Integrand size = 16, antiderivative size = 203 \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a+b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a+b}\right ) \]
1/4*ln(-b*(1-x)/(1-a-b))*ln(-b*x-a+1)-1/4*ln(b*(1+x)/(1-a+b))*ln(-b*x-a+1) -1/4*ln(b*(1-x)/(1+a+b))*ln(b*x+a+1)+1/4*ln(-b*(1+x)/(1+a-b))*ln(b*x+a+1)+ 1/4*polylog(2,(-b*x-a+1)/(1-a-b))-1/4*polylog(2,(-b*x-a+1)/(1-a+b))+1/4*po lylog(2,(b*x+a+1)/(1+a-b))-1/4*polylog(2,(b*x+a+1)/(1+a+b))
Time = 0.03 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a+b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a+b}\right ) \]
(Log[-((b*(1 - x))/(1 - a - b))]*Log[1 - a - b*x])/4 - (Log[(b*(1 + x))/(1 - a + b)]*Log[1 - a - b*x])/4 - (Log[(b*(1 - x))/(1 + a + b)]*Log[1 + a + b*x])/4 + (Log[-((b*(1 + x))/(1 + a - b))]*Log[1 + a + b*x])/4 + PolyLog[ 2, (1 - a - b*x)/(1 - a - b)]/4 - PolyLog[2, (1 - a - b*x)/(1 - a + b)]/4 + PolyLog[2, (1 + a + b*x)/(1 + a - b)]/4 - PolyLog[2, (1 + a + b*x)/(1 + a + b)]/4
Time = 0.54 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6665, 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.
\(\displaystyle \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx\) |
\(\Big \downarrow \) 6665 |
\(\displaystyle \frac {1}{2} \int \frac {\log (a+b x+1)}{1-x^2}dx-\frac {1}{2} \int \frac {\log (-a-b x+1)}{1-x^2}dx\) |
\(\Big \downarrow \) 2856 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\log (a+b x+1)}{2 (1-x)}+\frac {\log (a+b x+1)}{2 (x+1)}\right )dx-\frac {1}{2} \int \left (\frac {\log (-a-b x+1)}{2 (1-x)}+\frac {\log (-a-b x+1)}{2 (x+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {-a-b x+1}{-a-b+1}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {-a-b x+1}{-a+b+1}\right )+\frac {1}{2} \log \left (-\frac {b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac {1}{2} \log \left (\frac {b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)\right )+\frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {a+b x+1}{a-b+1}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {a+b x+1}{a+b+1}\right )-\frac {1}{2} \log \left (\frac {b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac {1}{2} \log \left (-\frac {b (x+1)}{a-b+1}\right ) \log (a+b x+1)\right )\) |
((Log[-((b*(1 - x))/(1 - a - b))]*Log[1 - a - b*x])/2 - (Log[(b*(1 + x))/( 1 - a + b)]*Log[1 - a - b*x])/2 + PolyLog[2, (1 - a - b*x)/(1 - a - b)]/2 - PolyLog[2, (1 - a - b*x)/(1 - a + b)]/2)/2 + (-1/2*(Log[(b*(1 - x))/(1 + a + b)]*Log[1 + a + b*x]) + (Log[-((b*(1 + x))/(1 + a - b))]*Log[1 + a + b*x])/2 + PolyLog[2, (1 + a + b*x)/(1 + a - b)]/2 - PolyLog[2, (1 + a + b* x)/(1 + a + b)]/2)/2
3.6.7.3.1 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_) + (d_.)*(x_)]/((e_) + (f_.)*(x_)^(n_.)), x_Symbol] :> Simp [1/2 Int[Log[1 + c + d*x]/(e + f*x^n), x], x] - Simp[1/2 Int[Log[1 - c - d*x]/(e + f*x^n), x], x] /; FreeQ[{c, d, e, f}, x] && RationalQ[n]
Time = 0.80 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {\ln \left (-b x -a +1\right ) \ln \left (\frac {-b x +b}{b -1+a}\right )}{4}+\frac {\operatorname {dilog}\left (\frac {-b x +b}{b -1+a}\right )}{4}-\frac {\ln \left (-b x -a +1\right ) \ln \left (\frac {-b x -b}{-b -1+a}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {-b x -b}{-b -1+a}\right )}{4}+\frac {\ln \left (b x +a +1\right ) \ln \left (\frac {b x +b}{-1+b -a}\right )}{4}+\frac {\operatorname {dilog}\left (\frac {b x +b}{-1+b -a}\right )}{4}-\frac {\ln \left (b x +a +1\right ) \ln \left (\frac {b x -b}{-1-b -a}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {b x -b}{-1-b -a}\right )}{4}\) | \(184\) |
parts | \(\operatorname {arctanh}\left (x \right ) \operatorname {arctanh}\left (b x +a \right )-b \left (\frac {\operatorname {arctanh}\left (x \right ) \ln \left (b x +a +1\right )}{2 b}-\frac {\operatorname {arctanh}\left (x \right ) \ln \left (b x +a -1\right )}{2 b}-\frac {\ln \left (\frac {b x -b}{1-a -b}\right ) \ln \left (b x +a -1\right )}{4 b}-\frac {\operatorname {dilog}\left (\frac {b x -b}{1-a -b}\right )}{4 b}+\frac {\ln \left (\frac {b x +b}{1-a +b}\right ) \ln \left (b x +a -1\right )}{4 b}+\frac {\operatorname {dilog}\left (\frac {b x +b}{1-a +b}\right )}{4 b}-\frac {\ln \left (b x +a +1\right ) \ln \left (\frac {b x +b}{-1+b -a}\right )}{4 b}-\frac {\operatorname {dilog}\left (\frac {b x +b}{-1+b -a}\right )}{4 b}+\frac {\ln \left (b x +a +1\right ) \ln \left (\frac {b x -b}{-1-b -a}\right )}{4 b}+\frac {\operatorname {dilog}\left (\frac {b x -b}{-1-b -a}\right )}{4 b}\right )\) | \(247\) |
derivativedivides | \(\frac {\frac {\operatorname {arctanh}\left (b x +a \right ) b \ln \left (-b x -b \right )}{2}-\frac {\operatorname {arctanh}\left (b x +a \right ) b \ln \left (-b x +b \right )}{2}+\frac {b^{2} \left (\frac {\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a +b}\right )}{2}+\frac {\ln \left (-b x -b \right ) \ln \left (\frac {-b x -a +1}{1-a +b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1+b -a}\right )}{2}-\frac {\ln \left (-b x -b \right ) \ln \left (\frac {-b x -a -1}{-1+b -a}\right )}{2}}{b}-\frac {\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a -b}\right )}{2}+\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -a +1}{1-a -b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-b -a}\right )}{2}-\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -a -1}{-1-b -a}\right )}{2}}{b}\right )}{2}}{b}\) | \(265\) |
default | \(\frac {\frac {\operatorname {arctanh}\left (b x +a \right ) b \ln \left (-b x -b \right )}{2}-\frac {\operatorname {arctanh}\left (b x +a \right ) b \ln \left (-b x +b \right )}{2}+\frac {b^{2} \left (\frac {\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a +b}\right )}{2}+\frac {\ln \left (-b x -b \right ) \ln \left (\frac {-b x -a +1}{1-a +b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1+b -a}\right )}{2}-\frac {\ln \left (-b x -b \right ) \ln \left (\frac {-b x -a -1}{-1+b -a}\right )}{2}}{b}-\frac {\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a -b}\right )}{2}+\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -a +1}{1-a -b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-b -a}\right )}{2}-\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -a -1}{-1-b -a}\right )}{2}}{b}\right )}{2}}{b}\) | \(265\) |
1/4*ln(-b*x-a+1)*ln((-b*x+b)/(b-1+a))+1/4*dilog((-b*x+b)/(b-1+a))-1/4*ln(- b*x-a+1)*ln((-b*x-b)/(-b-1+a))-1/4*dilog((-b*x-b)/(-b-1+a))+1/4*ln(b*x+a+1 )*ln((b*x+b)/(-1+b-a))+1/4*dilog((b*x+b)/(-1+b-a))-1/4*ln(b*x+a+1)*ln((b*x -b)/(-1-b-a))-1/4*dilog((b*x-b)/(-1-b-a))
\[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\int { -\frac {\operatorname {artanh}\left (b x + a\right )}{x^{2} - 1} \,d x } \]
\[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=- \int \frac {\operatorname {atanh}{\left (a + b x \right )}}{x^{2} - 1}\, dx \]
Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\frac {1}{4} \, b {\left (\frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{a + b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{a + b + 1}\right )}{b} - \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{a + b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{a + b - 1}\right )}{b} - \frac {\log \left (x + 1\right ) \log \left (\frac {b x + b}{a - b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x + b}{a - b + 1}\right )}{b} + \frac {\log \left (x + 1\right ) \log \left (\frac {b x + b}{a - b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x + b}{a - b - 1}\right )}{b}\right )} + \frac {1}{2} \, {\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \operatorname {artanh}\left (b x + a\right ) \]
1/4*b*((log(x - 1)*log((b*x - b)/(a + b + 1) + 1) + dilog(-(b*x - b)/(a + b + 1)))/b - (log(x - 1)*log((b*x - b)/(a + b - 1) + 1) + dilog(-(b*x - b) /(a + b - 1)))/b - (log(x + 1)*log((b*x + b)/(a - b + 1) + 1) + dilog(-(b* x + b)/(a - b + 1)))/b + (log(x + 1)*log((b*x + b)/(a - b - 1) + 1) + dilo g(-(b*x + b)/(a - b - 1)))/b) + 1/2*(log(x + 1) - log(x - 1))*arctanh(b*x + a)
\[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\int { -\frac {\operatorname {artanh}\left (b x + a\right )}{x^{2} - 1} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=-\int \frac {\mathrm {atanh}\left (a+b\,x\right )}{x^2-1} \,d x \]