\(\int \frac {\coth (c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [186]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^2 d}+\frac {\log (\tanh (c+d x))}{a^2 d}-\frac {b (2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 (a+b)^2 d}+\frac {b}{2 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \] Output:

ln(cosh(d*x+c))/(a+b)^2/d+ln(tanh(d*x+c))/a^2/d-1/2*b*(2*a+b)*ln(a+b*tanh( 
d*x+c)^2)/a^2/(a+b)^2/d+1/2*b/a/(a+b)/d/(a+b*tanh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.87 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {2 \log (\cosh (c+d x))}{(a+b)^2}+\frac {2 \log (\tanh (c+d x))+\frac {b \left (-\left ((2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )\right )+\frac {a (a+b)}{a+b \tanh ^2(c+d x)}\right )}{(a+b)^2}}{a^2}}{2 d} \] Input:

Integrate[Coth[c + d*x]/(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

((2*Log[Cosh[c + d*x]])/(a + b)^2 + (2*Log[Tanh[c + d*x]] + (b*(-((2*a + b 
)*Log[a + b*Tanh[c + d*x]^2]) + (a*(a + b))/(a + b*Tanh[c + d*x]^2)))/(a + 
 b)^2)/a^2)/(2*d)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 4153, 26, 354, 93, 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.

\(\displaystyle \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i}{\tan (i c+i d x) \left (a-b \tan (i c+i d x)^2\right )^2}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {1}{\tan (i c+i d x) \left (a-b \tan (i c+i d x)^2\right )^2}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \frac {i \int -\frac {i \coth (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\int \frac {\coth (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {\int \frac {\coth (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 93

\(\displaystyle \frac {\int \left (-\frac {(2 a+b) b^2}{a^2 (a+b)^2 \left (b \tanh ^2(c+d x)+a\right )}-\frac {b^2}{a (a+b) \left (b \tanh ^2(c+d x)+a\right )^2}+\frac {\coth (c+d x)}{a^2}-\frac {1}{(a+b)^2 \left (\tanh ^2(c+d x)-1\right )}\right )d\tanh ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 2009

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

Input:

Int[Coth[c + d*x]/(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

(Log[Tanh[c + d*x]^2]/a^2 - Log[1 - Tanh[c + d*x]^2]/(a + b)^2 - (b*(2*a + 
 b)*Log[a + b*Tanh[c + d*x]^2])/(a^2*(a + b)^2) + b/(a*(a + b)*(a + b*Tanh 
[c + d*x]^2)))/(2*d)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 93

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre 
eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(-\frac {\frac {b^{2} \left (\frac {\left (2 a +b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}-\frac {a \left (a +b \right )}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2} a^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )\right )}{a^{2}}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(110\)
default	\(-\frac {\frac {b^{2} \left (\frac {\left (2 a +b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}-\frac {a \left (a +b \right )}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2} a^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )\right )}{a^{2}}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(110\)
parallelrisch	\(\frac {-2 \left (a +\frac {b}{2}\right ) \left (a +b \tanh \left (d x +c \right )^{2}\right ) b \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )+\left (-2 \tanh \left (d x +c \right )^{2} a^{2} b -2 a^{3}\right ) \ln \left (1-\tanh \left (d x +c \right )\right )+2 \left (a +b \right )^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\left (-2 a^{2} b d x -a \,b^{2}-b^{3}\right ) \tanh \left (d x +c \right )^{2}-2 a^{3} d x}{2 \left (a +b \tanh \left (d x +c \right )^{2}\right ) d \left (a +b \right )^{2} a^{2}}\)	\(156\)
risch	\(\frac {x}{a^{2}+2 a b +b^{2}}-\frac {2 x}{a^{2}}-\frac {2 c}{a^{2} d}+\frac {4 b x}{a \left (a^{2}+2 a b +b^{2}\right )}+\frac {4 b c}{a d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 b^{2} x}{a^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 b^{2} c}{a^{2} d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 b^{2} {\mathrm e}^{2 d x +2 c}}{a d \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{a^{2} d}-\frac {b \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{a d \left (a^{2}+2 a b +b^{2}\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 a^{2} d \left (a^{2}+2 a b +b^{2}\right )}\)	\(329\)

Input:

int(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/d*(1/2*b^2/(a+b)^2/a^2*((2*a+b)/b*ln(a+b*tanh(d*x+c)^2)-a*(a+b)/b/(a+b* 
tanh(d*x+c)^2))-1/a^2*ln(tanh(d*x+c))+1/2/(a+b)^2*ln(1+tanh(d*x+c))+1/2/(a 
+b)^2*ln(-1+tanh(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1148 vs. \(2 (91) = 182\).

Time = 0.23 (sec) , antiderivative size = 1148, normalized size of antiderivative = 12.08 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

-1/2*(2*(a^3 + a^2*b)*d*x*cosh(d*x + c)^4 + 8*(a^3 + a^2*b)*d*x*cosh(d*x + 
 c)*sinh(d*x + c)^3 + 2*(a^3 + a^2*b)*d*x*sinh(d*x + c)^4 + 2*(a^3 + a^2*b 
)*d*x - 4*(a*b^2 - (a^3 - a^2*b)*d*x)*cosh(d*x + c)^2 + 4*(3*(a^3 + a^2*b) 
*d*x*cosh(d*x + c)^2 - a*b^2 + (a^3 - a^2*b)*d*x)*sinh(d*x + c)^2 + ((2*a^ 
2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 4*(2*a^2*b + 3*a*b^2 + b^3)*cosh(d* 
x + c)*sinh(d*x + c)^3 + (2*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^4 + 2*a^2 
*b + 3*a*b^2 + b^3 + 2*(2*a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(2*a^2* 
b - a*b^2 - b^3 + 3*(2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + 
c)^2 + 4*((2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (2*a^2*b - a*b^2 - b 
^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*((a + b)*cosh(d*x + c)^2 + (a + b) 
*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) 
 + sinh(d*x + c)^2)) - 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 
+ 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 
 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 
+ 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b - a*b^2 - 
 b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 
+ 4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^ 
2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) 
- sinh(d*x + c))) + 8*((a^3 + a^2*b)*d*x*cosh(d*x + c)^3 - (a*b^2 - (a^3 - 
 a^2*b)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^5 + 3*a^4*b + 3*a^3*b^2 ...
 

Sympy [F]

\[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {\coth {\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)**2)**2,x)
 

Output:

Integral(coth(c + d*x)/(a + b*tanh(c + d*x)**2)**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (91) = 182\).

Time = 0.06 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.47 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {2 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} + 2 \, {\left (a^{4} + a^{3} b - a^{2} b^{2} - a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {{\left (2 \, a b + b^{2}\right )} \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

2*b^2*e^(-2*d*x - 2*c)/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 + 2*(a^4 + a^3* 
b - a^2*b^2 - a*b^3)*e^(-2*d*x - 2*c) + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 
)*e^(-4*d*x - 4*c))*d) - 1/2*(2*a*b + b^2)*log(2*(a - b)*e^(-2*d*x - 2*c) 
+ (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^4 + 2*a^3*b + a^2*b^2)*d) + (d*x + 
 c)/((a^2 + 2*a*b + b^2)*d) + log(e^(-d*x - c) + 1)/(a^2*d) + log(e^(-d*x 
- c) - 1)/(a^2*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (91) = 182\).

Time = 0.18 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.05 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (2 \, a b + b^{2}\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{a^{4} + 2 \, a^{3} b + a^{2} b^{2}} + \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} - \frac {4 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} {\left (a + b\right )}^{2} a} - \frac {2 \, \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a^{2}}}{2 \, d} \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(coth(c + d*x)/(a + b*tanh(c + d*x)^2)^2,x)
 

Output:

int(coth(c + d*x)/(a + b*tanh(c + d*x)^2)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1760, normalized size of antiderivative = 18.53 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x)
 

Output:

(2*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*a**4 + 4*e**(4*c + 4*d*x)*log(e* 
*(c + d*x) - 1)*a**3*b - 4*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*a*b**3 - 
 2*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*b**4 + 2*e**(4*c + 4*d*x)*log(e* 
*(c + d*x) + 1)*a**4 + 4*e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*a**3*b - 4 
*e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*a*b**3 - 2*e**(4*c + 4*d*x)*log(e* 
*(c + d*x) + 1)*b**4 - 2*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) 
 + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a**3*b - e**(4*c + 4*d*x)*log(e** 
(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a**2*b** 
2 + 2*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2* 
e**(c + d*x)*sqrt(b))*a*b**3 + e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt( 
a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*b**4 - 2*e**(4*c + 4*d*x)*l 
og(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*a* 
*3*b - e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2 
*e**(c + d*x)*sqrt(b))*a**2*b**2 + 2*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) 
*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*a*b**3 + e**(4*c + 4* 
d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt( 
b))*b**4 - 2*e**(4*c + 4*d*x)*a**4*d*x + 2*e**(4*c + 4*d*x)*a**2*b**2*d*x 
- 2*e**(4*c + 4*d*x)*a**2*b**2 - 2*e**(4*c + 4*d*x)*a*b**3 + 4*e**(2*c + 2 
*d*x)*log(e**(c + d*x) - 1)*a**4 - 8*e**(2*c + 2*d*x)*log(e**(c + d*x) - 1 
)*a**2*b**2 + 4*e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*b**4 + 4*e**(2*c...