3.85 \(\int \frac{\text{csch}^2(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=115 \[ -\frac{2 b^2 \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \left (a^2+b^2\right )^{3/2}}-\frac{\left (a^2+2 b^2\right ) \coth (x)}{a^2 \left (a^2+b^2\right )}+\frac{b^2 \coth (x)}{a \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{2 b \tanh ^{-1}(\cosh (x))}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.371979, antiderivative size = 115, 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 = {2802, 3055, 3001, 3770, 2660, 618, 206} \[ -\frac{2 b^2 \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \left (a^2+b^2\right )^{3/2}}-\frac{\left (a^2+2 b^2\right ) \coth (x)}{a^2 \left (a^2+b^2\right )}+\frac{b^2 \coth (x)}{a \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{2 b \tanh ^{-1}(\cosh (x))}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^2/(a + b*Sinh[x])^2,x]

[Out]

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

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*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] + Dist[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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

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

Mathematica [A]  time = 0.648951, size = 118, normalized size = 1.03 \[ -\frac{\frac{4 b^2 \left (3 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\frac{2 a b^3 \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+a \tanh \left (\frac{x}{2}\right )+a \coth \left (\frac{x}{2}\right )+4 b \log \left (\tanh \left (\frac{x}{2}\right )\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^2/(a + b*Sinh[x])^2,x]

[Out]

-((4*b^2*(3*a^2 + 2*b^2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/(-a^2 - b^2)^(3/2) + a*Coth[x/2] + 4*b*Lo
g[Tanh[x/2]] + (2*a*b^3*Cosh[x])/((a^2 + b^2)*(a + b*Sinh[x])) + a*Tanh[x/2])/(2*a^3)

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Maple [A]  time = 0.047, size = 193, normalized size = 1.7 \begin{align*} -{\frac{1}{2\,{a}^{2}}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{2\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-2\,{\frac{b\ln \left ( \tanh \left ( x/2 \right ) \right ) }{{a}^{3}}}+2\,{\frac{{b}^{4}\tanh \left ( x/2 \right ) }{{a}^{3} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}+2\,{\frac{{b}^{3}}{{a}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}+6\,{\frac{{b}^{2}}{a \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+4\,{\frac{{b}^{4}}{{a}^{3} \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\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(csch(x)^2/(a+b*sinh(x))^2,x)

[Out]

-1/2/a^2*tanh(1/2*x)-1/2/a^2/tanh(1/2*x)-2/a^3*b*ln(tanh(1/2*x))+2/a^3*b^4/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)
/(a^2+b^2)*tanh(1/2*x)+2/a^2*b^3/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)+6/a*b^2/(a^2+b^2)^(3/2)*arctanh
(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))+4/a^3*b^4/(a^2+b^2)^(3/2)*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(csch(x)^2/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.55821, size = 4055, normalized size = 35.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^2/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

(2*a^5*b + 6*a^3*b^3 + 4*a*b^5 + 2*(a^4*b^2 + a^2*b^4)*cosh(x)^3 + 2*(a^4*b^2 + a^2*b^4)*sinh(x)^3 - 2*(a^5*b
+ 3*a^3*b^3 + 2*a*b^5)*cosh(x)^2 - 2*(a^5*b + 3*a^3*b^3 + 2*a*b^5 - 3*(a^4*b^2 + a^2*b^4)*cosh(x))*sinh(x)^2 +
(3*a^2*b^3 + 2*b^5 + (3*a^2*b^3 + 2*b^5)*cosh(x)^4 + (3*a^2*b^3 + 2*b^5)*sinh(x)^4 + 2*(3*a^3*b^2 + 2*a*b^4)*
cosh(x)^3 + 2*(3*a^3*b^2 + 2*a*b^4 + 2*(3*a^2*b^3 + 2*b^5)*cosh(x))*sinh(x)^3 - 2*(3*a^2*b^3 + 2*b^5)*cosh(x)^
2 - 2*(3*a^2*b^3 + 2*b^5 - 3*(3*a^2*b^3 + 2*b^5)*cosh(x)^2 - 3*(3*a^3*b^2 + 2*a*b^4)*cosh(x))*sinh(x)^2 - 2*(3
*a^3*b^2 + 2*a*b^4)*cosh(x) - 2*(3*a^3*b^2 + 2*a*b^4 - 2*(3*a^2*b^3 + 2*b^5)*cosh(x)^3 - 3*(3*a^3*b^2 + 2*a*b^
4)*cosh(x)^2 + 2*(3*a^2*b^3 + 2*b^5)*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)) - 2*(2*a^6 + 5*a^4*b^2 + 3*a^2*b^4)*co
sh(x) + 2*(a^4*b^2 + 2*a^2*b^4 + b^6 + (a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^4 + (a^4*b^2 + 2*a^2*b^4 + b^6)*sin
h(x)^4 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*cosh(x)^3 + 2*(a^5*b + 2*a^3*b^3 + a*b^5 + 2*(a^4*b^2 + 2*a^2*b^4 + b^6
)*cosh(x))*sinh(x)^3 - 2*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^2 - 2*(a^4*b^2 + 2*a^2*b^4 + b^6 - 3*(a^4*b^2 + 2
*a^2*b^4 + b^6)*cosh(x)^2 - 3*(a^5*b + 2*a^3*b^3 + a*b^5)*cosh(x))*sinh(x)^2 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*c
osh(x) - 2*(a^5*b + 2*a^3*b^3 + a*b^5 - 2*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^3 - 3*(a^5*b + 2*a^3*b^3 + a*b^5
)*cosh(x)^2 + 2*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) - 2*(a^4*b^2 + 2*a^2*
b^4 + b^6 + (a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^4 + (a^4*b^2 + 2*a^2*b^4 + b^6)*sinh(x)^4 + 2*(a^5*b + 2*a^3*b
^3 + a*b^5)*cosh(x)^3 + 2*(a^5*b + 2*a^3*b^3 + a*b^5 + 2*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x))*sinh(x)^3 - 2*(a
^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^2 - 2*(a^4*b^2 + 2*a^2*b^4 + b^6 - 3*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^2 -
3*(a^5*b + 2*a^3*b^3 + a*b^5)*cosh(x))*sinh(x)^2 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*cosh(x) - 2*(a^5*b + 2*a^3*b
^3 + a*b^5 - 2*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^3 - 3*(a^5*b + 2*a^3*b^3 + a*b^5)*cosh(x)^2 + 2*(a^4*b^2 +
2*a^2*b^4 + b^6)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) - 1) - 2*(2*a^6 + 5*a^4*b^2 + 3*a^2*b^4 - 3*(a^4*b^2
+ a^2*b^4)*cosh(x)^2 + 2*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*cosh(x))*sinh(x))/(a^7*b + 2*a^5*b^3 + a^3*b^5 + (a^7*b
+ 2*a^5*b^3 + a^3*b^5)*cosh(x)^4 + (a^7*b + 2*a^5*b^3 + a^3*b^5)*sinh(x)^4 + 2*(a^8 + 2*a^6*b^2 + a^4*b^4)*co
sh(x)^3 + 2*(a^8 + 2*a^6*b^2 + a^4*b^4 + 2*(a^7*b + 2*a^5*b^3 + a^3*b^5)*cosh(x))*sinh(x)^3 - 2*(a^7*b + 2*a^5
*b^3 + a^3*b^5)*cosh(x)^2 - 2*(a^7*b + 2*a^5*b^3 + a^3*b^5 - 3*(a^7*b + 2*a^5*b^3 + a^3*b^5)*cosh(x)^2 - 3*(a^
8 + 2*a^6*b^2 + a^4*b^4)*cosh(x))*sinh(x)^2 - 2*(a^8 + 2*a^6*b^2 + a^4*b^4)*cosh(x) - 2*(a^8 + 2*a^6*b^2 + a^4
*b^4 - 2*(a^7*b + 2*a^5*b^3 + a^3*b^5)*cosh(x)^3 - 3*(a^8 + 2*a^6*b^2 + a^4*b^4)*cosh(x)^2 + 2*(a^7*b + 2*a^5*
b^3 + a^3*b^5)*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{2}{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**2/(a+b*sinh(x))**2,x)

[Out]

Integral(csch(x)**2/(a + b*sinh(x))**2, x)

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Giac [A]  time = 1.4016, size = 277, normalized size = 2.41 \begin{align*} \frac{{\left (3 \, a^{2} b^{2} + 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 )}{{\left (a^{5} + a^{3} b^{2}\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (a b^{2} e^{\left (3 \, x\right )} - a^{2} b e^{\left (2 \, x\right )} - 2 \, b^{3} e^{\left (2 \, x\right )} - 2 \, a^{3} e^{x} - 3 \, a b^{2} e^{x} + a^{2} b + 2 \, b^{3}\right )}}{{\left (a^{4} + a^{2} b^{2}\right )}{\left (b e^{\left (4 \, x\right )} + 2 \, a e^{\left (3 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} - 2 \, a e^{x} + b\right )}} + \frac{2 \, b \log \left (e^{x} + 1\right )}{a^{3}} - \frac{2 \, b \log \left ({\left | e^{x} - 1 \right |}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^2/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

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