Integrand size = 14, antiderivative size = 171 \[ \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-b}\right ) \log (1-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1+b}\right ) \log (1-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+b}\right ) \log (1+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1-b}\right ) \log (1+b x)+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-b x}{1-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-b x}{1+b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+b x}{1-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+b x}{1+b}\right ) \]
1/4*ln(-b*(1-x)/(1-b))*ln(-b*x+1)-1/4*ln(b*(1+x)/(1+b))*ln(-b*x+1)-1/4*ln( b*(1-x)/(1+b))*ln(b*x+1)+1/4*ln(-b*(1+x)/(1-b))*ln(b*x+1)+1/4*polylog(2,(- b*x+1)/(1-b))-1/4*polylog(2,(-b*x+1)/(1+b))+1/4*polylog(2,(b*x+1)/(1-b))-1 /4*polylog(2,(b*x+1)/(1+b))
Result contains complex when optimal does not.
Time = 0.72 (sec) , antiderivative size = 576, normalized size of antiderivative = 3.37 \[ \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx=-\frac {b \left (2 i \arccos \left (\frac {1+b^2}{1-b^2}\right ) \arctan \left (\frac {b x}{\sqrt {-b^2}}\right )-4 \arctan \left (\frac {\sqrt {-b^2}}{b x}\right ) \text {arctanh}(b x)-\left (\arccos \left (\frac {1+b^2}{1-b^2}\right )-2 \arctan \left (\frac {b x}{\sqrt {-b^2}}\right )\right ) \log \left (\frac {2 b \left (-i+\sqrt {-b^2}\right ) (-1+b x)}{\left (-1+b^2\right ) \left (-i b+\sqrt {-b^2} x\right )}\right )-\left (\arccos \left (\frac {1+b^2}{1-b^2}\right )+2 \arctan \left (\frac {b x}{\sqrt {-b^2}}\right )\right ) \log \left (\frac {2 b \left (i+\sqrt {-b^2}\right ) (1+b x)}{\left (-1+b^2\right ) \left (-i b+\sqrt {-b^2} x\right )}\right )+\left (\arccos \left (\frac {1+b^2}{1-b^2}\right )-2 \left (\arctan \left (\frac {\sqrt {-b^2}}{b x}\right )+\arctan \left (\frac {b x}{\sqrt {-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-b^2} e^{-\text {arctanh}(b x)}}{\sqrt {-1+b^2} \sqrt {1+b^2+\left (-1+b^2\right ) \cosh (2 \text {arctanh}(b x))}}\right )+\left (\arccos \left (\frac {1+b^2}{1-b^2}\right )+2 \left (\arctan \left (\frac {\sqrt {-b^2}}{b x}\right )+\arctan \left (\frac {b x}{\sqrt {-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-b^2} e^{\text {arctanh}(b x)}}{\sqrt {-1+b^2} \sqrt {1+b^2+\left (-1+b^2\right ) \cosh (2 \text {arctanh}(b x))}}\right )+i \left (\operatorname {PolyLog}\left (2,\frac {\left (1+b^2-2 i \sqrt {-b^2}\right ) \left (b-i \sqrt {-b^2} x\right )}{\left (-1+b^2\right ) \left (b+i \sqrt {-b^2} x\right )}\right )-\operatorname {PolyLog}\left (2,\frac {\left (1+b^2+2 i \sqrt {-b^2}\right ) \left (b-i \sqrt {-b^2} x\right )}{\left (-1+b^2\right ) \left (b+i \sqrt {-b^2} x\right )}\right )\right )\right )}{4 \sqrt {-b^2}} \]
-1/4*(b*((2*I)*ArcCos[(1 + b^2)/(1 - b^2)]*ArcTan[(b*x)/Sqrt[-b^2]] - 4*Ar cTan[Sqrt[-b^2]/(b*x)]*ArcTanh[b*x] - (ArcCos[(1 + b^2)/(1 - b^2)] - 2*Arc Tan[(b*x)/Sqrt[-b^2]])*Log[(2*b*(-I + Sqrt[-b^2])*(-1 + b*x))/((-1 + b^2)* ((-I)*b + Sqrt[-b^2]*x))] - (ArcCos[(1 + b^2)/(1 - b^2)] + 2*ArcTan[(b*x)/ Sqrt[-b^2]])*Log[(2*b*(I + Sqrt[-b^2])*(1 + b*x))/((-1 + b^2)*((-I)*b + Sq rt[-b^2]*x))] + (ArcCos[(1 + b^2)/(1 - b^2)] - 2*(ArcTan[Sqrt[-b^2]/(b*x)] + ArcTan[(b*x)/Sqrt[-b^2]]))*Log[(Sqrt[2]*Sqrt[-b^2])/(Sqrt[-1 + b^2]*E^A rcTanh[b*x]*Sqrt[1 + b^2 + (-1 + b^2)*Cosh[2*ArcTanh[b*x]]])] + (ArcCos[(1 + b^2)/(1 - b^2)] + 2*(ArcTan[Sqrt[-b^2]/(b*x)] + ArcTan[(b*x)/Sqrt[-b^2] ]))*Log[(Sqrt[2]*Sqrt[-b^2]*E^ArcTanh[b*x])/(Sqrt[-1 + b^2]*Sqrt[1 + b^2 + (-1 + b^2)*Cosh[2*ArcTanh[b*x]]])] + I*(PolyLog[2, ((1 + b^2 - (2*I)*Sqrt [-b^2])*(b - I*Sqrt[-b^2]*x))/((-1 + b^2)*(b + I*Sqrt[-b^2]*x))] - PolyLog [2, ((1 + b^2 + (2*I)*Sqrt[-b^2])*(b - I*Sqrt[-b^2]*x))/((-1 + b^2)*(b + I *Sqrt[-b^2]*x))])))/Sqrt[-b^2]
Time = 0.50 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6534, 2856, 2009}
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.
\(\displaystyle \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx\) |
\(\Big \downarrow \) 6534 |
\(\displaystyle \frac {1}{2} \int \frac {\log (b x+1)}{1-x^2}dx-\frac {1}{2} \int \frac {\log (1-b x)}{1-x^2}dx\) |
\(\Big \downarrow \) 2856 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\log (b x+1)}{2 (1-x)}+\frac {\log (b x+1)}{2 (x+1)}\right )dx-\frac {1}{2} \int \left (\frac {\log (1-b x)}{2 (1-x)}+\frac {\log (1-b x)}{2 (x+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {1-b x}{1-b}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {1-b x}{b+1}\right )+\frac {1}{2} \log \left (-\frac {b (1-x)}{1-b}\right ) \log (1-b x)-\frac {1}{2} \log \left (\frac {b (x+1)}{b+1}\right ) \log (1-b x)\right )+\frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {b x+1}{1-b}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {b x+1}{b+1}\right )-\frac {1}{2} \log \left (\frac {b (1-x)}{b+1}\right ) \log (b x+1)+\frac {1}{2} \log \left (-\frac {b (x+1)}{1-b}\right ) \log (b x+1)\right )\) |
((Log[-((b*(1 - x))/(1 - b))]*Log[1 - b*x])/2 - (Log[(b*(1 + x))/(1 + b)]* Log[1 - b*x])/2 + PolyLog[2, (1 - b*x)/(1 - b)]/2 - PolyLog[2, (1 - b*x)/( 1 + b)]/2)/2 + (-1/2*(Log[(b*(1 - x))/(1 + b)]*Log[1 + b*x]) + (Log[-((b*( 1 + x))/(1 - b))]*Log[1 + b*x])/2 + PolyLog[2, (1 + b*x)/(1 - b)]/2 - Poly Log[2, (1 + b*x)/(1 + b)]/2)/2
3.6.6.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) ^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
Int[ArcTanh[(c_.)*(x_)]/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Simp[1/2 Int [Log[1 + c*x]/(d + e*x^2), x], x] - Simp[1/2 Int[Log[1 - c*x]/(d + e*x^2) , x], x] /; FreeQ[{c, d, e}, x]
Time = 0.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {\ln \left (-b x +1\right ) \ln \left (\frac {-b x -b}{-b -1}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {-b x -b}{-b -1}\right )}{4}+\frac {\ln \left (-b x +1\right ) \ln \left (\frac {-b x +b}{-1+b}\right )}{4}+\frac {\operatorname {dilog}\left (\frac {-b x +b}{-1+b}\right )}{4}-\frac {\ln \left (b x +1\right ) \ln \left (\frac {b x -b}{-b -1}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {b x -b}{-b -1}\right )}{4}+\frac {\ln \left (b x +1\right ) \ln \left (\frac {b x +b}{-1+b}\right )}{4}+\frac {\operatorname {dilog}\left (\frac {b x +b}{-1+b}\right )}{4}\) | \(160\) |
derivativedivides | \(\frac {-\frac {\operatorname {arctanh}\left (b x \right ) b \ln \left (-b x +b \right )}{2}+\frac {\operatorname {arctanh}\left (b x \right ) b \ln \left (b x +b \right )}{2}-\frac {b^{2} \left (\frac {\frac {\operatorname {dilog}\left (\frac {-b x +1}{1-b}\right )}{2}+\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x +1}{1-b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -1}{-b -1}\right )}{2}-\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -1}{-b -1}\right )}{2}}{b}-\frac {-\frac {\operatorname {dilog}\left (\frac {b x +1}{1-b}\right )}{2}-\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x +1}{1-b}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {b x -1}{-b -1}\right )}{2}+\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x -1}{-b -1}\right )}{2}}{b}\right )}{2}}{b}\) | \(208\) |
default | \(\frac {-\frac {\operatorname {arctanh}\left (b x \right ) b \ln \left (-b x +b \right )}{2}+\frac {\operatorname {arctanh}\left (b x \right ) b \ln \left (b x +b \right )}{2}-\frac {b^{2} \left (\frac {\frac {\operatorname {dilog}\left (\frac {-b x +1}{1-b}\right )}{2}+\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x +1}{1-b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -1}{-b -1}\right )}{2}-\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -1}{-b -1}\right )}{2}}{b}-\frac {-\frac {\operatorname {dilog}\left (\frac {b x +1}{1-b}\right )}{2}-\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x +1}{1-b}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {b x -1}{-b -1}\right )}{2}+\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x -1}{-b -1}\right )}{2}}{b}\right )}{2}}{b}\) | \(208\) |
parts | \(\operatorname {arctanh}\left (x \right ) \operatorname {arctanh}\left (b x \right )-b \left (\frac {\operatorname {arctanh}\left (x \right ) \ln \left (b x +1\right )}{2 b}-\frac {\operatorname {arctanh}\left (x \right ) \ln \left (b x -1\right )}{2 b}+\frac {\ln \left (b x -1\right ) \ln \left (\frac {b x +b}{1+b}\right )}{4 b}+\frac {\operatorname {dilog}\left (\frac {b x +b}{1+b}\right )}{4 b}-\frac {\ln \left (\frac {b x -b}{1-b}\right ) \ln \left (b x -1\right )}{4 b}-\frac {\operatorname {dilog}\left (\frac {b x -b}{1-b}\right )}{4 b}+\frac {\ln \left (\frac {b x -b}{-b -1}\right ) \ln \left (b x +1\right )}{4 b}+\frac {\operatorname {dilog}\left (\frac {b x -b}{-b -1}\right )}{4 b}-\frac {\ln \left (\frac {b x +b}{-1+b}\right ) \ln \left (b x +1\right )}{4 b}-\frac {\operatorname {dilog}\left (\frac {b x +b}{-1+b}\right )}{4 b}\right )\) | \(215\) |
-1/4*ln(-b*x+1)*ln((-b*x-b)/(-b-1))-1/4*dilog((-b*x-b)/(-b-1))+1/4*ln(-b*x +1)*ln((-b*x+b)/(-1+b))+1/4*dilog((-b*x+b)/(-1+b))-1/4*ln(b*x+1)*ln((b*x-b )/(-b-1))-1/4*dilog((b*x-b)/(-b-1))+1/4*ln(b*x+1)*ln((b*x+b)/(-1+b))+1/4*d ilog((b*x+b)/(-1+b))
\[ \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx=\int { -\frac {\operatorname {artanh}\left (b x\right )}{x^{2} - 1} \,d x } \]
\[ \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx=- \int \frac {\operatorname {atanh}{\left (b x \right )}}{x^{2} - 1}\, dx \]
Time = 0.19 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05 \[ \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx=\frac {1}{4} \, b {\left (\frac {\log \left (x + 1\right ) \log \left (-\frac {b x + b}{b + 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + b}{b + 1}\right )}{b} + \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{b + 1}\right )}{b} - \frac {\log \left (x + 1\right ) \log \left (-\frac {b x + b}{b - 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + b}{b - 1}\right )}{b} - \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{b - 1}\right )}{b}\right )} + \frac {1}{2} \, {\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \operatorname {artanh}\left (b x\right ) \]
1/4*b*((log(x + 1)*log(-(b*x + b)/(b + 1) + 1) + dilog((b*x + b)/(b + 1))) /b + (log(x - 1)*log((b*x - b)/(b + 1) + 1) + dilog(-(b*x - b)/(b + 1)))/b - (log(x + 1)*log(-(b*x + b)/(b - 1) + 1) + dilog((b*x + b)/(b - 1)))/b - (log(x - 1)*log((b*x - b)/(b - 1) + 1) + dilog(-(b*x - b)/(b - 1)))/b) + 1/2*(log(x + 1) - log(x - 1))*arctanh(b*x)
\[ \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx=\int { -\frac {\operatorname {artanh}\left (b x\right )}{x^{2} - 1} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}(b x)}{1-x^2} \, dx=-\int \frac {\mathrm {atanh}\left (b\,x\right )}{x^2-1} \,d x \]