3.2.1 \(\int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx\) [101]

3.2.1.1 Optimal result

Integrand size = 13, antiderivative size = 143 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {3 b^2}{(b x-\text {arctanh}(\tanh (a+b x)))^3 \text {arctanh}(\tanh (a+b x))}+\frac {3 b}{2 x (b x-\text {arctanh}(\tanh (a+b x)))^2 \text {arctanh}(\tanh (a+b x))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{(b x-\text {arctanh}(\tanh (a+b x)))^4}-\frac {3 b^2 \log (\text {arctanh}(\tanh (a+b x)))}{(b x-\text {arctanh}(\tanh (a+b x)))^4} \]

output

-3*b^2/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))+3/2*b/x/(b*x-arct 
anh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))+1/2/x^2/(b*x-arctanh(tanh(b*x+a)) 
)/arctanh(tanh(b*x+a))+3*b^2*ln(x)/(b*x-arctanh(tanh(b*x+a)))^4-3*b^2*ln(a 
rctanh(tanh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))^4
 

3.2.1.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {2 b^3 x^3-6 b x \text {arctanh}(\tanh (a+b x))^2+\text {arctanh}(\tanh (a+b x))^3-3 b^2 x^2 \text {arctanh}(\tanh (a+b x)) (-1+2 \log (x)-2 \log (\text {arctanh}(\tanh (a+b x))))}{2 x^2 \text {arctanh}(\tanh (a+b x)) (-b x+\text {arctanh}(\tanh (a+b x)))^4} \]

input

Integrate[1/(x^3*ArcTanh[Tanh[a + b*x]]^2),x]
 

output

-1/2*(2*b^3*x^3 - 6*b*x*ArcTanh[Tanh[a + b*x]]^2 + ArcTanh[Tanh[a + b*x]]^ 
3 - 3*b^2*x^2*ArcTanh[Tanh[a + b*x]]*(-1 + 2*Log[x] - 2*Log[ArcTanh[Tanh[a 
 + b*x]]]))/(x^2*ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^ 
4)
 

3.2.1.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.31, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2602, 2602, 2594, 2591, 14, 2588, 14}

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.

\(\displaystyle \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx\)

\(\Big \downarrow \) 2602

\(\displaystyle \frac {3 b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))^2}dx}{2 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\)

\(\Big \downarrow \) 2602

\(\displaystyle \frac {3 b \left (\frac {2 b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{2 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\)

\(\Big \downarrow \) 2594

\(\displaystyle \frac {3 b \left (\frac {2 b \left (-\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{2 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\)

\(\Big \downarrow \) 2591

\(\displaystyle \frac {3 b \left (\frac {2 b \left (-\frac {\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\int \frac {1}{x}dx}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{2 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {3 b \left (\frac {2 b \left (-\frac {\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{2 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\)

\(\Big \downarrow \) 2588

\(\displaystyle \frac {3 b \left (\frac {2 b \left (-\frac {\frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\right )}{2 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}+\frac {3 b \left (\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}+\frac {2 b \left (-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}-\frac {\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{2 (b x-\text {arctanh}(\tanh (a+b x)))}\)

input

Int[1/(x^3*ArcTanh[Tanh[a + b*x]]^2),x]
 

output

1/(2*x^2*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]) + (3*b*(1/ 
(x*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]) + (2*b*(-(1/((b* 
x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]])) - (-(Log[x]/(b*x - Ar 
cTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/(b*x - ArcTanh[Tanh[a 
 + b*x]]))/(b*x - ArcTanh[Tanh[a + b*x]])))/(b*x - ArcTanh[Tanh[a + b*x]]) 
))/(2*(b*x - ArcTanh[Tanh[a + b*x]]))
 

3.2.1.3.1 Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 2588

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c   Subst 
[Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, x]
 

rule 2591

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D 
[v, x]]}, Simp[b/(b*u - a*v)   Int[1/v, x], x] - Simp[a/(b*u - a*v)   Int[1 
/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]
 

rule 2594

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ 
D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 
 1)*(b*u - a*v)))   Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew 
iseLinearQ[u, v, x] && LtQ[n, -1]
 

rule 2602

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp 
lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] + S 
imp[b*((m + n + 2)/((m + 1)*(b*u - a*v)))   Int[u^(m + 1)*v^n, x], x] /; Ne 
Q[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m 
, -1]
 

3.2.1.4 Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81

\[-\frac {3 b^{2} \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}+\frac {b^{2}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}-\frac {1}{2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{2}}+\frac {3 b^{2} \ln \left (x \right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}+\frac {2 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x}\]

input

int(1/x^3/arctanh(tanh(b*x+a))^2,x)
 

output

-3/(arctanh(tanh(b*x+a))-b*x)^4*b^2*ln(arctanh(tanh(b*x+a)))+1/(arctanh(ta 
nh(b*x+a))-b*x)^3*b^2/arctanh(tanh(b*x+a))-1/2/(arctanh(tanh(b*x+a))-b*x)^ 
2/x^2+3/(arctanh(tanh(b*x+a))-b*x)^4*b^2*ln(x)+2/(arctanh(tanh(b*x+a))-b*x 
)^3*b/x
 

3.2.1.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {6 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3} - 6 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}} \]

input

integrate(1/x^3/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")
 

output

1/2*(6*a*b^2*x^2 + 3*a^2*b*x - a^3 - 6*(b^3*x^3 + a*b^2*x^2)*log(b*x + a) 
+ 6*(b^3*x^3 + a*b^2*x^2)*log(x))/(a^4*b*x^3 + a^5*x^2)
 

3.2.1.6 Sympy [F]

\[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {1}{x^{3} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

input

integrate(1/x**3/atanh(tanh(b*x+a))**2,x)
 

output

Integral(1/(x**3*atanh(tanh(a + b*x))**2), x)
 

3.2.1.7 Maxima [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {6 \, b^{2} x^{2} + 3 \, a b x - a^{2}}{2 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}} - \frac {3 \, b^{2} \log \left (b x + a\right )}{a^{4}} + \frac {3 \, b^{2} \log \left (x\right )}{a^{4}} \]

input

integrate(1/x^3/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")
 

output

1/2*(6*b^2*x^2 + 3*a*b*x - a^2)/(a^3*b*x^3 + a^4*x^2) - 3*b^2*log(b*x + a) 
/a^4 + 3*b^2*log(x)/a^4
 

3.2.1.8 Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {3 \, b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{4}} + \frac {3 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {6 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3}}{2 \, {\left (b x + a\right )} a^{4} x^{2}} \]

input

integrate(1/x^3/arctanh(tanh(b*x+a))^2,x, algorithm="giac")
 

output

-3*b^2*log(abs(b*x + a))/a^4 + 3*b^2*log(abs(x))/a^4 + 1/2*(6*a*b^2*x^2 + 
3*a^2*b*x - a^3)/((b*x + a)*a^4*x^2)
 

3.2.1.9 Mupad [B] (verification not implemented)

Time = 5.88 (sec) , antiderivative size = 660, normalized size of antiderivative = 4.62 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {6\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-6\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3-2\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3-32\,b^3\,x^3+24\,b\,x\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2+24\,b^2\,x^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-24\,b^2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+24\,b\,x\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-48\,b\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+b^2\,x^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,96{}\mathrm {i}-b^2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,96{}\mathrm {i}}{x^2\,\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4} \]

input

int(1/(x^3*atanh(tanh(a + b*x))^2),x)
 

output

-(6*log(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*e 
xp(2*b*x) + 1))^2 - 6*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log((exp(2*a)*exp 
(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*log(1/(exp(2*a)*exp(2*b*x) + 1))^3 
 - 2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3 - 32*b^3*x^3 + 
 24*b*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 + 24*b^2*x^ 
2*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 24*b^2*x^2*log((exp(2*a)*exp(2*b*x))/ 
(exp(2*a)*exp(2*b*x) + 1)) + 24*b*x*log(1/(exp(2*a)*exp(2*b*x) + 1))^2 + b 
^2*x^2*log(1/(exp(2*a)*exp(2*b*x) + 1))*atan((log((exp(2*a)*exp(2*b*x))/(e 
xp(2*a)*exp(2*b*x) + 1))*1i - log(1/(exp(2*a)*exp(2*b*x) + 1))*1i + b*x*2i 
)/(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)* 
exp(2*b*x) + 1)) + 2*b*x))*96i - b^2*x^2*log((exp(2*a)*exp(2*b*x))/(exp(2* 
a)*exp(2*b*x) + 1))*atan((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 
 1))*1i - log(1/(exp(2*a)*exp(2*b*x) + 1))*1i + b*x*2i)/(log(1/(exp(2*a)*e 
xp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2 
*b*x))*96i - 48*b*x*log(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b 
*x))/(exp(2*a)*exp(2*b*x) + 1)))/(x^2*(log(1/(exp(2*a)*exp(2*b*x) + 1)) - 
log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))*(log(1/(exp(2*a)*exp 
(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b 
*x)^4)