\right ) \log \left (1-e^{-2 i \arcsin \left (\sqrt {1+x^2}\right )}\right )+\log \left (1+x^2\right ) \log \left (\sqrt {-x^2}+i \sqrt {1+x^2}\right )+\operatorname {PolyLog}\left (2,e^{-2 i \arcsin \left (\sqrt {1+x^2}\right )}\right )\right )}{\sqrt {1+x^2}} \] Input:
Integrate[Log[1 + x^2]/Sqrt[-1 - x^2],x]
Output:
((-I)*Sqrt[1 + x^(-2)]*x*(ArcSin[Sqrt[1 + x^2]]^2 - (2*I)*ArcSin[Sqrt[1 + x^2]]*Log[1 - E^((-2*I)*ArcSin[Sqrt[1 + x^2]])] + Log[1 + x^2]*Log[Sqrt[-x ^2] + I*Sqrt[1 + x^2]] + PolyLog[2, E^((-2*I)*ArcSin[Sqrt[1 + x^2]])]))/Sq rt[1 + x^2]
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
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]
\[\int \frac {\ln \left (x^{2}+1\right )}{\sqrt {-x^{2}-1}}d x\]
Input:
int(ln(x^2+1)/(-x^2-1)^(1/2),x)
Output:
int(ln(x^2+1)/(-x^2-1)^(1/2),x)
\[ \int \frac {\log \left (1+x^2\right )}{\sqrt {-1-x^2}} \, dx=\int { \frac {\log \left (x^{2} + 1\right )}{\sqrt {-x^{2} - 1}} \,d x } \] 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)
\[ \int \frac {\log \left (1+x^2\right )}{\sqrt {-1-x^2}} \, dx=\int \frac {\log {\left (x^{2} + 1 \right )}}{\sqrt {- x^{2} - 1}}\, dx \] Input:
integrate(ln(x**2+1)/(-x**2-1)**(1/2),x)
Output:
Integral(log(x**2 + 1)/sqrt(-x**2 - 1), x)
\[ \int \frac {\log \left (1+x^2\right )}{\sqrt {-1-x^2}} \, dx=\int { \frac {\log \left (x^{2} + 1\right )}{\sqrt {-x^{2} - 1}} \,d x } \] 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)
\[ \int \frac {\log \left (1+x^2\right )}{\sqrt {-1-x^2}} \, dx=\int { \frac {\log \left (x^{2} + 1\right )}{\sqrt {-x^{2} - 1}} \,d x } \] 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)
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)
\[ \int \frac {\log \left (1+x^2\right )}{\sqrt {-1-x^2}} \, dx=-\left (\int \frac {\mathrm {log}\left (x^{2}+1\right )}{\sqrt {x^{2}+1}}d x \right ) i \] Input:
int(log(x^2+1)/(-x^2-1)^(1/2),x)
Output:
- int(log(x**2 + 1)/sqrt(x**2 + 1),x)*i