Integrand size = 21, antiderivative size = 101 \[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=-\arctan \left (\sqrt {-1+x^2}\right )+\frac {\sqrt {-1+x^2} \log \left (1-x^2\right )}{2 x^2}+\frac {1}{2} \arctan \left (\sqrt {-1+x^2}\right ) \log \left (1-x^2\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,-i \sqrt {-1+x^2}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,i \sqrt {-1+x^2}\right ) \] Output:
-arctan((x^2-1)^(1/2))+1/2*(x^2-1)^(1/2)*ln(-x^2+1)/x^2+1/2*arctan((x^2-1) ^(1/2))*ln(-x^2+1)-1/2*I*polylog(2,-I*(x^2-1)^(1/2))+1/2*I*polylog(2,I*(x^ 2-1)^(1/2))
Time = 0.10 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.57 \[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=\frac {1}{4} \left (\frac {2 \sqrt {-1+x^2} \log \left (1-x^2\right )}{x^2}+2 i \log \left (i-\sqrt {-1+x^2}\right )+i \log \left (1-x^2\right ) \log \left (1-i \sqrt {-1+x^2}\right )-i \log \left (1-x^2\right ) \log \left (1+i \sqrt {-1+x^2}\right )-2 i \log \left (i+\sqrt {-1+x^2}\right )-2 i \operatorname {PolyLog}\left (2,-i \sqrt {-1+x^2}\right )+2 i \operatorname {PolyLog}\left (2,i \sqrt {-1+x^2}\right )\right ) \] Input:
Integrate[Log[1 - x^2]/(x^3*Sqrt[-1 + x^2]),x]
Output:
((2*Sqrt[-1 + x^2]*Log[1 - x^2])/x^2 + (2*I)*Log[I - Sqrt[-1 + x^2]] + I*L og[1 - x^2]*Log[1 - I*Sqrt[-1 + x^2]] - I*Log[1 - x^2]*Log[1 + I*Sqrt[-1 + x^2]] - (2*I)*Log[I + Sqrt[-1 + x^2]] - (2*I)*PolyLog[2, (-I)*Sqrt[-1 + x ^2]] + (2*I)*PolyLog[2, I*Sqrt[-1 + x^2]])/4
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.
\(\displaystyle \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {x^2-1}} \, dx\) |
\(\Big \downarrow \) 2929 |
\(\displaystyle \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {x^2-1}}dx\) |
Input:
Int[Log[1 - x^2]/(x^3*Sqrt[-1 + x^2]),x]
Output:
$Aborted
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]
\[\int \frac {\ln \left (-x^{2}+1\right )}{x^{3} \sqrt {x^{2}-1}}d x\]
Input:
int(ln(-x^2+1)/x^3/(x^2-1)^(1/2),x)
Output:
int(ln(-x^2+1)/x^3/(x^2-1)^(1/2),x)
\[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=\int { \frac {\log \left (-x^{2} + 1\right )}{\sqrt {x^{2} - 1} x^{3}} \,d x } \] Input:
integrate(log(-x^2+1)/x^3/(x^2-1)^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(x^2 - 1)*log(-x^2 + 1)/(x^5 - x^3), x)
\[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=\int \frac {\log {\left (1 - x^{2} \right )}}{x^{3} \sqrt {\left (x - 1\right ) \left (x + 1\right )}}\, dx \] Input:
integrate(ln(-x**2+1)/x**3/(x**2-1)**(1/2),x)
Output:
Integral(log(1 - x**2)/(x**3*sqrt((x - 1)*(x + 1))), x)
\[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=\int { \frac {\log \left (-x^{2} + 1\right )}{\sqrt {x^{2} - 1} x^{3}} \,d x } \] Input:
integrate(log(-x^2+1)/x^3/(x^2-1)^(1/2),x, algorithm="maxima")
Output:
integrate(log(-x^2 + 1)/(sqrt(x^2 - 1)*x^3), x)
\[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=\int { \frac {\log \left (-x^{2} + 1\right )}{\sqrt {x^{2} - 1} x^{3}} \,d x } \] Input:
integrate(log(-x^2+1)/x^3/(x^2-1)^(1/2),x, algorithm="giac")
Output:
integrate(log(-x^2 + 1)/(sqrt(x^2 - 1)*x^3), x)
Timed out. \[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=\int \frac {\ln \left (1-x^2\right )}{x^3\,\sqrt {x^2-1}} \,d x \] Input:
int(log(1 - x^2)/(x^3*(x^2 - 1)^(1/2)),x)
Output:
int(log(1 - x^2)/(x^3*(x^2 - 1)^(1/2)), x)
\[ \int \frac {\log \left (1-x^2\right )}{x^3 \sqrt {-1+x^2}} \, dx=\int \frac {\mathrm {log}\left (-x^{2}+1\right )}{\sqrt {x^{2}-1}\, x^{3}}d x \] Input:
int(log(-x^2+1)/x^3/(x^2-1)^(1/2),x)
Output:
int(log( - x**2 + 1)/(sqrt(x**2 - 1)*x**3),x)