3.447 \(\int \frac{\text{csch}(c+d x) \text{sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=160 \[ -\frac{b^4 \log (a+b \sinh (c+d x))}{a d \left (a^2+b^2\right )^2}-\frac{b^3 \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{b \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )}-\frac{a \left (a^2+2 b^2\right ) \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 d \left (a^2+b^2\right )}+\frac{\log (\sinh (c+d x))}{a d} \]

[Out]

-((b^3*ArcTan[Sinh[c + d*x]])/((a^2 + b^2)^2*d)) - (b*ArcTan[Sinh[c + d*x]])/(2*(a^2 + b^2)*d) - (a*(a^2 + 2*b
^2)*Log[Cosh[c + d*x]])/((a^2 + b^2)^2*d) + Log[Sinh[c + d*x]]/(a*d) - (b^4*Log[a + b*Sinh[c + d*x]])/(a*(a^2
+ b^2)^2*d) + (Sech[c + d*x]^2*(a - b*Sinh[c + d*x]))/(2*(a^2 + b^2)*d)

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Rubi [A]  time = 0.270547, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2837, 12, 894, 639, 203, 635, 260} \[ -\frac{b^4 \log (a+b \sinh (c+d x))}{a d \left (a^2+b^2\right )^2}-\frac{b^3 \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{b \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )}-\frac{a \left (a^2+2 b^2\right ) \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 d \left (a^2+b^2\right )}+\frac{\log (\sinh (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^3*ArcTan[Sinh[c + d*x]])/((a^2 + b^2)^2*d)) - (b*ArcTan[Sinh[c + d*x]])/(2*(a^2 + b^2)*d) - (a*(a^2 + 2*b
^2)*Log[Cosh[c + d*x]])/((a^2 + b^2)^2*d) + Log[Sinh[c + d*x]]/(a*d) - (b^4*Log[a + b*Sinh[c + d*x]])/(a*(a^2
+ b^2)^2*d) + (Sech[c + d*x]^2*(a - b*Sinh[c + d*x]))/(2*(a^2 + b^2)*d)

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 203

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

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\text{csch}(c+d x) \text{sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{b}{x (a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{x (a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=\frac{b^4 \operatorname{Subst}\left (\int \left (\frac{1}{a b^4 x}-\frac{1}{a \left (a^2+b^2\right )^2 (a+x)}+\frac{-b^2-a x}{b^2 \left (a^2+b^2\right ) \left (b^2+x^2\right )^2}+\frac{-b^4-a \left (a^2+2 b^2\right ) x}{b^4 \left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=\frac{\log (\sinh (c+d x))}{a d}-\frac{b^4 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-b^4-a \left (a^2+2 b^2\right ) x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right )^2 d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{-b^2-a x}{\left (b^2+x^2\right )^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{\log (\sinh (c+d x))}{a d}-\frac{b^4 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right )^2 d}+\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right )^2 d}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac{\left (a \left (a^2+2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac{b^3 \tan ^{-1}(\sinh (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{b \tan ^{-1}(\sinh (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac{a \left (a^2+2 b^2\right ) \log (\cosh (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac{\log (\sinh (c+d x))}{a d}-\frac{b^4 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right )^2 d}+\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 0.793089, size = 196, normalized size = 1.22 \[ -\frac{-a^2 \left (a^2+b^2\right ) \text{sech}^2(c+d x)-2 \left (a^2+b^2\right )^2 \log (\sinh (c+d x))+a \left (a^3+2 a b^2+\left (-b^2\right )^{3/2}\right ) \log \left (\sqrt{-b^2}-b \sinh (c+d x)\right )+a \left (a^3+2 a b^2-\left (-b^2\right )^{3/2}\right ) \log \left (\sqrt{-b^2}+b \sinh (c+d x)\right )+a b \left (a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))+a b \left (a^2+b^2\right ) \tanh (c+d x) \text{sech}(c+d x)+2 b^4 \log (a+b \sinh (c+d x))}{2 a d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*b*(a^2 + b^2)*ArcTan[Sinh[c + d*x]] - 2*(a^2 + b^2)^2*Log[Sinh[c + d*x]] + a*(a^3 + 2*a*b^2 + (-b^2)^(3/2)
)*Log[Sqrt[-b^2] - b*Sinh[c + d*x]] + 2*b^4*Log[a + b*Sinh[c + d*x]] + a*(a^3 + 2*a*b^2 - (-b^2)^(3/2))*Log[Sq
rt[-b^2] + b*Sinh[c + d*x]] - a^2*(a^2 + b^2)*Sech[c + d*x]^2 + a*b*(a^2 + b^2)*Sech[c + d*x]*Tanh[c + d*x])/(
2*a*(a^2 + b^2)^2*d)

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Maple [B]  time = 0.003, size = 530, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.81423, size = 358, normalized size = 2.24 \begin{align*} -\frac{b^{4} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d} + \frac{{\left (a^{2} b + 3 \, b^{3}\right )} \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac{{\left (a^{3} + 2 \, a b^{2}\right )} \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac{b e^{\left (-d x - c\right )} - 2 \, a e^{\left (-2 \, d x - 2 \, c\right )} - b e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{2} + b^{2} + 2 \,{\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a^{2} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^4*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^5 + 2*a^3*b^2 + a*b^4)*d) + (a^2*b + 3*b^3)*arctan(e^
(-d*x - c))/((a^4 + 2*a^2*b^2 + b^4)*d) - (a^3 + 2*a*b^2)*log(e^(-2*d*x - 2*c) + 1)/((a^4 + 2*a^2*b^2 + b^4)*d
) - (b*e^(-d*x - c) - 2*a*e^(-2*d*x - 2*c) - b*e^(-3*d*x - 3*c))/((a^2 + b^2 + 2*(a^2 + b^2)*e^(-2*d*x - 2*c)
+ (a^2 + b^2)*e^(-4*d*x - 4*c))*d) + log(e^(-d*x - c) + 1)/(a*d) + log(e^(-d*x - c) - 1)/(a*d)

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Fricas [B]  time = 4.2508, size = 3066, normalized size = 19.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^3*b + a*b^3)*cosh(d*x + c)^3 + (a^3*b + a*b^3)*sinh(d*x + c)^3 - 2*(a^4 + a^2*b^2)*cosh(d*x + c)^2 - (2*a
^4 + 2*a^2*b^2 - 3*(a^3*b + a*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + ((a^3*b + 3*a*b^3)*cosh(d*x + c)^4 + 4*(a^
3*b + 3*a*b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b + 3*a*b^3)*sinh(d*x + c)^4 + a^3*b + 3*a*b^3 + 2*(a^3*b
+ 3*a*b^3)*cosh(d*x + c)^2 + 2*(a^3*b + 3*a*b^3 + 3*(a^3*b + 3*a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a
^3*b + 3*a*b^3)*cosh(d*x + c)^3 + (a^3*b + 3*a*b^3)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(
d*x + c)) - (a^3*b + a*b^3)*cosh(d*x + c) + (b^4*cosh(d*x + c)^4 + 4*b^4*cosh(d*x + c)*sinh(d*x + c)^3 + b^4*s
inh(d*x + c)^4 + 2*b^4*cosh(d*x + c)^2 + b^4 + 2*(3*b^4*cosh(d*x + c)^2 + b^4)*sinh(d*x + c)^2 + 4*(b^4*cosh(d
*x + c)^3 + b^4*cosh(d*x + c))*sinh(d*x + c))*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + (
(a^4 + 2*a^2*b^2)*cosh(d*x + c)^4 + 4*(a^4 + 2*a^2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 + 2*a^2*b^2)*sinh
(d*x + c)^4 + a^4 + 2*a^2*b^2 + 2*(a^4 + 2*a^2*b^2)*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + 3*(a^4 + 2*a^2*b^2)
*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^4 + 2*a^2*b^2)*cosh(d*x + c)^3 + (a^4 + 2*a^2*b^2)*cosh(d*x + c))*si
nh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - ((a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)^4 +
4*(a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 + 2*a^2*b^2 + b^4)*sinh(d*x + c)^4 + a^4 + 2*a
^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4
)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)^3 + (a^4 + 2*a^2*b^2 + b^4)*cosh
(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - (a^3*b + a*b^3 - 3*(a^3*b + a
*b^3)*cosh(d*x + c)^2 + 4*(a^4 + a^2*b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x
+ c)^4 + 4*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + 2*a^3*b^2 + a*b^4)*d*sinh(d*x +
c)^4 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c)^2 + 2*(3*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c)^2 + (a^5
+ 2*a^3*b^2 + a*b^4)*d)*sinh(d*x + c)^2 + (a^5 + 2*a^3*b^2 + a*b^4)*d + 4*((a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d
*x + c)^3 + (a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c))*sinh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.76585, size = 485, normalized size = 3.03 \begin{align*} -\frac{b^{5} \log \left ({\left | b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{5} b d + 2 \, a^{3} b^{3} d + a b^{5} d} - \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )}{\left (a^{2} b + 3 \, b^{3}\right )}}{4 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac{{\left (a^{3} + 2 \, a b^{2}\right )} \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{2 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} + \frac{a^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 2 \, a b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 2 \, a^{2} b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 2 \, b^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 8 \, a^{3} + 12 \, a b^{2}}{2 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}} + \frac{\log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^5*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^5*b*d + 2*a^3*b^3*d + a*b^5*d) - 1/4*(pi + 2*arctan(1/2
*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(a^2*b + 3*b^3)/(a^4*d + 2*a^2*b^2*d + b^4*d) - 1/2*(a^3 + 2*a*b^2)*log(
(e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^4*d + 2*a^2*b^2*d + b^4*d) + 1/2*(a^3*(e^(d*x + c) - e^(-d*x - c))^2 +
2*a*b^2*(e^(d*x + c) - e^(-d*x - c))^2 - 2*a^2*b*(e^(d*x + c) - e^(-d*x - c)) - 2*b^3*(e^(d*x + c) - e^(-d*x -
c)) + 8*a^3 + 12*a*b^2)/((a^4*d + 2*a^2*b^2*d + b^4*d)*((e^(d*x + c) - e^(-d*x - c))^2 + 4)) + log(abs(e^(d*x
+ c) - e^(-d*x - c)))/(a*d)