\(\int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx\) [174]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 99 \[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\frac {1}{4 (a+b) (1-\cosh (x))}-\frac {1}{4 (a-b) (1+\cosh (x))}-\frac {(a+2 b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac {(a-2 b) \log (1+\cosh (x))}{4 (a-b)^2}+\frac {b^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01 \[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\frac {1}{8} \left (-\frac {\text {csch}^2\left (\frac {x}{2}\right )}{a+b}+\frac {4 (a-2 b) \log \left (\cosh \left (\frac {x}{2}\right )\right )}{(a-b)^2}+\frac {8 b^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}-\frac {4 (a+2 b) \log \left (\sinh \left (\frac {x}{2}\right )\right )}{(a+b)^2}-\frac {\text {sech}^2\left (\frac {x}{2}\right )}{a-b}\right ) \] Input:

Integrate[Csch[x]^3/(a + b*Cosh[x]),x]
 

Output:

(-(Csch[x/2]^2/(a + b)) + (4*(a - 2*b)*Log[Cosh[x/2]])/(a - b)^2 + (8*b^3* 
Log[a + b*Cosh[x]])/(a^2 - b^2)^2 - (4*(a + 2*b)*Log[Sinh[x/2]])/(a + b)^2 
 - Sech[x/2]^2/(a - b))/8
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 26, 3147, 477, 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.

\(\displaystyle \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i}{\cos \left (-\frac {\pi }{2}+i x\right )^3 \left (a-b \sin \left (-\frac {\pi }{2}+i x\right )\right )}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {1}{\cos \left (i x-\frac {\pi }{2}\right )^3 \left (a-b \sin \left (i x-\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 477

\(\displaystyle \frac {\int \left (\frac {b^4}{\left (a^2-b^2\right )^2 (a+b \cosh (x))}+\frac {b^2}{4 (a+b) (b-b \cosh (x))^2}+\frac {b^2}{4 (a-b) (\cosh (x) b+b)^2}+\frac {(a+2 b) b}{4 (a+b)^2 (b-b \cosh (x))}+\frac {(a-2 b) b}{4 (a-b)^2 (\cosh (x) b+b)}\right )d(b \cosh (x))}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b^4 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}+\frac {b^2}{4 (a+b) (b-b \cosh (x))}-\frac {b^2}{4 (a-b) (b \cosh (x)+b)}-\frac {b (a+2 b) \log (b-b \cosh (x))}{4 (a+b)^2}+\frac {b (a-2 b) \log (b \cosh (x)+b)}{4 (a-b)^2}}{b}\)

Input:

Int[Csch[x]^3/(a + b*Cosh[x]),x]
 

Output:

(b^2/(4*(a + b)*(b - b*Cosh[x])) - b^2/(4*(a - b)*(b + b*Cosh[x])) - (b*(a 
 + 2*b)*Log[b - b*Cosh[x]])/(4*(a + b)^2) + (b^4*Log[a + b*Cosh[x]])/(a^2 
- b^2)^2 + ((a - 2*b)*b*Log[b + b*Cosh[x]])/(4*(a - b)^2))/b
 

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 477
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
a^p   Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 
]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & 
& NiceSqrtQ[-b/a] &&  !FractionalPowerFactorQ[Rt[-b/a, 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 3147
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(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] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 
Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91

method result size
default \(\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -b \tanh \left (\frac {x}{2}\right )^{2}-a -b \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right )^{2}}{8 a -8 b}-\frac {1}{8 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\left (-2 a -4 b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{4 \left (a +b \right )^{2}}\) \(90\)
risch \(\frac {a x}{2 a^{2}+4 a b +2 b^{2}}+\frac {x b}{a^{2}+2 a b +b^{2}}-\frac {x a}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {x b}{a^{2}-2 a b +b^{2}}-\frac {2 x \,b^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {{\mathrm e}^{x} \left ({\mathrm e}^{2 x} a -2 \,{\mathrm e}^{x} b +a \right )}{\left ({\mathrm e}^{2 x}-1\right )^{2} \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{x}-1\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{x}-1\right ) b}{a^{2}+2 a b +b^{2}}+\frac {a \ln \left ({\mathrm e}^{x}+1\right )}{2 a^{2}-4 a b +2 b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+1\right ) b}{a^{2}-2 a b +b^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) \(247\)

Input:

int(csch(x)^3/(a+b*cosh(x)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 818 vs. \(2 (89) = 178\).

Time = 0.11 (sec) , antiderivative size = 818, normalized size of antiderivative = 8.26 \[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \] Input:

integrate(csch(x)^3/(a+b*cosh(x)),x, algorithm="fricas")
 

Output:

-1/2*(2*(a^3 - a*b^2)*cosh(x)^3 + 2*(a^3 - a*b^2)*sinh(x)^3 - 4*(a^2*b - b 
^3)*cosh(x)^2 - 2*(2*a^2*b - 2*b^3 - 3*(a^3 - a*b^2)*cosh(x))*sinh(x)^2 + 
2*(a^3 - a*b^2)*cosh(x) - 2*(b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3 
*sinh(x)^4 - 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 - b^3)*sinh(x)^2 + 
 4*(b^3*cosh(x)^3 - b^3*cosh(x))*sinh(x))*log(2*(b*cosh(x) + a)/(cosh(x) - 
 sinh(x))) - ((a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^4 + 4*(a^3 - 3*a*b^2 - 2*b^3 
)*cosh(x)*sinh(x)^3 + (a^3 - 3*a*b^2 - 2*b^3)*sinh(x)^4 + a^3 - 3*a*b^2 - 
2*b^3 - 2*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2 - 2*(a^3 - 3*a*b^2 - 2*b^3 - 3 
*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^3 - 3*a*b^2 - 2*b^3) 
*cosh(x)^3 - (a^3 - 3*a*b^2 - 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) + sinh( 
x) + 1) + ((a^3 - 3*a*b^2 + 2*b^3)*cosh(x)^4 + 4*(a^3 - 3*a*b^2 + 2*b^3)*c 
osh(x)*sinh(x)^3 + (a^3 - 3*a*b^2 + 2*b^3)*sinh(x)^4 + a^3 - 3*a*b^2 + 2*b 
^3 - 2*(a^3 - 3*a*b^2 + 2*b^3)*cosh(x)^2 - 2*(a^3 - 3*a*b^2 + 2*b^3 - 3*(a 
^3 - 3*a*b^2 + 2*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^3 - 3*a*b^2 + 2*b^3)*co 
sh(x)^3 - (a^3 - 3*a*b^2 + 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) 
- 1) + 2*(a^3 - a*b^2 + 3*(a^3 - a*b^2)*cosh(x)^2 - 4*(a^2*b - b^3)*cosh(x 
))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4 + 4*(a^4 - 2*a^2*b^2 + b^4) 
*cosh(x)*sinh(x)^3 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^4 + a^4 - 2*a^2*b^2 + 
 b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^4 - 2*a^2*b^2 + b^4 - 3* 
(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 - 2*a^2*b^2 + b^...
 

Sympy [F]

\[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\int \frac {\operatorname {csch}^{3}{\left (x \right )}}{a + b \cosh {\left (x \right )}}\, dx \] Input:

integrate(csch(x)**3/(a+b*cosh(x)),x)
 

Output:

Integral(csch(x)**3/(a + b*cosh(x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.56 \[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\frac {b^{3} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a - 2 \, b\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (a + 2 \, b\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a e^{\left (-x\right )} - 2 \, b e^{\left (-2 \, x\right )} + a e^{\left (-3 \, x\right )}}{a^{2} - b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} - b^{2}\right )} e^{\left (-4 \, x\right )}} \] Input:

integrate(csch(x)^3/(a+b*cosh(x)),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (89) = 178\).

Time = 0.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.81 \[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\frac {b^{4} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac {{\left (a - 2 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (a + 2 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 2 \, a^{3} {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )} + 4 \, a^{2} b - 8 \, b^{3}}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}} \] Input:

integrate(csch(x)^3/(a+b*cosh(x)),x, algorithm="giac")
 

Output:

b^4*log(abs(b*(e^(-x) + e^x) + 2*a))/(a^4*b - 2*a^2*b^3 + b^5) + 1/4*(a - 
2*b)*log(e^(-x) + e^x + 2)/(a^2 - 2*a*b + b^2) - 1/4*(a + 2*b)*log(e^(-x) 
+ e^x - 2)/(a^2 + 2*a*b + b^2) + 1/2*(b^3*(e^(-x) + e^x)^2 - 2*a^3*(e^(-x) 
 + e^x) + 2*a*b^2*(e^(-x) + e^x) + 4*a^2*b - 8*b^3)/((a^4 - 2*a^2*b^2 + b^ 
4)*((e^(-x) + e^x)^2 - 4))
 

Mupad [B] (verification not implemented)

Time = 2.55 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.94 \[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\frac {\frac {2\,\left (a^2\,b-b^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {{\mathrm {e}}^x\,\left (a\,b^2-a^3\right )}{{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{2\,x}-1}+\frac {\frac {2\,b}{a^2-b^2}-\frac {2\,a\,{\mathrm {e}}^x}{a^2-b^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}+\frac {b^3\,\ln \left (16\,b^7\,{\mathrm {e}}^{2\,x}-a^6\,b+16\,b^7-9\,a^2\,b^5+6\,a^4\,b^3-2\,a^7\,{\mathrm {e}}^x-9\,a^2\,b^5\,{\mathrm {e}}^{2\,x}+6\,a^4\,b^3\,{\mathrm {e}}^{2\,x}+32\,a\,b^6\,{\mathrm {e}}^x-a^6\,b\,{\mathrm {e}}^{2\,x}-18\,a^3\,b^4\,{\mathrm {e}}^x+12\,a^5\,b^2\,{\mathrm {e}}^x\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left ({\mathrm {e}}^x-1\right )\,\left (a+2\,b\right )}{2\,a^2+4\,a\,b+2\,b^2}+\frac {\ln \left ({\mathrm {e}}^x+1\right )\,\left (a-2\,b\right )}{2\,a^2-4\,a\,b+2\,b^2} \] Input:

int(1/(sinh(x)^3*(a + b*cosh(x))),x)
 

Output:

((2*(a^2*b - b^3))/(a^2 - b^2)^2 + (exp(x)*(a*b^2 - a^3))/(a^2 - b^2)^2)/( 
exp(2*x) - 1) + ((2*b)/(a^2 - b^2) - (2*a*exp(x))/(a^2 - b^2))/(exp(4*x) - 
 2*exp(2*x) + 1) + (b^3*log(16*b^7*exp(2*x) - a^6*b + 16*b^7 - 9*a^2*b^5 + 
 6*a^4*b^3 - 2*a^7*exp(x) - 9*a^2*b^5*exp(2*x) + 6*a^4*b^3*exp(2*x) + 32*a 
*b^6*exp(x) - a^6*b*exp(2*x) - 18*a^3*b^4*exp(x) + 12*a^5*b^2*exp(x)))/(a^ 
4 + b^4 - 2*a^2*b^2) - (log(exp(x) - 1)*(a + 2*b))/(4*a*b + 2*a^2 + 2*b^2) 
 + (log(exp(x) + 1)*(a - 2*b))/(2*a^2 - 4*a*b + 2*b^2)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 494, normalized size of antiderivative = 4.99 \[ \int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx=\frac {-2 b^{3}-\mathrm {log}\left (e^{x}-1\right ) a^{3}+\mathrm {log}\left (e^{x}+1\right ) a^{3}-2 e^{3 x} a^{3}-2 e^{x} a^{3}-2 \,\mathrm {log}\left (e^{x}-1\right ) b^{3}-2 \,\mathrm {log}\left (e^{x}+1\right ) b^{3}+2 \,\mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) b^{3}+2 e^{x} a \,b^{2}+2 e^{4 x} a^{2} b +3 e^{4 x} \mathrm {log}\left (e^{x}-1\right ) a \,b^{2}-3 e^{4 x} \mathrm {log}\left (e^{x}+1\right ) a \,b^{2}-6 e^{2 x} \mathrm {log}\left (e^{x}-1\right ) a \,b^{2}+6 e^{2 x} \mathrm {log}\left (e^{x}+1\right ) a \,b^{2}-e^{4 x} \mathrm {log}\left (e^{x}-1\right ) a^{3}+e^{4 x} \mathrm {log}\left (e^{x}+1\right ) a^{3}-2 e^{4 x} \mathrm {log}\left (e^{x}-1\right ) b^{3}-2 e^{4 x} \mathrm {log}\left (e^{x}+1\right ) b^{3}+2 e^{4 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) b^{3}+2 e^{3 x} a \,b^{2}+2 e^{2 x} \mathrm {log}\left (e^{x}-1\right ) a^{3}+4 e^{2 x} \mathrm {log}\left (e^{x}-1\right ) b^{3}-2 e^{2 x} \mathrm {log}\left (e^{x}+1\right ) a^{3}+4 e^{2 x} \mathrm {log}\left (e^{x}+1\right ) b^{3}-4 e^{2 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) b^{3}+3 \,\mathrm {log}\left (e^{x}-1\right ) a \,b^{2}-3 \,\mathrm {log}\left (e^{x}+1\right ) a \,b^{2}-2 e^{4 x} b^{3}+2 a^{2} b}{2 e^{4 x} a^{4}-4 e^{4 x} a^{2} b^{2}+2 e^{4 x} b^{4}-4 e^{2 x} a^{4}+8 e^{2 x} a^{2} b^{2}-4 e^{2 x} b^{4}+2 a^{4}-4 a^{2} b^{2}+2 b^{4}} \] Input:

int(csch(x)^3/(a+b*cosh(x)),x)
 

Output:

( - e**(4*x)*log(e**x - 1)*a**3 + 3*e**(4*x)*log(e**x - 1)*a*b**2 - 2*e**( 
4*x)*log(e**x - 1)*b**3 + e**(4*x)*log(e**x + 1)*a**3 - 3*e**(4*x)*log(e** 
x + 1)*a*b**2 - 2*e**(4*x)*log(e**x + 1)*b**3 + 2*e**(4*x)*log(e**(2*x)*b 
+ 2*e**x*a + b)*b**3 + 2*e**(4*x)*a**2*b - 2*e**(4*x)*b**3 - 2*e**(3*x)*a* 
*3 + 2*e**(3*x)*a*b**2 + 2*e**(2*x)*log(e**x - 1)*a**3 - 6*e**(2*x)*log(e* 
*x - 1)*a*b**2 + 4*e**(2*x)*log(e**x - 1)*b**3 - 2*e**(2*x)*log(e**x + 1)* 
a**3 + 6*e**(2*x)*log(e**x + 1)*a*b**2 + 4*e**(2*x)*log(e**x + 1)*b**3 - 4 
*e**(2*x)*log(e**(2*x)*b + 2*e**x*a + b)*b**3 - 2*e**x*a**3 + 2*e**x*a*b** 
2 - log(e**x - 1)*a**3 + 3*log(e**x - 1)*a*b**2 - 2*log(e**x - 1)*b**3 + l 
og(e**x + 1)*a**3 - 3*log(e**x + 1)*a*b**2 - 2*log(e**x + 1)*b**3 + 2*log( 
e**(2*x)*b + 2*e**x*a + b)*b**3 + 2*a**2*b - 2*b**3)/(2*(e**(4*x)*a**4 - 2 
*e**(4*x)*a**2*b**2 + e**(4*x)*b**4 - 2*e**(2*x)*a**4 + 4*e**(2*x)*a**2*b* 
*2 - 2*e**(2*x)*b**4 + a**4 - 2*a**2*b**2 + b**4))