3.6.7 \(\int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx\) [507]

3.6.7.1 Optimal result

Integrand size = 16, antiderivative size = 203 \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a+b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a+b}\right ) \]

output

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))
 

3.6.7.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-a-b x}{1-a+b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a-b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+a+b x}{1+a+b}\right ) \]

input

Integrate[ArcTanh[a + b*x]/(1 - x^2),x]
 

output

(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
 

3.6.7.3 Rubi [A] (verified)

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

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {-a-b x+1}{-a-b+1}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {-a-b x+1}{-a+b+1}\right )+\frac {1}{2} \log \left (-\frac {b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac {1}{2} \log \left (\frac {b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)\right )+\frac {1}{2} \left (\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {a+b x+1}{a-b+1}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {a+b x+1}{a+b+1}\right )-\frac {1}{2} \log \left (\frac {b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac {1}{2} \log \left (-\frac {b (x+1)}{a-b+1}\right ) \log (a+b x+1)\right )\)

input

Int[ArcTanh[a + b*x]/(1 - x^2),x]
 

output

((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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2856

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]))
 

rule 6665

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]
 

3.6.7.4 Maple [A] (verified)

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	\(\frac {\frac {\operatorname {arctanh}\left (b x +a \right ) b \ln \left (-b x -b \right )}{2}-\frac {\operatorname {arctanh}\left (b x +a \right ) b \ln \left (-b x +b \right )}{2}+\frac {b^{2} \left (\frac {\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a +b}\right )}{2}+\frac {\ln \left (-b x -b \right ) \ln \left (\frac {-b x -a +1}{1-a +b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1+b -a}\right )}{2}-\frac {\ln \left (-b x -b \right ) \ln \left (\frac {-b x -a -1}{-1+b -a}\right )}{2}}{b}-\frac {\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a -b}\right )}{2}+\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -a +1}{1-a -b}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-b -a}\right )}{2}-\frac {\ln \left (-b x +b \right ) \ln \left (\frac {-b x -a -1}{-1-b -a}\right )}{2}}{b}\right )}{2}}{b}\)	\(265\)

input

int(arctanh(b*x+a)/(-x^2+1),x,method=_RETURNVERBOSE)
 

output

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))
 

3.6.7.5 Fricas [F]

\[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\int { -\frac {\operatorname {artanh}\left (b x + a\right )}{x^{2} - 1} \,d x } \]

input

integrate(arctanh(b*x+a)/(-x^2+1),x, algorithm="fricas")
 

output

integral(-arctanh(b*x + a)/(x^2 - 1), x)
 

3.6.7.6 Sympy [F]

\[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=- \int \frac {\operatorname {atanh}{\left (a + b x \right )}}{x^{2} - 1}\, dx \]

input

integrate(atanh(b*x+a)/(-x**2+1),x)
 

output

-Integral(atanh(a + b*x)/(x**2 - 1), x)
 

3.6.7.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\frac {1}{4} \, b {\left (\frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{a + b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{a + b + 1}\right )}{b} - \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{a + b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{a + b - 1}\right )}{b} - \frac {\log \left (x + 1\right ) \log \left (\frac {b x + b}{a - b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x + b}{a - b + 1}\right )}{b} + \frac {\log \left (x + 1\right ) \log \left (\frac {b x + b}{a - b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x + b}{a - b - 1}\right )}{b}\right )} + \frac {1}{2} \, {\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \operatorname {artanh}\left (b x + a\right ) \]

input

integrate(arctanh(b*x+a)/(-x^2+1),x, algorithm="maxima")
 

output

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)
 

3.6.7.8 Giac [F]

\[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=\int { -\frac {\operatorname {artanh}\left (b x + a\right )}{x^{2} - 1} \,d x } \]

input

integrate(arctanh(b*x+a)/(-x^2+1),x, algorithm="giac")
 

output

integrate(-arctanh(b*x + a)/(x^2 - 1), x)
 

3.6.7.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(a+b x)}{1-x^2} \, dx=-\int \frac {\mathrm {atanh}\left (a+b\,x\right )}{x^2-1} \,d x \]

input

int(-atanh(a + b*x)/(x^2 - 1),x)
 

output

-int(atanh(a + b*x)/(x^2 - 1), x)