Integrand size = 23, antiderivative size = 57 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tanh (c+d x)}{d}-\frac {2 a (a-b) \tanh ^3(c+d x)}{3 d}+\frac {(a-b)^2 \tanh ^5(c+d x)}{5 d} \]
Time = 1.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx=\frac {\left (8 a^2+4 a b+3 b^2+2 \left (2 a^2+a b-3 b^2\right ) \text {sech}^2(c+d x)+3 (a-b)^2 \text {sech}^4(c+d x)\right ) \tanh (c+d x)}{15 d} \]
((8*a^2 + 4*a*b + 3*b^2 + 2*(2*a^2 + a*b - 3*b^2)*Sech[c + d*x]^2 + 3*(a - b)^2*Sech[c + d*x]^4)*Tanh[c + d*x])/(15*d)
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3670, 210, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-b \sin (i c+i d x)^2\right )^2}{\cos (i c+i d x)^6}dx\) |
| \(\Big \downarrow \) 3670 |
| \(\displaystyle \frac {\int \left (a-(a-b) \tanh ^2(c+d x)\right )^2d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 210 |
| \(\displaystyle \frac {\int \left ((a-b)^2 \tanh ^4(c+d x)-2 a (a-b) \tanh ^2(c+d x)+a^2\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \tanh (c+d x)+\frac {1}{5} (a-b)^2 \tanh ^5(c+d x)-\frac {2}{3} a (a-b) \tanh ^3(c+d x)}{d}\) |
3.4.2.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(128\) vs. \(2(53)=106\).
Time = 109.80 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.26
| method | result | size |
| risch | \(-\frac {2 \left (15 b^{2} {\mathrm e}^{8 d x +8 c}+60 \,{\mathrm e}^{6 d x +6 c} a b +80 \,{\mathrm e}^{4 d x +4 c} a^{2}-20 \,{\mathrm e}^{4 d x +4 c} a b +30 b^{2} {\mathrm e}^{4 d x +4 c}+40 \,{\mathrm e}^{2 d x +2 c} a^{2}+20 \,{\mathrm e}^{2 d x +2 c} b a +8 a^{2}+4 a b +3 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(129\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(158\) |
| default | \(\frac {a^{2} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(158\) |
-2/15*(15*b^2*exp(8*d*x+8*c)+60*exp(6*d*x+6*c)*a*b+80*exp(4*d*x+4*c)*a^2-2 0*exp(4*d*x+4*c)*a*b+30*b^2*exp(4*d*x+4*c)+40*exp(2*d*x+2*c)*a^2+20*exp(2* d*x+2*c)*b*a+8*a^2+4*a*b+3*b^2)/d/(exp(2*d*x+2*c)+1)^5
Leaf count of result is larger than twice the leaf count of optimal. 403 vs. \(2 (53) = 106\).
Time = 0.27 (sec) , antiderivative size = 403, normalized size of antiderivative = 7.07 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx=-\frac {4 \, {\left ({\left (4 \, a^{2} + 2 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} - 8 \, {\left (2 \, a^{2} + a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (4 \, a^{2} + 2 \, a b + 9 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (4 \, a^{2} + 2 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 10 \, a^{2} + 20 \, a b\right )} \sinh \left (d x + c\right )^{2} + 40 \, a^{2} - 10 \, a b + 15 \, b^{2} - 8 \, {\left ({\left (2 \, a^{2} + a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (a^{2} - a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} + 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{4} + 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} + 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 10 \, d\right )}} \]
-4/15*((4*a^2 + 2*a*b + 9*b^2)*cosh(d*x + c)^4 - 8*(2*a^2 + a*b - 3*b^2)*c osh(d*x + c)*sinh(d*x + c)^3 + (4*a^2 + 2*a*b + 9*b^2)*sinh(d*x + c)^4 + 2 0*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(4*a^2 + 2*a*b + 9*b^2)*cosh(d*x + c)^2 + 10*a^2 + 20*a*b)*sinh(d*x + c)^2 + 40*a^2 - 10*a*b + 15*b^2 - 8*((2 *a^2 + a*b - 3*b^2)*cosh(d*x + c)^3 + 5*(a^2 - a*b)*cosh(d*x + c))*sinh(d* x + c))/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d* x + c)^6 + 6*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 + 4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 15*d* cosh(d*x + c)^2 + 3*(5*d*cosh(d*x + c)^4 + 12*d*cosh(d*x + c)^2 + 5*d)*sin h(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 + 8*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c) + 10*d)
Timed out. \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 698 vs. \(2 (53) = 106\).
Time = 0.20 (sec) , antiderivative size = 698, normalized size of antiderivative = 12.25 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx=\frac {16}{15} \, a^{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 {10 \, 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 {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 )} + \frac {8}{15} \, a b {\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 )} + \frac {2}{5} \, b^{2} {\left (\frac {10 \, 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 {5 \, e^{\left (-8 \, d x - 8 \, 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 )} \]
16/15*a^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)) + 1 0*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)) + 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))) + 8/15*a*b*(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))) + 2/5*b^2*(10*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)) + 5*e^(-8*d*x - 8*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)))
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (53) = 106\).
Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.25 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (15 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 20 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 30 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 20 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]
-2/15*(15*b^2*e^(8*d*x + 8*c) + 60*a*b*e^(6*d*x + 6*c) + 80*a^2*e^(4*d*x + 4*c) - 20*a*b*e^(4*d*x + 4*c) + 30*b^2*e^(4*d*x + 4*c) + 40*a^2*e^(2*d*x + 2*c) + 20*a*b*e^(2*d*x + 2*c) + 8*a^2 + 4*a*b + 3*b^2)/(d*(e^(2*d*x + 2* c) + 1)^5)
Time = 1.61 (sec) , antiderivative size = 464, normalized size of antiderivative = 8.14 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx=-\frac {\frac {2\,\left (8\,a^2-8\,a\,b+3\,b^2\right )}{15\,d}+\frac {2\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,b^2}{5\,d}+\frac {2\,b^2\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (8\,a^2-8\,a\,b+3\,b^2\right )}{5\,d}+\frac {8\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-b\right )}{5\,d}+\frac {8\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,b\,\left (2\,a-b\right )}{5\,d}+\frac {2\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,b\,\left (2\,a-b\right )}{5\,d}+\frac {2\,b^2\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (8\,a^2-8\,a\,b+3\,b^2\right )}{5\,d}+\frac {6\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {2\,b^2}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]
- ((2*(8*a^2 - 8*a*b + 3*b^2))/(15*d) + (2*b^2*exp(4*c + 4*d*x))/(5*d) + ( 4*b*exp(2*c + 2*d*x)*(2*a - b))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4 *d*x) + exp(6*c + 6*d*x) + 1) - ((2*b^2)/(5*d) + (2*b^2*exp(8*c + 8*d*x))/ (5*d) + (4*exp(4*c + 4*d*x)*(8*a^2 - 8*a*b + 3*b^2))/(5*d) + (8*b*exp(2*c + 2*d*x)*(2*a - b))/(5*d) + (8*b*exp(6*c + 6*d*x)*(2*a - b))/(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*b*(2*a - b))/(5*d) + (2*b^2*exp(2*c + 2*d*x))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*b*(2*a - b))/(5*d) + (2*b^2*exp(6*c + 6*d*x))/(5*d) + (2*exp(2*c + 2*d*x)*(8*a^2 - 8*a*b + 3*b^2))/(5*d) + (6*b*exp(4*c + 4*d*x)*(2*a - b))/(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) - (2*b^2)/(5*d*(exp(2*c + 2*d*x) + 1))