\(\int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx\) [182]

Optimal result

Integrand size = 13, antiderivative size = 131 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {3 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2202, 2194, 2191, 2188, 29} \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {3 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4} \]

[In]

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

[Out]

(-3*b)/(2*(b*x - ArcCoth[Tanh[a + b*x]])^2*ArcCoth[Tanh[a + b*x]]^2) + 1/(x*(b*x - ArcCoth[Tanh[a + b*x]])*Arc
Coth[Tanh[a + b*x]]^2) + (3*b)/((b*x - ArcCoth[Tanh[a + b*x]])^3*ArcCoth[Tanh[a + b*x]]) - (3*b*Log[x])/(b*x -
ArcCoth[Tanh[a + b*x]])^4 + (3*b*Log[ArcCoth[Tanh[a + b*x]]])/(b*x - ArcCoth[Tanh[a + b*x]])^4

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2188

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

Rule 2191

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

Rule 2194

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] - Dist[a*((n + 1)/((n + 1)*(b*u - a*v))), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rule 2202

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}-\frac {(3 b) \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^3} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))} \\ & = -\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {(3 b) \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )} \\ & = -\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {(3 b) \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \\ & = -\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {(3 b) \int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\left (3 b^2\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \\ & = -\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \\ & = -\frac {3 b}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))^2}+\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}-\frac {3 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {3 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {b^3 x^3-6 b^2 x^2 \coth ^{-1}(\tanh (a+b x))+2 \coth ^{-1}(\tanh (a+b x))^3+3 b x \coth ^{-1}(\tanh (a+b x))^2 \left (1+2 \log (x)-2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )}{2 x \coth ^{-1}(\tanh (a+b x))^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^4} \]

[In]

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

[Out]

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

Maple [F(-1)]

Timed out.

\[\int \frac {1}{x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.60 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {8 \, {\left (24 \, a b^{2} x^{2} + 36 \, a^{2} b x - i \, \pi ^{3} - 3 \, \pi ^{2} {\left (3 \, b x + 2 \, a\right )} + 8 \, a^{3} + 12 i \, \pi {\left (b^{2} x^{2} + 3 \, a b x + a^{2}\right )} - 6 \, {\left (4 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} - \pi ^{2} b x + 4 \, a^{2} b x + 4 i \, \pi {\left (b^{2} x^{2} + a b x\right )}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right ) + 6 \, {\left (4 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} - \pi ^{2} b x + 4 \, a^{2} b x + 4 i \, \pi {\left (b^{2} x^{2} + a b x\right )}\right )} \log \left (x\right )\right )}}{64 \, a^{4} b^{2} x^{3} + 128 \, a^{5} b x^{2} - \pi ^{6} x + 64 \, a^{6} x + 4 i \, \pi ^{5} {\left (b x^{2} + 3 \, a x\right )} + 4 \, \pi ^{4} {\left (b^{2} x^{3} + 10 \, a b x^{2} + 15 \, a^{2} x\right )} - 32 i \, \pi ^{3} {\left (a b^{2} x^{3} + 5 \, a^{2} b x^{2} + 5 \, a^{3} x\right )} - 16 \, \pi ^{2} {\left (6 \, a^{2} b^{2} x^{3} + 20 \, a^{3} b x^{2} + 15 \, a^{4} x\right )} + 64 i \, \pi {\left (2 \, a^{3} b^{2} x^{3} + 5 \, a^{4} b x^{2} + 3 \, a^{5} x\right )}} \]

[In]

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

[Out]

-8*(24*a*b^2*x^2 + 36*a^2*b*x - I*pi^3 - 3*pi^2*(3*b*x + 2*a) + 8*a^3 + 12*I*pi*(b^2*x^2 + 3*a*b*x + a^2) - 6*
(4*b^3*x^3 + 8*a*b^2*x^2 - pi^2*b*x + 4*a^2*b*x + 4*I*pi*(b^2*x^2 + a*b*x))*log(I*pi + 2*b*x + 2*a) + 6*(4*b^3
*x^3 + 8*a*b^2*x^2 - pi^2*b*x + 4*a^2*b*x + 4*I*pi*(b^2*x^2 + a*b*x))*log(x))/(64*a^4*b^2*x^3 + 128*a^5*b*x^2
- pi^6*x + 64*a^6*x + 4*I*pi^5*(b*x^2 + 3*a*x) + 4*pi^4*(b^2*x^3 + 10*a*b*x^2 + 15*a^2*x) - 32*I*pi^3*(a*b^2*x
^3 + 5*a^2*b*x^2 + 5*a^3*x) - 16*pi^2*(6*a^2*b^2*x^3 + 20*a^3*b*x^2 + 15*a^4*x) + 64*I*pi*(2*a^3*b^2*x^3 + 5*a
^4*b*x^2 + 3*a^5*x))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.87 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.87 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {48 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} - \frac {48 \, b \log \left (x\right )}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} - \frac {8 \, {\left (12 \, b^{2} x^{2} - \pi ^{2} - 4 i \, \pi a + 4 \, a^{2} - 9 \, {\left (i \, \pi b - 2 \, a b\right )} x\right )}}{4 \, {\left (i \, \pi ^{3} b^{2} - 6 \, \pi ^{2} a b^{2} - 12 i \, \pi a^{2} b^{2} + 8 \, a^{3} b^{2}\right )} x^{3} + 4 \, {\left (\pi ^{4} b + 8 i \, \pi ^{3} a b - 24 \, \pi ^{2} a^{2} b - 32 i \, \pi a^{3} b + 16 \, a^{4} b\right )} x^{2} - {\left (i \, \pi ^{5} - 10 \, \pi ^{4} a - 40 i \, \pi ^{3} a^{2} + 80 \, \pi ^{2} a^{3} + 80 i \, \pi a^{4} - 32 \, a^{5}\right )} x} \]

[In]

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

[Out]

48*b*log(-I*pi + 2*b*x + 2*a)/(pi^4 + 8*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi*a^3 + 16*a^4) - 48*b*log(x)/(pi^4 + 8
*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi*a^3 + 16*a^4) - 8*(12*b^2*x^2 - pi^2 - 4*I*pi*a + 4*a^2 - 9*(I*pi*b - 2*a*b)
*x)/(4*(I*pi^3*b^2 - 6*pi^2*a*b^2 - 12*I*pi*a^2*b^2 + 8*a^3*b^2)*x^3 + 4*(pi^4*b + 8*I*pi^3*a*b - 24*pi^2*a^2*
b - 32*I*pi*a^3*b + 16*a^4*b)*x^2 - (I*pi^5 - 10*pi^4*a - 40*I*pi^3*a^2 + 80*pi^2*a^3 + 80*I*pi*a^4 - 32*a^5)*
x)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.97 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {48 \, b \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{4} - 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} + 32 i \, \pi a^{3} + 16 \, a^{4}} - \frac {48 \, b \log \left (x\right )}{\pi ^{4} - 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} + 32 i \, \pi a^{3} + 16 \, a^{4}} + \frac {16 \, {\left (8 \, b^{2} x + 5 i \, \pi b + 10 \, a b\right )}}{8 i \, \pi ^{3} b^{2} x^{2} + 48 \, \pi ^{2} a b^{2} x^{2} - 96 i \, \pi a^{2} b^{2} x^{2} - 64 \, a^{3} b^{2} x^{2} - 8 \, \pi ^{4} b x + 64 i \, \pi ^{3} a b x + 192 \, \pi ^{2} a^{2} b x - 256 i \, \pi a^{3} b x - 128 \, a^{4} b x - 2 i \, \pi ^{5} - 20 \, \pi ^{4} a + 80 i \, \pi ^{3} a^{2} + 160 \, \pi ^{2} a^{3} - 160 i \, \pi a^{4} - 64 \, a^{5}} + \frac {8}{i \, \pi ^{3} x + 6 \, \pi ^{2} a x - 12 i \, \pi a^{2} x - 8 \, a^{3} x} \]

[In]

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

[Out]

48*b*log(I*pi + 2*b*x + 2*a)/(pi^4 - 8*I*pi^3*a - 24*pi^2*a^2 + 32*I*pi*a^3 + 16*a^4) - 48*b*log(x)/(pi^4 - 8*
I*pi^3*a - 24*pi^2*a^2 + 32*I*pi*a^3 + 16*a^4) + 16*(8*b^2*x + 5*I*pi*b + 10*a*b)/(8*I*pi^3*b^2*x^2 + 48*pi^2*
a*b^2*x^2 - 96*I*pi*a^2*b^2*x^2 - 64*a^3*b^2*x^2 - 8*pi^4*b*x + 64*I*pi^3*a*b*x + 192*pi^2*a^2*b*x - 256*I*pi*
a^3*b*x - 128*a^4*b*x - 2*I*pi^5 - 20*pi^4*a + 80*I*pi^3*a^2 + 160*pi^2*a^3 - 160*I*pi*a^4 - 64*a^5) + 8/(I*pi
^3*x + 6*pi^2*a*x - 12*I*pi*a^2*x - 8*a^3*x)

Mupad [B] (verification not implemented)

Time = 8.29 (sec) , antiderivative size = 1074, normalized size of antiderivative = 8.20 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \]

[In]

int(1/(x^2*acoth(tanh(a + b*x))^3),x)

[Out]

(8/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x) - (72*
b*x)/((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*
x)^2 - 4*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) +
2*b*x) + 4*a^2) + (96*b^2*x^2)/((log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex
p(2*b*x) - 1)) + 2*b*x)*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(
2*b*x) - 1)) + 2*b*x)^2 - 4*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)
*exp(2*b*x) - 1)) + 2*b*x) + 4*a^2)))/(x*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(
-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 - 4*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) +
log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x) + 4*a^2) + x^2*(8*a*b - 4*b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)) + 4*b^2*x^3) + (96*b*atanh(((2*a - log((
2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^4 + 24*a^2*(2*a
- log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 + 16*
a^4 - 8*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) +
2*b*x)^3 - 32*a^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x)
- 1)) + 2*b*x))/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) +
2*b*x)^4 - (4*b*x*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x)
- 1)) + 2*b*x)^2 - 4*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2
*b*x) - 1)) + 2*b*x) + 4*a^2))/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(
2*b*x) - 1)) + 2*b*x)^3))/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x
) - 1)) + 2*b*x)^4