3.2.14 \(\int (a+b \text {sech}^2(c+d x))^2 \tanh ^2(c+d x) \, dx\) [114]

Optimal. Leaf size=59 \[ a^2 x-\frac {a^2 \tanh (c+d x)}{d}+\frac {b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

[Out]

a^2*x-a^2*tanh(d*x+c)/d+1/3*b*(2*a+b)*tanh(d*x+c)^3/d-1/5*b^2*tanh(d*x+c)^5/d

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4226, 1816, 212} \begin {gather*} -\frac {a^2 \tanh (c+d x)}{d}+a^2 x+\frac {b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^5(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*x - (a^2*Tanh[c + d*x])/d + (b*(2*a + b)*Tanh[c + d*x]^3)/(3*d) - (b^2*Tanh[c + d*x]^5)/(5*d)

Rule 212

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

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4226

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2
*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

\begin {align*} \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^2(c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \left (a+b \left (1-x^2\right )\right )^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-a^2+b (2 a+b) x^2-b^2 x^4+\frac {a^2}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {a^2 \tanh (c+d x)}{d}+\frac {b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^5(c+d x)}{5 d}+\frac {a^2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=a^2 x-\frac {a^2 \tanh (c+d x)}{d}+\frac {b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^5(c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(281\) vs. \(2(59)=118\).
time = 0.59, size = 281, normalized size = 4.76 \begin {gather*} \frac {\text {sech}(c) \text {sech}^5(c+d x) \left (150 a^2 d x \cosh (d x)+150 a^2 d x \cosh (2 c+d x)+75 a^2 d x \cosh (2 c+3 d x)+75 a^2 d x \cosh (4 c+3 d x)+15 a^2 d x \cosh (4 c+5 d x)+15 a^2 d x \cosh (6 c+5 d x)-180 a^2 \sinh (d x)+80 a b \sinh (d x)-20 b^2 \sinh (d x)+120 a^2 \sinh (2 c+d x)-120 a b \sinh (2 c+d x)-60 b^2 \sinh (2 c+d x)-120 a^2 \sinh (2 c+3 d x)+40 a b \sinh (2 c+3 d x)+20 b^2 \sinh (2 c+3 d x)+30 a^2 \sinh (4 c+3 d x)-60 a b \sinh (4 c+3 d x)-30 a^2 \sinh (4 c+5 d x)+20 a b \sinh (4 c+5 d x)+4 b^2 \sinh (4 c+5 d x)\right )}{480 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[c]*Sech[c + d*x]^5*(150*a^2*d*x*Cosh[d*x] + 150*a^2*d*x*Cosh[2*c + d*x] + 75*a^2*d*x*Cosh[2*c + 3*d*x] +
75*a^2*d*x*Cosh[4*c + 3*d*x] + 15*a^2*d*x*Cosh[4*c + 5*d*x] + 15*a^2*d*x*Cosh[6*c + 5*d*x] - 180*a^2*Sinh[d*x
] + 80*a*b*Sinh[d*x] - 20*b^2*Sinh[d*x] + 120*a^2*Sinh[2*c + d*x] - 120*a*b*Sinh[2*c + d*x] - 60*b^2*Sinh[2*c
+ d*x] - 120*a^2*Sinh[2*c + 3*d*x] + 40*a*b*Sinh[2*c + 3*d*x] + 20*b^2*Sinh[2*c + 3*d*x] + 30*a^2*Sinh[4*c + 3
*d*x] - 60*a*b*Sinh[4*c + 3*d*x] - 30*a^2*Sinh[4*c + 5*d*x] + 20*a*b*Sinh[4*c + 5*d*x] + 4*b^2*Sinh[4*c + 5*d*
x]))/(480*d)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(189\) vs. \(2(55)=110\).
time = 2.23, size = 190, normalized size = 3.22

method	result	size
risch	\(a^{2} x +\frac {2 a^{2} {\mathrm e}^{8 d x +8 c}-4 a b \,{\mathrm e}^{8 d x +8 c}+8 a^{2} {\mathrm e}^{6 d x +6 c}-8 a b \,{\mathrm e}^{6 d x +6 c}-4 b^{2} {\mathrm e}^{6 d x +6 c}+12 a^{2} {\mathrm e}^{4 d x +4 c}-\frac {16 a b \,{\mathrm e}^{4 d x +4 c}}{3}+\frac {4 b^{2} {\mathrm e}^{4 d x +4 c}}{3}+8 a^{2} {\mathrm e}^{2 d x +2 c}-\frac {8 a b \,{\mathrm e}^{2 d x +2 c}}{3}-\frac {4 b^{2} {\mathrm e}^{2 d x +2 c}}{3}+2 a^{2}-\frac {4 a b}{3}-\frac {4 b^{2}}{15}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )^{5}}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sech(d*x+c)^2)^2*tanh(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

a^2*x+2/15*(15*a^2*exp(8*d*x+8*c)-30*a*b*exp(8*d*x+8*c)+60*a^2*exp(6*d*x+6*c)-60*a*b*exp(6*d*x+6*c)-30*b^2*exp
(6*d*x+6*c)+90*a^2*exp(4*d*x+4*c)-40*a*b*exp(4*d*x+4*c)+10*b^2*exp(4*d*x+4*c)+60*a^2*exp(2*d*x+2*c)-20*a*b*exp
(2*d*x+2*c)-10*b^2*exp(2*d*x+2*c)+15*a^2-10*a*b-2*b^2)/d/(1+exp(2*d*x+2*c))^5

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (55) = 110\).
time = 0.28, size = 325, normalized size = 5.51 \begin {gather*} \frac {2 \, a b \tanh \left (d x + c\right )^{3}}{3 \, d} + a^{2} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + \frac {4}{15} \, b^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac {5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a*b*tanh(d*x + c)^3/d + a^2*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + 4/15*b^2*(5*e^(-2*d*x - 2*c)/(d*(5*
e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) -
5*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) +
e^(-10*d*x - 10*c) + 1)) + 15*e^(-6*d*x - 6*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6
*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-
6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)))

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 435 vs. \(2 (55) = 110\).
time = 0.35, size = 435, normalized size = 7.37 \begin {gather*} \frac {{\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (2 \, {\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} + 2 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (9 \, a^{2} + 2 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a^{2} + 4 \, a b + 8 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c)^5 + 5*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d
*x + c)*sinh(d*x + c)^4 - (15*a^2 - 10*a*b - 2*b^2)*sinh(d*x + c)^5 + 5*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)
*cosh(d*x + c)^3 - 5*(2*(15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c)^2 + 9*a^2 + 2*a*b - 2*b^2)*sinh(d*x + c)^3 + 5
*(2*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c)^3 + 3*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x
+ c))*sinh(d*x + c)^2 + 10*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c) - 5*((15*a^2 - 10*a*b - 2*b^2
)*cosh(d*x + c)^4 + 3*(9*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c)^2 + 6*a^2 + 4*a*b + 8*b^2)*sinh(d*x + c))/(d*cosh(
d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x +
c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2} \tanh ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (55) = 110\).
time = 0.43, size = 196, normalized size = 3.32 \begin {gather*} \frac {15 \, {\left (d x + c\right )} a^{2} + \frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} - 30 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 30 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 20 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(d*x + c)*a^2 + 2*(15*a^2*e^(8*d*x + 8*c) - 30*a*b*e^(8*d*x + 8*c) + 60*a^2*e^(6*d*x + 6*c) - 60*a*b*
e^(6*d*x + 6*c) - 30*b^2*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) - 40*a*b*e^(4*d*x + 4*c) + 10*b^2*e^(4*d*x +
4*c) + 60*a^2*e^(2*d*x + 2*c) - 20*a*b*e^(2*d*x + 2*c) - 10*b^2*e^(2*d*x + 2*c) + 15*a^2 - 10*a*b - 2*b^2)/(e
^(2*d*x + 2*c) + 1)^5)/d

________________________________________________________________________________________

Mupad [B]
time = 0.14, size = 513, normalized size = 8.69 \begin {gather*} \frac {\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2-b^2\right )}{5\,d}-\frac {2\,\left (2\,a\,b-a^2\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2-b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (2\,a\,b-a^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+2\,a\,b+4\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2+2\,a\,b+4\,b^2\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2-b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a\,b-a^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {\frac {2\,\left (a^2-b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a\,b-a^2\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\frac {2\,\left (a^2-b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2-b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (2\,a\,b-a^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+2\,a\,b+4\,b^2\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+a^2\,x-\frac {2\,\left (2\,a\,b-a^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8*exp(2*c + 2*d*x)*(a^2 - b^2))/(5*d) - (2*(2*a*b - a^2))/(5*d) + (8*exp(6*c + 6*d*x)*(a^2 - b^2))/(5*d) - (
2*exp(8*c + 8*d*x)*(2*a*b - a^2))/(5*d) + (4*exp(4*c + 4*d*x)*(2*a*b + 3*a^2 + 4*b^2))/(5*d))/(5*exp(2*c + 2*d
*x) + 10*exp(4*c + 4*d*x) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) + ((2*(2*a*b +
3*a^2 + 4*b^2))/(15*d) + (4*exp(2*c + 2*d*x)*(a^2 - b^2))/(5*d) - (2*exp(4*c + 4*d*x)*(2*a*b - a^2))/(5*d))/(3
*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) + ((2*(a^2 - b^2))/(5*d) - (2*exp(2*c + 2*d*x)*
(2*a*b - a^2))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) + ((2*(a^2 - b^2))/(5*d) + (6*exp(4*c + 4*d*
x)*(a^2 - b^2))/(5*d) - (2*exp(6*c + 6*d*x)*(2*a*b - a^2))/(5*d) + (2*exp(2*c + 2*d*x)*(2*a*b + 3*a^2 + 4*b^2)
)/(5*d))/(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) + a^2*x - (2*(2
*a*b - a^2))/(5*d*(exp(2*c + 2*d*x) + 1))

________________________________________________________________________________________