\(\int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) [489]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b \text {csch}(c+d x)}{a^2 d}-\frac {\text {csch}^2(c+d x)}{2 a d}+\frac {\left (a^2+b^2\right ) \log (\sinh (c+d x))}{a^3 d}-\frac {\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^3 d} \] Output:

b*csch(d*x+c)/a^2/d-1/2*csch(d*x+c)^2/a/d+(a^2+b^2)*ln(sinh(d*x+c))/a^3/d- 
(a^2+b^2)*ln(a+b*sinh(d*x+c))/a^3/d
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 a b \text {csch}(c+d x)-a^2 \text {csch}^2(c+d x)+2 \left (a^2+b^2\right ) (\log (\sinh (c+d x))-\log (a+b \sinh (c+d x)))}{2 a^3 d} \] Input:

Integrate[Coth[c + d*x]^3/(a + b*Sinh[c + d*x]),x]
 

Output:

(2*a*b*Csch[c + d*x] - a^2*Csch[c + d*x]^2 + 2*(a^2 + b^2)*(Log[Sinh[c + d 
*x]] - Log[a + b*Sinh[c + d*x]]))/(2*a^3*d)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 26, 3200, 25, 522, 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.

\(\displaystyle \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i}{\tan (i c+i d x)^3 (a-i b \sin (i c+i d x))}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {1}{(a-i b \sin (i c+i d x)) \tan (i c+i d x)^3}dx\)

\(\Big \downarrow \) 3200

\(\displaystyle -\frac {\int -\frac {\text {csch}^3(c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{b^3 (a+b \sinh (c+d x))}d(b \sinh (c+d x))}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\text {csch}^3(c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{b^3 (a+b \sinh (c+d x))}d(b \sinh (c+d x))}{d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {\int \left (\frac {\text {csch}^3(c+d x)}{a b}-\frac {\text {csch}^2(c+d x)}{a^2}+\frac {\left (a^2+b^2\right ) \text {csch}(c+d x)}{a^3 b}+\frac {-a^2-b^2}{a^3 (a+b \sinh (c+d x))}\right )d(b \sinh (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {b \text {csch}(c+d x)}{a^2}-\frac {\left (a^2+b^2\right ) \log (b \sinh (c+d x))}{a^3}+\frac {\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^3}+\frac {\text {csch}^2(c+d x)}{2 a}}{d}\)

Input:

Int[Coth[c + d*x]^3/(a + b*Sinh[c + d*x]),x]
 

Output:

-((-((b*Csch[c + d*x])/a^2) + Csch[c + d*x]^2/(2*a) - ((a^2 + b^2)*Log[b*S 
inh[c + d*x]])/a^3 + ((a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/a^3)/d)
 

Defintions of rubi rules used

rule 25

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

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 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

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 3200

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b 
^2, 0] && IntegerQ[(p + 1)/2]
 

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.79

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}+2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (4 a^{2}+4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\left (-4 a^{2}-4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{4 a^{3}}}{d}\)	\(143\)
default	\(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}+2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (4 a^{2}+4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\left (-4 a^{2}-4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{4 a^{3}}}{d}\)	\(143\)
risch	\(-\frac {2 \,{\mathrm e}^{d x +c} \left (-b \,{\mathrm e}^{2 d x +2 c}+a \,{\mathrm e}^{d x +c}+b \right )}{a^{2} d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{a d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{a d}-\frac {b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{a^{3} d}\)	\(159\)

Input:

int(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/4/a^2*(1/2*tanh(1/2*d*x+1/2*c)^2*a+2*b*tanh(1/2*d*x+1/2*c))-1/8/a/ 
tanh(1/2*d*x+1/2*c)^2+1/4/a^3*(4*a^2+4*b^2)*ln(tanh(1/2*d*x+1/2*c))+1/2*b/ 
a^2/tanh(1/2*d*x+1/2*c)+1/4/a^3*(-4*a^2-4*b^2)*ln(tanh(1/2*d*x+1/2*c)^2*a- 
2*b*tanh(1/2*d*x+1/2*c)-a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 617 vs. \(2 (78) = 156\).

Time = 0.13 (sec) , antiderivative size = 617, normalized size of antiderivative = 7.71 \[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="fricas")
 

Output:

(2*a*b*cosh(d*x + c)^3 + 2*a*b*sinh(d*x + c)^3 - 2*a^2*cosh(d*x + c)^2 - 2 
*a*b*cosh(d*x + c) + 2*(3*a*b*cosh(d*x + c) - a^2)*sinh(d*x + c)^2 - ((a^2 
 + b^2)*cosh(d*x + c)^4 + 4*(a^2 + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a 
^2 + b^2)*sinh(d*x + c)^4 - 2*(a^2 + b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + b^ 
2)*cosh(d*x + c)^2 - a^2 - b^2)*sinh(d*x + c)^2 + a^2 + b^2 + 4*((a^2 + b^ 
2)*cosh(d*x + c)^3 - (a^2 + b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*(b*si 
nh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + ((a^2 + b^2)*cosh(d*x 
+ c)^4 + 4*(a^2 + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + b^2)*sinh(d* 
x + c)^4 - 2*(a^2 + b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + b^2)*cosh(d*x + c)^ 
2 - a^2 - b^2)*sinh(d*x + c)^2 + a^2 + b^2 + 4*((a^2 + b^2)*cosh(d*x + c)^ 
3 - (a^2 + b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d* 
x + c) - sinh(d*x + c))) + 2*(3*a*b*cosh(d*x + c)^2 - 2*a^2*cosh(d*x + c) 
- a*b)*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x + c)*sinh( 
d*x + c)^3 + a^3*d*sinh(d*x + c)^4 - 2*a^3*d*cosh(d*x + c)^2 + a^3*d + 2*( 
3*a^3*d*cosh(d*x + c)^2 - a^3*d)*sinh(d*x + c)^2 + 4*(a^3*d*cosh(d*x + c)^ 
3 - a^3*d*cosh(d*x + c))*sinh(d*x + c))
 

Sympy [F]

\[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\coth ^{3}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \] Input:

integrate(coth(d*x+c)**3/(a+b*sinh(d*x+c)),x)
 

Output:

Integral(coth(c + d*x)**3/(a + b*sinh(c + d*x)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (78) = 156\).

Time = 0.04 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.16 \[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, {\left (b e^{\left (-d x - c\right )} - a e^{\left (-2 \, d x - 2 \, c\right )} - b e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{{\left (2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a^{2} e^{\left (-4 \, d x - 4 \, c\right )} - a^{2}\right )} d} - \frac {{\left (a^{2} + b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a^{3} d} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{3} d} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{3} d} \] Input:

integrate(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (78) = 156\).

Time = 0.17 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.30 \[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a^{3}} - \frac {2 \, {\left (a^{2} b + b^{3}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{3} b} - \frac {3 \, a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 3 \, b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 4 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 4 \, a^{2}}{a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2}}}{2 \, d} \] Input:

integrate(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.04 (sec) , antiderivative size = 1329, normalized size of antiderivative = 16.61 \[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

int(coth(c + d*x)^3/(a + b*sinh(c + d*x)),x)
 

Output:

((2*atan((a^2*(-a^6*d^2)^(1/2)*(a^4 + b^4 + 2*a^2*b^2)^(1/2) + 2*b^2*(-a^6 
*d^2)^(1/2)*(a^4 + b^4 + 2*a^2*b^2)^(1/2))/(2*a^3*d*(a^2 + b^2)^2) + ((a^7 
*d + a^5*b^2*d)*(-a^6*d^2)^(1/2))/(2*a^6*d^2*((a^2 + b^2)^2)^(1/2)*(a^2 + 
b^2)) - (a^6*b^2*exp(2*c)*exp(2*d*x)*(-a^6*d^2)^(1/2)*((4*(a^2 + 2*b^2)*(a 
^4 + b^4 + 2*a^2*b^2))/(a^9*b^2*d*(a^2 + b^2)^2) + (2*(2*a^4*b^3*d + 2*a^6 
*b*d)*(a^4 + b^4 + 2*a^2*b^2)^(1/2))/(a^11*b^3*d^2*((a^2 + b^2)^2)^(1/2)*( 
a^2 + b^2)) + (4*(a^2*(-a^6*d^2)^(1/2)*(a^4 + b^4 + 2*a^2*b^2)^(1/2) + 2*b 
^2*(-a^6*d^2)^(1/2)*(a^4 + b^4 + 2*a^2*b^2)^(1/2))*(a^4 + b^4 + 2*a^2*b^2) 
^(1/2))/(a^9*b^2*d*(a^2 + b^2)^2*(-a^6*d^2)^(1/2)) + (4*(a^7*d + a^5*b^2*d 
)*(a^4 + b^4 + 2*a^2*b^2)^(1/2))/(a^12*b^2*d^2*((a^2 + b^2)^2)^(1/2)*(a^2 
+ b^2))))/(8*(a^4 + b^4 + 2*a^2*b^2)^(1/2)) + (a^6*b^2*exp(3*c)*exp(3*d*x) 
*((2*(a^7*d + a^5*b^2*d)*(a^4 + b^4 + 2*a^2*b^2)^(1/2))/(a^11*b^3*d^2*((a^ 
2 + b^2)^2)^(1/2)*(a^2 + b^2)) - (2*(a^2 + 2*b^2)*(a^2*(-a^6*d^2)^(1/2)*(a 
^4 + b^4 + 2*a^2*b^2)^(1/2) + 2*b^2*(-a^6*d^2)^(1/2)*(a^4 + b^4 + 2*a^2*b^ 
2)^(1/2))*(a^4 + b^4 + 2*a^2*b^2)^(1/2))/(a^10*b^3*d*(a^2 + b^2)^2*(-a^6*d 
^2)^(1/2)))*(-a^6*d^2)^(1/2))/(8*(a^4 + b^4 + 2*a^2*b^2)^(1/2)) - (a^6*b^2 
*exp(d*x)*exp(c)*(-a^6*d^2)^(1/2)*((8*(a^4 + b^4 + 2*a^2*b^2))/(a^8*b*d*(a 
^2 + b^2)^2) - (4*(2*a^4*b^3*d + 2*a^6*b*d)*(a^4 + b^4 + 2*a^2*b^2)^(1/2)) 
/(a^12*b^2*d^2*((a^2 + b^2)^2)^(1/2)*(a^2 + b^2)) + (2*(a^7*d + a^5*b^2*d) 
*(a^4 + b^4 + 2*a^2*b^2)^(1/2))/(a^11*b^3*d^2*((a^2 + b^2)^2)^(1/2)*(a^...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 563, normalized size of antiderivative = 7.04 \[ \int \frac {\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{2}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{2}-e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) a^{2}-e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) b^{2}-e^{4 d x +4 c} a^{2}+2 e^{3 d x +3 c} a b -2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{2}+2 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) a^{2}+2 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) b^{2}-2 e^{d x +c} a b +\mathrm {log}\left (e^{d x +c}-1\right ) a^{2}+\mathrm {log}\left (e^{d x +c}-1\right ) b^{2}+\mathrm {log}\left (e^{d x +c}+1\right ) a^{2}+\mathrm {log}\left (e^{d x +c}+1\right ) b^{2}-\mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) a^{2}-\mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) b^{2}-a^{2}}{a^{3} d \left (e^{4 d x +4 c}-2 e^{2 d x +2 c}+1\right )} \] Input:

int(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x)
 

Output:

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