3.360 \(\int \frac{\text{sech}^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=122 \[ -\frac{a b^2 \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{b \left (a^2-b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}+\frac{a b^2 \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 d \left (a^2+b^2\right )} \]

[Out]

-(b*(a^2 - b^2)*ArcTan[Sinh[c + d*x]])/(2*(a^2 + b^2)^2*d) + (a*b^2*Log[Cosh[c + d*x]])/((a^2 + b^2)^2*d) - (a
*b^2*Log[a + b*Sinh[c + d*x]])/((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 = 0.199132, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2837, 12, 823, 801, 635, 203, 260} \[ -\frac{a b^2 \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{b \left (a^2-b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}+\frac{a b^2 \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[c + d*x]^2*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

-(b*(a^2 - b^2)*ArcTan[Sinh[c + d*x]])/(2*(a^2 + b^2)^2*d) + (a*b^2*Log[Cosh[c + d*x]])/((a^2 + b^2)^2*d) - (a
*b^2*Log[a + b*Sinh[c + d*x]])/((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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
 && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 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 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{sech}^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{x}{b (a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=\frac{b^2 \operatorname{Subst}\left (\int \frac{x}{(a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=-\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}+\frac{\operatorname{Subst}\left (\int \frac{a b^2-b^2 x}{(a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=-\frac{\text{sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}+\frac{\operatorname{Subst}\left (\int \left (-\frac{2 a b^2}{\left (a^2+b^2\right ) (a+x)}+\frac{b^2 \left (-a^2+b^2+2 a x\right )}{\left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=-\frac{a b^2 \log (a+b \sinh (c+d x))}{\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^2 \operatorname{Subst}\left (\int \frac{-a^2+b^2+2 a x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=-\frac{a b^2 \log (a+b \sinh (c+d x))}{\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{\left (a b^2\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{\left (b^2 \left (a^2-b^2\right )\right ) \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 )^2 d}\\ &=-\frac{b \left (a^2-b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 \left (a^2+b^2\right )^2 d}+\frac{a b^2 \log (\cosh (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{a b^2 \log (a+b \sinh (c+d x))}{\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 = 0.208201, size = 105, normalized size = 0.86 \[ \frac{-a \left (a^2+b^2\right ) \text{sech}^2(c+d x)+b \left (a^2+b^2\right ) \tanh (c+d x) \text{sech}(c+d x)+2 b \left (\left (b^2-a^2\right ) \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+a b (\log (\cosh (c+d x))-\log (a+b \sinh (c+d x)))\right )}{2 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[c + d*x]^2*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

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

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Maple [B]  time = 0.003, size = 474, normalized size = 3.9 \begin{align*} -2\,{\frac{a{b}^{2}\ln \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }{d \left ( 2\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) }}-{\frac{{a}^{2}b}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-{\frac{{b}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+2\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}{a}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a{b}^{2}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{{a}^{2}b}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{{b}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{a{b}^{2}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{{a}^{2}b}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\arctan \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\arctan \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

-2/d*a*b^2/(2*a^4+4*a^2*b^2+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)/(tanh(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)*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+1/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 = 1.58781, size = 294, normalized size = 2.41 \begin{align*} -\frac{a b^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} + \frac{a b^{2} \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{{\left (a^{2} b - 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{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.46903, size = 2240, normalized size = 18.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

((a^2*b + b^3)*cosh(d*x + c)^3 + (a^2*b + b^3)*sinh(d*x + c)^3 - 2*(a^3 + a*b^2)*cosh(d*x + c)^2 - (2*a^3 + 2*
a*b^2 - 3*(a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - ((a^2*b - b^3)*cosh(d*x + c)^4 + 4*(a^2*b - b^3)*cosh
(d*x + c)*sinh(d*x + c)^3 + (a^2*b - b^3)*sinh(d*x + c)^4 + a^2*b - b^3 + 2*(a^2*b - b^3)*cosh(d*x + c)^2 + 2*
(a^2*b - b^3 + 3*(a^2*b - b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^2*b - b^3)*cosh(d*x + c)^3 + (a^2*b -
b^3)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - (a^2*b + b^3)*cosh(d*x + c) - (a*b^
2*cosh(d*x + c)^4 + 4*a*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + a*b^2*sinh(d*x + c)^4 + 2*a*b^2*cosh(d*x + c)^2 +
a*b^2 + 2*(3*a*b^2*cosh(d*x + c)^2 + a*b^2)*sinh(d*x + c)^2 + 4*(a*b^2*cosh(d*x + c)^3 + a*b^2*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*b^2*cosh(d*x + c)^4 + 4*a*b^2
*cosh(d*x + c)*sinh(d*x + c)^3 + a*b^2*sinh(d*x + c)^4 + 2*a*b^2*cosh(d*x + c)^2 + a*b^2 + 2*(3*a*b^2*cosh(d*x
 + c)^2 + a*b^2)*sinh(d*x + c)^2 + 4*(a*b^2*cosh(d*x + c)^3 + a*b^2*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d
*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - (a^2*b + b^3 - 3*(a^2*b + b^3)*cosh(d*x + c)^2 + 4*(a^3 + a*b^2)*co
sh(d*x + c))*sinh(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^4 + 4*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x
+ c)*sinh(d*x + c)^3 + (a^4 + 2*a^2*b^2 + b^4)*d*sinh(d*x + c)^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^2
 + 2*(3*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^2 + (a^4 + 2*a^2*b^2 + b^4)*d)*sinh(d*x + c)^2 + (a^4 + 2*a^2*
b^2 + b^4)*d + 4*((a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^3 + (a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c))*sinh(d*
x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral(tanh(c + d*x)*sech(c + d*x)**2/(a + b*sinh(c + d*x)), x)

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Giac [A]  time = 1.35872, size = 304, normalized size = 2.49 \begin{align*} \frac{\frac{a b^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{a b^{2} \log \left ({\left | -b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a e^{\left (d x + c\right )} + b \right |}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a^{2} b e^{c} - b^{3} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} b e^{\left (3 \, c\right )} + b^{3} e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} - 2 \,{\left (a^{3} e^{\left (2 \, c\right )} + a b^{2} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} -{\left (a^{2} b e^{c} + b^{3} e^{c}\right )} e^{\left (d x\right )}}{{\left (a^{2} + b^{2}\right )}^{2}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

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