\(\int \frac {\text {arctanh}(a+b x)^2}{x^2} \, dx\) [6]

Optimal result

Integrand size = 12, antiderivative size = 251 \[ \int \frac {\text {arctanh}(a+b x)^2}{x^2} \, dx=-\frac {\text {arctanh}(a+b x)^2}{x}+\frac {b \text {arctanh}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{1-a}+\frac {b \text {arctanh}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1+a}-\frac {2 b \text {arctanh}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1-a^2}+\frac {2 b \text {arctanh}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{1-a^2}+\frac {b \operatorname {PolyLog}\left (2,-\frac {1+a+b x}{1-a-b x}\right )}{2 (1-a)}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+a+b x}\right )}{2 (1+a)}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+a+b x}\right )}{1-a^2}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (1+a+b x)}\right )}{1-a^2} \] Output:

-arctanh(b*x+a)^2/x+b*arctanh(b*x+a)*ln(2/(-b*x-a+1))/(1-a)+b*arctanh(b*x+ 
a)*ln(2/(b*x+a+1))/(1+a)-2*b*arctanh(b*x+a)*ln(2/(b*x+a+1))/(-a^2+1)+2*b*a 
rctanh(b*x+a)*ln(2*b*x/(1-a)/(b*x+a+1))/(-a^2+1)+b*polylog(2,-(b*x+a+1)/(- 
b*x-a+1))/(2-2*a)-b*polylog(2,1-2/(b*x+a+1))/(2+2*a)+b*polylog(2,1-2/(b*x+ 
a+1))/(-a^2+1)-b*polylog(2,1-2*b*x/(1-a)/(b*x+a+1))/(-a^2+1)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.93 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.83 \[ \int \frac {\text {arctanh}(a+b x)^2}{x^2} \, dx=\frac {-\left (\left (-a+a^3+a^2 b x+b \left (-1+\sqrt {1-a^2} e^{\text {arctanh}(a)}\right ) x\right ) \text {arctanh}(a+b x)^2\right )+a b x \text {arctanh}(a+b x) \left (-i \pi +2 \text {arctanh}(a)-2 \log \left (1-e^{2 \text {arctanh}(a)-2 \text {arctanh}(a+b x)}\right )\right )+a b x \left (i \pi \left (\log \left (1+e^{2 \text {arctanh}(a+b x)}\right )-\log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )\right )+2 \text {arctanh}(a) \left (\log \left (1-e^{2 \text {arctanh}(a)-2 \text {arctanh}(a+b x)}\right )-\log (-i \sinh (\text {arctanh}(a)-\text {arctanh}(a+b x)))\right )\right )+a b x \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}(a)-2 \text {arctanh}(a+b x)}\right )}{a \left (-1+a^2\right ) x} \] Input:

Integrate[ArcTanh[a + b*x]^2/x^2,x]
 

Output:

(-((-a + a^3 + a^2*b*x + b*(-1 + Sqrt[1 - a^2]*E^ArcTanh[a])*x)*ArcTanh[a 
+ b*x]^2) + a*b*x*ArcTanh[a + b*x]*((-I)*Pi + 2*ArcTanh[a] - 2*Log[1 - E^( 
2*ArcTanh[a] - 2*ArcTanh[a + b*x])]) + a*b*x*(I*Pi*(Log[1 + E^(2*ArcTanh[a 
 + b*x])] - Log[1/Sqrt[1 - (a + b*x)^2]]) + 2*ArcTanh[a]*(Log[1 - E^(2*Arc 
Tanh[a] - 2*ArcTanh[a + b*x])] - Log[(-I)*Sinh[ArcTanh[a] - ArcTanh[a + b* 
x]]])) + a*b*x*PolyLog[2, E^(2*ArcTanh[a] - 2*ArcTanh[a + b*x])])/(a*(-1 + 
 a^2)*x)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6659, 7292, 6671, 25, 27, 7276, 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.

\(\displaystyle \int \frac {\text {arctanh}(a+b x)^2}{x^2} \, dx\)

\(\Big \downarrow \) 6659

\(\displaystyle 2 b \int \frac {\text {arctanh}(a+b x)}{x \left (1-(a+b x)^2\right )}dx-\frac {\text {arctanh}(a+b x)^2}{x}\)

\(\Big \downarrow \) 7292

\(\displaystyle 2 b \int \frac {\text {arctanh}(a+b x)}{x \left (-a^2-2 b x a-b^2 x^2+1\right )}dx-\frac {\text {arctanh}(a+b x)^2}{x}\)

\(\Big \downarrow \) 6671

\(\displaystyle 2 \int \frac {\text {arctanh}(a+b x)}{x \left (1-(a+b x)^2\right )}d(a+b x)-\frac {\text {arctanh}(a+b x)^2}{x}\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \int -\frac {\text {arctanh}(a+b x)}{x \left (1-(a+b x)^2\right )}d(a+b x)-\frac {\text {arctanh}(a+b x)^2}{x}\)

\(\Big \downarrow \) 27

\(\displaystyle -2 b \int -\frac {\text {arctanh}(a+b x)}{b x \left (1-(a+b x)^2\right )}d(a+b x)-\frac {\text {arctanh}(a+b x)^2}{x}\)

\(\Big \downarrow \) 7276

\(\displaystyle -2 b \int \left (\frac {\text {arctanh}(a+b x)}{\left (a^2-1\right ) b x}-\frac {\text {arctanh}(a+b x)}{2 (a-1) (a+b x-1)}+\frac {\text {arctanh}(a+b x)}{2 (a+1) (a+b x+1)}\right )d(a+b x)-\frac {\text {arctanh}(a+b x)^2}{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 b \left (\frac {\text {arctanh}(a+b x) \log \left (\frac {2}{a+b x+1}\right )}{1-a^2}-\frac {\text {arctanh}(a+b x) \log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right )}{1-a^2}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )}{2 \left (1-a^2\right )}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (a+b x+1)}\right )}{2 \left (1-a^2\right )}-\frac {\text {arctanh}(a+b x) \log \left (\frac {2}{-a-b x+1}\right )}{2 (1-a)}-\frac {\text {arctanh}(a+b x) \log \left (\frac {2}{a+b x+1}\right )}{2 (a+1)}-\frac {\operatorname {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{4 (1-a)}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )}{4 (a+1)}\right )-\frac {\text {arctanh}(a+b x)^2}{x}\)

Input:

Int[ArcTanh[a + b*x]^2/x^2,x]
 

Output:

-(ArcTanh[a + b*x]^2/x) - 2*b*(-1/2*(ArcTanh[a + b*x]*Log[2/(1 - a - b*x)] 
)/(1 - a) - (ArcTanh[a + b*x]*Log[2/(1 + a + b*x)])/(2*(1 + a)) + (ArcTanh 
[a + b*x]*Log[2/(1 + a + b*x)])/(1 - a^2) - (ArcTanh[a + b*x]*Log[(2*b*x)/ 
((1 - a)*(1 + a + b*x))])/(1 - a^2) - PolyLog[2, -((1 + a + b*x)/(1 - a - 
b*x))]/(4*(1 - a)) + PolyLog[2, 1 - 2/(1 + a + b*x)]/(4*(1 + a)) - PolyLog 
[2, 1 - 2/(1 + a + b*x)]/(2*(1 - a^2)) + PolyLog[2, 1 - (2*b*x)/((1 - a)*( 
1 + a + b*x))]/(2*(1 - a^2)))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 6659

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( 
m_), x_Symbol] :> Simp[(e + f*x)^(m + 1)*((a + b*ArcTanh[c + d*x])^p/(f*(m 
+ 1))), x] - Simp[b*d*(p/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*((a + b*ArcTa 
nh[c + d*x])^(p - 1)/(1 - (c + d*x)^2)), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x] && IGtQ[p, 0] && ILtQ[m, -1]
 

rule 6671

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( 
m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/d   Sub 
st[Int[((d*e - c*f)/d + f*(x/d))^m*(-C/d^2 + (C/d^2)*x^2)^q*(a + b*ArcTanh[ 
x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x 
] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.20

method	result	size
parts	\(-\frac {\operatorname {arctanh}\left (b x +a \right )^{2}}{x}+2 b \left (-\frac {\operatorname {arctanh}\left (b x +a \right ) \ln \left (-b x \right )}{\left (a -1\right ) \left (a +1\right )}-\frac {\operatorname {arctanh}\left (b x +a \right ) \ln \left (b x +a +1\right )}{2 a +2}+\frac {\operatorname {arctanh}\left (b x +a \right ) \ln \left (b x +a -1\right )}{2 a -2}-\frac {-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-a}\right )}{2}-\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a -1}{-1-a}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a}\right )}{2}+\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a +1}{1-a}\right )}{2}}{\left (a -1\right ) \left (a +1\right )}+\frac {\frac {\ln \left (b x +a -1\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}}{2 a -2}-\frac {-\frac {\ln \left (b x +a +1\right )^{2}}{4}+\frac {\left (\ln \left (b x +a +1\right )-\ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}}{2 \left (a +1\right )}\right )\)	\(301\)
derivativedivides	\(b \left (-\frac {\operatorname {arctanh}\left (b x +a \right )^{2}}{b x}-\frac {2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (-b x \right )}{\left (a -1\right ) \left (a +1\right )}-\frac {2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (b x +a +1\right )}{2 a +2}+\frac {2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (b x +a -1\right )}{2 a -2}+\frac {\frac {\ln \left (b x +a -1\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}}{a -1}-\frac {-\frac {\ln \left (b x +a +1\right )^{2}}{4}+\frac {\left (\ln \left (b x +a +1\right )-\ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}}{a +1}-\frac {2 \left (-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-a}\right )}{2}-\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a -1}{-1-a}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a}\right )}{2}+\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a +1}{1-a}\right )}{2}\right )}{\left (a -1\right ) \left (a +1\right )}\right )\)	\(302\)
default	\(b \left (-\frac {\operatorname {arctanh}\left (b x +a \right )^{2}}{b x}-\frac {2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (-b x \right )}{\left (a -1\right ) \left (a +1\right )}-\frac {2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (b x +a +1\right )}{2 a +2}+\frac {2 \,\operatorname {arctanh}\left (b x +a \right ) \ln \left (b x +a -1\right )}{2 a -2}+\frac {\frac {\ln \left (b x +a -1\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}}{a -1}-\frac {-\frac {\ln \left (b x +a +1\right )^{2}}{4}+\frac {\left (\ln \left (b x +a +1\right )-\ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{2}}{a +1}-\frac {2 \left (-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-a}\right )}{2}-\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a -1}{-1-a}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a}\right )}{2}+\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a +1}{1-a}\right )}{2}\right )}{\left (a -1\right ) \left (a +1\right )}\right )\)	\(302\)

Input:

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

Output:

-arctanh(b*x+a)^2/x+2*b*(-arctanh(b*x+a)/(a-1)/(a+1)*ln(-b*x)-arctanh(b*x+ 
a)/(2*a+2)*ln(b*x+a+1)+arctanh(b*x+a)/(2*a-2)*ln(b*x+a-1)-1/(a-1)/(a+1)*(- 
1/2*dilog((-b*x-a-1)/(-1-a))-1/2*ln(-b*x)*ln((-b*x-a-1)/(-1-a))+1/2*dilog( 
1/(1-a)*(-b*x-a+1))+1/2*ln(-b*x)*ln(1/(1-a)*(-b*x-a+1)))+1/2/(a-1)*(1/4*ln 
(b*x+a-1)^2-1/2*dilog(1/2*b*x+1/2*a+1/2)-1/2*ln(b*x+a-1)*ln(1/2*b*x+1/2*a+ 
1/2))-1/2/(a+1)*(-1/4*ln(b*x+a+1)^2+1/2*(ln(b*x+a+1)-ln(1/2*b*x+1/2*a+1/2) 
)*ln(-1/2*b*x-1/2*a+1/2)-1/2*dilog(1/2*b*x+1/2*a+1/2)))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.97 \[ \int \frac {\text {arctanh}(a+b x)^2}{x^2} \, dx=\frac {1}{4} \, b^{2} {\left (\frac {{\left (a - 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a - 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a + 1\right )} \log \left (b x + a - 1\right )^{2}}{a^{2} b - b} - \frac {4 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )}}{a^{2} b - b} + \frac {4 \, {\left (\log \left (\frac {b x}{a + 1} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a + 1}\right )\right )}}{a^{2} b - b} - \frac {4 \, {\left (\log \left (\frac {b x}{a - 1} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a - 1}\right )\right )}}{a^{2} b - b}\right )} - b {\left (\frac {\log \left (b x + a + 1\right )}{a + 1} - \frac {\log \left (b x + a - 1\right )}{a - 1} + \frac {2 \, \log \left (x\right )}{a^{2} - 1}\right )} \operatorname {artanh}\left (b x + a\right ) - \frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x} \] Input:

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

Output:

1/4*b^2*(((a - 1)*log(b*x + a + 1)^2 - 2*(a - 1)*log(b*x + a + 1)*log(b*x 
+ a - 1) + (a + 1)*log(b*x + a - 1)^2)/(a^2*b - b) - 4*(log(b*x + a - 1)*l 
og(1/2*b*x + 1/2*a + 1/2) + dilog(-1/2*b*x - 1/2*a + 1/2))/(a^2*b - b) + 4 
*(log(b*x/(a + 1) + 1)*log(x) + dilog(-b*x/(a + 1)))/(a^2*b - b) - 4*(log( 
b*x/(a - 1) + 1)*log(x) + dilog(-b*x/(a - 1)))/(a^2*b - b)) - b*(log(b*x + 
 a + 1)/(a + 1) - log(b*x + a - 1)/(a - 1) + 2*log(x)/(a^2 - 1))*arctanh(b 
*x + a) - arctanh(b*x + a)^2/x
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\text {arctanh}(a+b x)^2}{x^2} \, dx=\frac {-2 \mathit {atanh} \left (b x +a \right )^{2} a^{3}-\mathit {atanh} \left (b x +a \right )^{2} a^{2} b x +2 \mathit {atanh} \left (b x +a \right )^{2} a +\mathit {atanh} \left (b x +a \right )^{2} b x +2 \mathit {atanh} \left (b x +a \right ) a^{2}+2 \mathit {atanh} \left (b x +a \right ) a b x -2 \mathit {atanh} \left (b x +a \right ) b x -2 \mathit {atanh} \left (b x +a \right )+2 \left (\int \frac {\mathit {atanh} \left (b x +a \right )}{b^{2} x^{4}+2 a b \,x^{3}+a^{2} x^{2}-x^{2}}d x \right ) a^{4} x -4 \left (\int \frac {\mathit {atanh} \left (b x +a \right )}{b^{2} x^{4}+2 a b \,x^{3}+a^{2} x^{2}-x^{2}}d x \right ) a^{2} x +2 \left (\int \frac {\mathit {atanh} \left (b x +a \right )}{b^{2} x^{4}+2 a b \,x^{3}+a^{2} x^{2}-x^{2}}d x \right ) x -2 \,\mathrm {log}\left (b x +a -1\right ) b x +2 \,\mathrm {log}\left (x \right ) b x}{2 a x \left (a^{2}-1\right )} \] Input:

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

Output:

( - 2*atanh(a + b*x)**2*a**3 - atanh(a + b*x)**2*a**2*b*x + 2*atanh(a + b* 
x)**2*a + atanh(a + b*x)**2*b*x + 2*atanh(a + b*x)*a**2 + 2*atanh(a + b*x) 
*a*b*x - 2*atanh(a + b*x)*b*x - 2*atanh(a + b*x) + 2*int(atanh(a + b*x)/(a 
**2*x**2 + 2*a*b*x**3 + b**2*x**4 - x**2),x)*a**4*x - 4*int(atanh(a + b*x) 
/(a**2*x**2 + 2*a*b*x**3 + b**2*x**4 - x**2),x)*a**2*x + 2*int(atanh(a + b 
*x)/(a**2*x**2 + 2*a*b*x**3 + b**2*x**4 - x**2),x)*x - 2*log(a + b*x - 1)* 
b*x + 2*log(x)*b*x)/(2*a*x*(a**2 - 1))