3.186 \(\int \frac{\coth ^4(x)}{a+b \cosh (x)} \, dx\)
Optimal. Leaf size=137 \[ -\frac{a \coth ^3(x)}{3 \left (a^2-b^2\right )}-\frac{a^3 \coth (x)}{\left (a^2-b^2\right )^2}+\frac{b \text{csch}^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^2 b \text{csch}(x)}{\left (a^2-b^2\right )^2}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{2 a^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]
[Out]
(2*a^4*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) - (a^3*Coth[x])/(a^2 - b^2)
^2 - (a*Coth[x]^3)/(3*(a^2 - b^2)) + (a^2*b*Csch[x])/(a^2 - b^2)^2 + (b*Csch[x])/(a^2 - b^2) + (b*Csch[x]^3)/(
3*(a^2 - b^2))
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Rubi [A] time = 0.19578, antiderivative size = 137, normalized size of antiderivative = 1.,
number of steps used = 12, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used
= {2727, 2607, 30, 2606, 3767, 8, 2659, 208} \[ -\frac{a \coth ^3(x)}{3 \left (a^2-b^2\right )}-\frac{a^3 \coth (x)}{\left (a^2-b^2\right )^2}+\frac{b \text{csch}^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^2 b \text{csch}(x)}{\left (a^2-b^2\right )^2}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{2 a^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Coth[x]^4/(a + b*Cosh[x]),x]
[Out]
(2*a^4*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) - (a^3*Coth[x])/(a^2 - b^2)
^2 - (a*Coth[x]^3)/(3*(a^2 - b^2)) + (a^2*b*Csch[x])/(a^2 - b^2)^2 + (b*Csch[x])/(a^2 - b^2) + (b*Csch[x]^3)/(
3*(a^2 - b^2))
Rule 2727
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(a^2 - b^
2), Int[(g*Tan[e + f*x])^p/Sin[e + f*x]^2, x], x] + (-Dist[(b*g)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 1)/Cos
[e + f*x], x], x] - Dist[(a^2*g^2)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 2)/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*p] && GtQ[p, 1]
Rule 2607
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])
Rule 30
Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]
Rule 2606
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Rule 3767
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 2659
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\coth ^4(x)}{a+b \cosh (x)} \, dx &=\frac{a \int \coth ^2(x) \text{csch}^2(x) \, dx}{a^2-b^2}+\frac{a^2 \int \frac{\coth ^2(x)}{a+b \cosh (x)} \, dx}{a^2-b^2}-\frac{b \int \coth ^3(x) \text{csch}(x) \, dx}{a^2-b^2}\\ &=\frac{a^3 \int \text{csch}^2(x) \, dx}{\left (a^2-b^2\right )^2}+\frac{a^4 \int \frac{1}{a+b \cosh (x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{\left (a^2 b\right ) \int \coth (x) \text{csch}(x) \, dx}{\left (a^2-b^2\right )^2}-\frac{(i a) \operatorname{Subst}\left (\int x^2 \, dx,x,i \coth (x)\right )}{a^2-b^2}-\frac{(i b) \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,-i \text{csch}(x)\right )}{a^2-b^2}\\ &=-\frac{a \coth ^3(x)}{3 \left (a^2-b^2\right )}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{b \text{csch}^3(x)}{3 \left (a^2-b^2\right )}-\frac{\left (i a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \coth (x))}{\left (a^2-b^2\right )^2}+\frac{\left (2 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{\left (a^2-b^2\right )^2}+\frac{\left (i a^2 b\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \text{csch}(x))}{\left (a^2-b^2\right )^2}\\ &=\frac{2 a^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac{a^3 \coth (x)}{\left (a^2-b^2\right )^2}-\frac{a \coth ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^2 b \text{csch}(x)}{\left (a^2-b^2\right )^2}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{b \text{csch}^3(x)}{3 \left (a^2-b^2\right )}\\ \end{align*}
Mathematica [A] time = 0.507083, size = 131, normalized size = 0.96 \[ \frac{1}{24} \left (-\frac{48 a^4 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{2 (5 b-8 a) \tanh \left (\frac{x}{2}\right )}{(a-b)^2}-\frac{2 (8 a+5 b) \coth \left (\frac{x}{2}\right )}{(a+b)^2}+\frac{8 \sinh ^4\left (\frac{x}{2}\right ) \text{csch}^3(x)}{a-b}-\frac{\sinh (x) \text{csch}^4\left (\frac{x}{2}\right )}{2 (a+b)}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[Coth[x]^4/(a + b*Cosh[x]),x]
[Out]
((-48*a^4*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) - (2*(8*a + 5*b)*Coth[x/2])/(a + b)
^2 + (8*Csch[x]^3*Sinh[x/2]^4)/(a - b) - (Csch[x/2]^4*Sinh[x])/(2*(a + b)) + (2*(-8*a + 5*b)*Tanh[x/2])/(a - b
)^2)/24
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Maple [A] time = 0.029, size = 127, normalized size = 0.9 \begin{align*} -{\frac{1}{8\, \left ( a-b \right ) ^{2}} \left ({\frac{a}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{b}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+5\,a\tanh \left ( x/2 \right ) -3\,\tanh \left ( x/2 \right ) b \right ) }+2\,{\frac{{a}^{4}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{24\,a+24\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{5\,a+3\,b}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)^4/(a+b*cosh(x)),x)
[Out]
-1/8/(a-b)^2*(1/3*a*tanh(1/2*x)^3-1/3*tanh(1/2*x)^3*b+5*a*tanh(1/2*x)-3*tanh(1/2*x)*b)+2/(a-b)^2/(a+b)^2*a^4/(
(a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-1/24/(a+b)/tanh(1/2*x)^3-1/8*(5*a+3*b)/(a+b)
^2/tanh(1/2*x)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^4/(a+b*cosh(x)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.30164, size = 5677, normalized size = 41.44 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^4/(a+b*cosh(x)),x, algorithm="fricas")
[Out]
[1/3*(6*(2*a^4*b - 3*a^2*b^3 + b^5)*cosh(x)^5 + 6*(2*a^4*b - 3*a^2*b^3 + b^5)*sinh(x)^5 - 8*a^5 + 10*a^3*b^2 -
2*a*b^4 - 6*(2*a^5 - 3*a^3*b^2 + a*b^4)*cosh(x)^4 - 6*(2*a^5 - 3*a^3*b^2 + a*b^4 - 5*(2*a^4*b - 3*a^2*b^3 + b
^5)*cosh(x))*sinh(x)^4 - 4*(4*a^4*b - 5*a^2*b^3 + b^5)*cosh(x)^3 - 4*(4*a^4*b - 5*a^2*b^3 + b^5 - 15*(2*a^4*b
- 3*a^2*b^3 + b^5)*cosh(x)^2 + 6*(2*a^5 - 3*a^3*b^2 + a*b^4)*cosh(x))*sinh(x)^3 + 12*(a^5 - a^3*b^2)*cosh(x)^2
+ 12*(a^5 - a^3*b^2 + 5*(2*a^4*b - 3*a^2*b^3 + b^5)*cosh(x)^3 - 3*(2*a^5 - 3*a^3*b^2 + a*b^4)*cosh(x)^2 - (4*
a^4*b - 5*a^2*b^3 + b^5)*cosh(x))*sinh(x)^2 + 3*(a^4*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 - 3*a
^4*cosh(x)^4 + 3*a^4*cosh(x)^2 + 3*(5*a^4*cosh(x)^2 - a^4)*sinh(x)^4 - a^4 + 4*(5*a^4*cosh(x)^3 - 3*a^4*cosh(x
))*sinh(x)^3 + 3*(5*a^4*cosh(x)^4 - 6*a^4*cosh(x)^2 + a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 - 2*a^4*cosh(x)^3 + a^
4*cosh(x))*sinh(x))*sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*
cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(
x) + 2*(b*cosh(x) + a)*sinh(x) + b)) + 6*(2*a^4*b - 3*a^2*b^3 + b^5)*cosh(x) + 6*(2*a^4*b - 3*a^2*b^3 + b^5 +
5*(2*a^4*b - 3*a^2*b^3 + b^5)*cosh(x)^4 - 4*(2*a^5 - 3*a^3*b^2 + a*b^4)*cosh(x)^3 - 2*(4*a^4*b - 5*a^2*b^3 + b
^5)*cosh(x)^2 + 4*(a^5 - a^3*b^2)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^6 + 6*(a^6 -
3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^5 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^6 - a^6 + 3*a^4*b
^2 - 3*a^2*b^4 + b^6 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6
- 5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x
)^3 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)
^2 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 - 6*(a^6 - 3*a^4*b
^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^5 - 2*(a^6 - 3*a^4
*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x)), 2/3*(3*(2*a^4*b - 3
*a^2*b^3 + b^5)*cosh(x)^5 + 3*(2*a^4*b - 3*a^2*b^3 + b^5)*sinh(x)^5 - 4*a^5 + 5*a^3*b^2 - a*b^4 - 3*(2*a^5 - 3
*a^3*b^2 + a*b^4)*cosh(x)^4 - 3*(2*a^5 - 3*a^3*b^2 + a*b^4 - 5*(2*a^4*b - 3*a^2*b^3 + b^5)*cosh(x))*sinh(x)^4
- 2*(4*a^4*b - 5*a^2*b^3 + b^5)*cosh(x)^3 - 2*(4*a^4*b - 5*a^2*b^3 + b^5 - 15*(2*a^4*b - 3*a^2*b^3 + b^5)*cosh
(x)^2 + 6*(2*a^5 - 3*a^3*b^2 + a*b^4)*cosh(x))*sinh(x)^3 + 6*(a^5 - a^3*b^2)*cosh(x)^2 + 6*(a^5 - a^3*b^2 + 5*
(2*a^4*b - 3*a^2*b^3 + b^5)*cosh(x)^3 - 3*(2*a^5 - 3*a^3*b^2 + a*b^4)*cosh(x)^2 - (4*a^4*b - 5*a^2*b^3 + b^5)*
cosh(x))*sinh(x)^2 - 3*(a^4*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 - 3*a^4*cosh(x)^4 + 3*a^4*cosh
(x)^2 + 3*(5*a^4*cosh(x)^2 - a^4)*sinh(x)^4 - a^4 + 4*(5*a^4*cosh(x)^3 - 3*a^4*cosh(x))*sinh(x)^3 + 3*(5*a^4*c
osh(x)^4 - 6*a^4*cosh(x)^2 + a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 - 2*a^4*cosh(x)^3 + a^4*cosh(x))*sinh(x))*sqrt(
-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 3*(2*a^4*b - 3*a^2*b^3 + b^5)*
cosh(x) + 3*(2*a^4*b - 3*a^2*b^3 + b^5 + 5*(2*a^4*b - 3*a^2*b^3 + b^5)*cosh(x)^4 - 4*(2*a^5 - 3*a^3*b^2 + a*b^
4)*cosh(x)^3 - 2*(4*a^4*b - 5*a^2*b^3 + b^5)*cosh(x)^2 + 4*(a^5 - a^3*b^2)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2
+ 3*a^2*b^4 - b^6)*cosh(x)^6 + 6*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^5 + (a^6 - 3*a^4*b^2 + 3
*a^2*b^4 - b^6)*sinh(x)^6 - a^6 + 3*a^4*b^2 - 3*a^2*b^4 + b^6 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^
4 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 - 5*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(
a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x)^3 + 3*(a
^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 5*(a^6 - 3*a^4*b^2 + 3*a^
2*b^4 - b^6)*cosh(x)^4 - 6*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 - 3*a^4*b^2 + 3*
a^2*b^4 - b^6)*cosh(x)^5 - 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^
6)*cosh(x))*sinh(x))]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{4}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)**4/(a+b*cosh(x)),x)
[Out]
Integral(coth(x)**4/(a + b*cosh(x)), x)
________________________________________________________________________________________
Giac [A] time = 1.21027, size = 232, normalized size = 1.69 \begin{align*} \frac{2 \, a^{4} \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (6 \, a^{2} b e^{\left (5 \, x\right )} - 3 \, b^{3} e^{\left (5 \, x\right )} - 6 \, a^{3} e^{\left (4 \, x\right )} + 3 \, a b^{2} e^{\left (4 \, x\right )} - 8 \, a^{2} b e^{\left (3 \, x\right )} + 2 \, b^{3} e^{\left (3 \, x\right )} + 6 \, a^{3} e^{\left (2 \, x\right )} + 6 \, a^{2} b e^{x} - 3 \, b^{3} e^{x} - 4 \, a^{3} + a b^{2}\right )}}{3 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^4/(a+b*cosh(x)),x, algorithm="giac")
[Out]
2*a^4*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + 2/3*(6*a^2*b*e^(5*x) -
3*b^3*e^(5*x) - 6*a^3*e^(4*x) + 3*a*b^2*e^(4*x) - 8*a^2*b*e^(3*x) + 2*b^3*e^(3*x) + 6*a^3*e^(2*x) + 6*a^2*b*e
^x - 3*b^3*e^x - 4*a^3 + a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(e^(2*x) - 1)^3)