3.235 \(\int \frac{\coth ^4(x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=108 \[ -\frac{2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^4}-\frac{\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}+\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a} \]

[Out]

(b*(3*a^2 + 2*b^2)*ArcTanh[Cosh[x]])/(2*a^4) - (2*(a^2 + b^2)^(3/2)*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]]
)/a^4 - ((4*a^2 + 3*b^2)*Coth[x])/(3*a^3) + (b*Coth[x]*Csch[x])/(2*a^2) - (Coth[x]*Csch[x]^2)/(3*a)

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Rubi [A]  time = 0.413013, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2725, 3055, 3001, 3770, 2660, 618, 206} \[ -\frac{2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^4}-\frac{\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}+\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^4/(a + b*Sinh[x]),x]

[Out]

(b*(3*a^2 + 2*b^2)*ArcTanh[Cosh[x]])/(2*a^4) - (2*(a^2 + b^2)^(3/2)*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]]
)/a^4 - ((4*a^2 + 3*b^2)*Coth[x])/(3*a^3) + (b*Coth[x]*Csch[x])/(2*a^2) - (Coth[x]*Csch[x]^2)/(3*a)

Rule 2725

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(3*a*f*Sin[e + f*x]^3), x] + (-Dist[1/(6*a^2), Int[((a + b*Sin[e + f*x])^m*Simp[8*
a^2 - b^2*(m - 1)*(m - 2) + a*b*m*Sin[e + f*x] - (6*a^2 - b^2*m*(m - 2))*Sin[e + f*x]^2, x])/Sin[e + f*x]^2, x
], x] - Simp[(b*(m - 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(6*a^2*f*Sin[e + f*x]^2), x]) /; FreeQ[{a,
b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1] && IntegerQ[2*m]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\coth ^4(x)}{a+b \sinh (x)} \, dx &=\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a}+\frac{\int \frac{\text{csch}^2(x) \left (2 \left (4 a^2+3 b^2\right )-a b \sinh (x)+3 \left (2 a^2+b^2\right ) \sinh ^2(x)\right )}{a+b \sinh (x)} \, dx}{6 a^2}\\ &=-\frac{\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a}+\frac{i \int \frac{\text{csch}(x) \left (3 i b \left (3 a^2+2 b^2\right )-3 i a \left (2 a^2+b^2\right ) \sinh (x)\right )}{a+b \sinh (x)} \, dx}{6 a^3}\\ &=-\frac{\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a}+\frac{\left (a^2+b^2\right )^2 \int \frac{1}{a+b \sinh (x)} \, dx}{a^4}-\frac{\left (b \left (3 a^2+2 b^2\right )\right ) \int \text{csch}(x) \, dx}{2 a^4}\\ &=\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}-\frac{\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a}+\frac{\left (2 \left (a^2+b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^4}\\ &=\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}-\frac{\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a}-\frac{\left (4 \left (a^2+b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )}{a^4}\\ &=\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^4}-\frac{2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^4}-\frac{\left (4 a^2+3 b^2\right ) \coth (x)}{3 a^3}+\frac{b \coth (x) \text{csch}(x)}{2 a^2}-\frac{\coth (x) \text{csch}^2(x)}{3 a}\\ \end{align*}

Mathematica [A]  time = 0.422632, size = 176, normalized size = 1.63 \[ \frac{-4 a \left (4 a^2+3 b^2\right ) \tanh \left (\frac{x}{2}\right )-4 a \left (4 a^2+3 b^2\right ) \coth \left (\frac{x}{2}\right )-12 b \left (3 a^2+2 b^2\right ) \log \left (\tanh \left (\frac{x}{2}\right )\right )+48 \left (-a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )+3 a^2 b \text{csch}^2\left (\frac{x}{2}\right )+3 a^2 b \text{sech}^2\left (\frac{x}{2}\right )+8 a^3 \sinh ^4\left (\frac{x}{2}\right ) \text{csch}^3(x)-\frac{1}{2} a^3 \sinh (x) \text{csch}^4\left (\frac{x}{2}\right )}{24 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^4/(a + b*Sinh[x]),x]

[Out]

(48*(-a^2 - b^2)^(3/2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]] - 4*a*(4*a^2 + 3*b^2)*Coth[x/2] + 3*a^2*b*Cs
ch[x/2]^2 - 12*b*(3*a^2 + 2*b^2)*Log[Tanh[x/2]] + 3*a^2*b*Sech[x/2]^2 + 8*a^3*Csch[x]^3*Sinh[x/2]^4 - (a^3*Csc
h[x/2]^4*Sinh[x])/2 - 4*a*(4*a^2 + 3*b^2)*Tanh[x/2])/(24*a^4)

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Maple [B]  time = 0.036, size = 232, normalized size = 2.2 \begin{align*} -{\frac{1}{24\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{b}{8\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{5}{8\,a}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{{b}^{2}}{2\,{a}^{3}}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{24\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{5}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{{b}^{2}}{2\,{a}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{b}{8\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}-{\frac{3\,b}{2\,{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+2\,{\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+4\,{\frac{{b}^{2}}{{a}^{2}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+2\,{\frac{{b}^{4}}{{a}^{4}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^4/(a+b*sinh(x)),x)

[Out]

-1/24/a*tanh(1/2*x)^3-1/8/a^2*b*tanh(1/2*x)^2-5/8/a*tanh(1/2*x)-1/2/a^3*b^2*tanh(1/2*x)-1/24/a/tanh(1/2*x)^3-5
/8/a/tanh(1/2*x)-1/2/a^3/tanh(1/2*x)*b^2+1/8/a^2*b/tanh(1/2*x)^2-3/2/a^2*b*ln(tanh(1/2*x))-1/a^4*b^3*ln(tanh(1
/2*x))+2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))+4*b^2/a^2/(a^2+b^2)^(1/2)*arctanh(
1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))+2/a^4*b^4/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^
2)^(1/2))

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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*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.47033, size = 3380, normalized size = 31.3 \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*sinh(x)),x, algorithm="fricas")

[Out]

1/6*(6*a^2*b*cosh(x)^5 + 6*a^2*b*sinh(x)^5 - 12*(2*a^3 + a*b^2)*cosh(x)^4 + 6*(5*a^2*b*cosh(x) - 4*a^3 - 2*a*b
^2)*sinh(x)^4 - 6*a^2*b*cosh(x) + 12*(5*a^2*b*cosh(x)^2 - 4*(2*a^3 + a*b^2)*cosh(x))*sinh(x)^3 - 16*a^3 - 12*a
*b^2 + 24*(a^3 + a*b^2)*cosh(x)^2 + 12*(5*a^2*b*cosh(x)^3 + 2*a^3 + 2*a*b^2 - 6*(2*a^3 + a*b^2)*cosh(x)^2)*sin
h(x)^2 + 6*((a^2 + b^2)*cosh(x)^6 + 6*(a^2 + b^2)*cosh(x)*sinh(x)^5 + (a^2 + b^2)*sinh(x)^6 - 3*(a^2 + b^2)*co
sh(x)^4 + 3*(5*(a^2 + b^2)*cosh(x)^2 - a^2 - b^2)*sinh(x)^4 + 4*(5*(a^2 + b^2)*cosh(x)^3 - 3*(a^2 + b^2)*cosh(
x))*sinh(x)^3 + 3*(a^2 + b^2)*cosh(x)^2 + 3*(5*(a^2 + b^2)*cosh(x)^4 - 6*(a^2 + b^2)*cosh(x)^2 + a^2 + b^2)*si
nh(x)^2 - a^2 - b^2 + 6*((a^2 + b^2)*cosh(x)^5 - 2*(a^2 + b^2)*cosh(x)^3 + (a^2 + b^2)*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)*si
nh(x) - b)) + 3*((3*a^2*b + 2*b^3)*cosh(x)^6 + 6*(3*a^2*b + 2*b^3)*cosh(x)*sinh(x)^5 + (3*a^2*b + 2*b^3)*sinh(
x)^6 - 3*(3*a^2*b + 2*b^3)*cosh(x)^4 - 3*(3*a^2*b + 2*b^3 - 5*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(3
*a^2*b + 2*b^3)*cosh(x)^3 - 3*(3*a^2*b + 2*b^3)*cosh(x))*sinh(x)^3 - 3*a^2*b - 2*b^3 + 3*(3*a^2*b + 2*b^3)*cos
h(x)^2 + 3*(5*(3*a^2*b + 2*b^3)*cosh(x)^4 + 3*a^2*b + 2*b^3 - 6*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^2 + 6*((3
*a^2*b + 2*b^3)*cosh(x)^5 - 2*(3*a^2*b + 2*b^3)*cosh(x)^3 + (3*a^2*b + 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) +
sinh(x) + 1) - 3*((3*a^2*b + 2*b^3)*cosh(x)^6 + 6*(3*a^2*b + 2*b^3)*cosh(x)*sinh(x)^5 + (3*a^2*b + 2*b^3)*sinh
(x)^6 - 3*(3*a^2*b + 2*b^3)*cosh(x)^4 - 3*(3*a^2*b + 2*b^3 - 5*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(
3*a^2*b + 2*b^3)*cosh(x)^3 - 3*(3*a^2*b + 2*b^3)*cosh(x))*sinh(x)^3 - 3*a^2*b - 2*b^3 + 3*(3*a^2*b + 2*b^3)*co
sh(x)^2 + 3*(5*(3*a^2*b + 2*b^3)*cosh(x)^4 + 3*a^2*b + 2*b^3 - 6*(3*a^2*b + 2*b^3)*cosh(x)^2)*sinh(x)^2 + 6*((
3*a^2*b + 2*b^3)*cosh(x)^5 - 2*(3*a^2*b + 2*b^3)*cosh(x)^3 + (3*a^2*b + 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) +
sinh(x) - 1) + 6*(5*a^2*b*cosh(x)^4 - 8*(2*a^3 + a*b^2)*cosh(x)^3 - a^2*b + 8*(a^3 + a*b^2)*cosh(x))*sinh(x))
/(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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{4}{\left (x \right )}}{a + b \sinh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**4/(a+b*sinh(x)),x)

[Out]

Integral(coth(x)**4/(a + b*sinh(x)), x)

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Giac [B]  time = 1.13321, size = 262, normalized size = 2.43 \begin{align*} \frac{{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (e^{x} + 1\right )}{2 \, a^{4}} - \frac{{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | e^{x} - 1 \right |}\right )}{2 \, a^{4}} + \frac{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{4}} + \frac{3 \, a b e^{\left (5 \, x\right )} - 12 \, a^{2} e^{\left (4 \, x\right )} - 6 \, b^{2} e^{\left (4 \, x\right )} + 12 \, a^{2} e^{\left (2 \, x\right )} + 12 \, b^{2} e^{\left (2 \, x\right )} - 3 \, a b e^{x} - 8 \, a^{2} - 6 \, b^{2}}{3 \, a^{3}{\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*sinh(x)),x, algorithm="giac")

[Out]

1/2*(3*a^2*b + 2*b^3)*log(e^x + 1)/a^4 - 1/2*(3*a^2*b + 2*b^3)*log(abs(e^x - 1))/a^4 + (a^4 + 2*a^2*b^2 + b^4)
*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^4) + 1/
3*(3*a*b*e^(5*x) - 12*a^2*e^(4*x) - 6*b^2*e^(4*x) + 12*a^2*e^(2*x) + 12*b^2*e^(2*x) - 3*a*b*e^x - 8*a^2 - 6*b^
2)/(a^3*(e^(2*x) - 1)^3)