Integrand size = 23, antiderivative size = 64 \[ \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=a^2 x-\frac {a^2 \coth (c+d x)}{d}-\frac {\left (a^2-b^2\right ) \coth ^3(c+d x)}{3 d}-\frac {(a+b)^2 \coth ^5(c+d x)}{5 d} \]
Leaf count is larger than twice the leaf count of optimal. \(256\) vs. \(2(64)=128\).
Time = 4.27 (sec) , antiderivative size = 256, normalized size of antiderivative = 4.00 \[ \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {\text {csch}(c) \text {csch}^5(c+d x) \left (-150 a^2 d x \cosh (d x)+150 a^2 d x \cosh (2 c+d x)+75 a^2 d x \cosh (2 c+3 d x)-75 a^2 d x \cosh (4 c+3 d x)-15 a^2 d x \cosh (4 c+5 d x)+15 a^2 d x \cosh (6 c+5 d x)+280 a^2 \sinh (d x)+120 a b \sinh (d x)+20 b^2 \sinh (d x)+180 a^2 \sinh (2 c+d x)-60 b^2 \sinh (2 c+d x)-140 a^2 \sinh (2 c+3 d x)+20 b^2 \sinh (2 c+3 d x)-90 a^2 \sinh (4 c+3 d x)-60 a b \sinh (4 c+3 d x)+46 a^2 \sinh (4 c+5 d x)+12 a b \sinh (4 c+5 d x)-4 b^2 \sinh (4 c+5 d x)\right )}{480 d} \]
(Csch[c]*Csch[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] + 280*a^2*Sinh[d*x] + 120*a*b*Sinh[d*x] + 20*b^2*Sinh[d*x] + 180*a^2*Sinh[2*c + d*x] - 60*b^2 *Sinh[2*c + d*x] - 140*a^2*Sinh[2*c + 3*d*x] + 20*b^2*Sinh[2*c + 3*d*x] - 90*a^2*Sinh[4*c + 3*d*x] - 60*a*b*Sinh[4*c + 3*d*x] + 46*a^2*Sinh[4*c + 5* d*x] + 12*a*b*Sinh[4*c + 5*d*x] - 4*b^2*Sinh[4*c + 5*d*x]))/(480*d)
Time = 0.33 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4629, 25, 2075, 364, 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.
| \(\displaystyle \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a+b \sec (i c+i d x)^2\right )^2}{\tan (i c+i d x)^6}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (b \sec (i c+i d x)^2+a\right )^2}{\tan (i c+i d x)^6}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle -\frac {\int -\frac {\coth ^6(c+d x) \left (a+b \left (1-\tanh ^2(c+d x)\right )\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\coth ^6(c+d x) \left (a+b \left (1-\tanh ^2(c+d x)\right )\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\coth ^6(c+d x) \left (-b \tanh ^2(c+d x)+a+b\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {\int \left ((a+b)^2 \coth ^6(c+d x)+\left (a^2-b^2\right ) \coth ^4(c+d x)+a^2 \coth ^2(c+d x)-\frac {a^2}{\tanh ^2(c+d x)-1}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-a^2 \text {arctanh}(\tanh (c+d x))+\frac {1}{3} \left (a^2-b^2\right ) \coth ^3(c+d x)+a^2 \coth (c+d x)+\frac {1}{5} (a+b)^2 \coth ^5(c+d x)}{d}\) |
-((-(a^2*ArcTanh[Tanh[c + d*x]]) + a^2*Coth[c + d*x] + ((a^2 - b^2)*Coth[c + d*x]^3)/3 + ((a + b)^2*Coth[c + d*x]^5)/5)/d)
3.2.22.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
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]}, Sim p[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] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(162\) vs. \(2(60)=120\).
Time = 34.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.55
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (d x +c -\coth \left (d x +c \right )-\frac {\coth \left (d x +c \right )^{3}}{3}-\frac {\coth \left (d x +c \right )^{5}}{5}\right )+2 a b \left (-\frac {\cosh \left (d x +c \right )^{3}}{2 \sinh \left (d x +c \right )^{5}}+\frac {3 \cosh \left (d x +c \right )}{8 \sinh \left (d x +c \right )^{5}}+\frac {3 \left (-\frac {8}{15}-\frac {\operatorname {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{8}\right )+b^{2} \left (-\frac {\cosh \left (d x +c \right )}{4 \sinh \left (d x +c \right )^{5}}-\frac {\left (-\frac {8}{15}-\frac {\operatorname {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{4}\right )}{d}\) | \(163\) |
| default | \(\frac {a^{2} \left (d x +c -\coth \left (d x +c \right )-\frac {\coth \left (d x +c \right )^{3}}{3}-\frac {\coth \left (d x +c \right )^{5}}{5}\right )+2 a b \left (-\frac {\cosh \left (d x +c \right )^{3}}{2 \sinh \left (d x +c \right )^{5}}+\frac {3 \cosh \left (d x +c \right )}{8 \sinh \left (d x +c \right )^{5}}+\frac {3 \left (-\frac {8}{15}-\frac {\operatorname {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{8}\right )+b^{2} \left (-\frac {\cosh \left (d x +c \right )}{4 \sinh \left (d x +c \right )^{5}}-\frac {\left (-\frac {8}{15}-\frac {\operatorname {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{4}\right )}{d}\) | \(163\) |
| risch | \(a^{2} x -\frac {2 \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+30 a b \,{\mathrm e}^{8 d x +8 c}-90 a^{2} {\mathrm e}^{6 d x +6 c}+30 b^{2} {\mathrm e}^{6 d x +6 c}+140 a^{2} {\mathrm e}^{4 d x +4 c}+60 a b \,{\mathrm e}^{4 d x +4 c}+10 \,{\mathrm e}^{4 d x +4 c} b^{2}-70 a^{2} {\mathrm e}^{2 d x +2 c}+10 \,{\mathrm e}^{2 d x +2 c} b^{2}+23 a^{2}+6 a b -2 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\) | \(164\) |
1/d*(a^2*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3-1/5*coth(d*x+c)^5)+2*a*b*(-1 /2/sinh(d*x+c)^5*cosh(d*x+c)^3+3/8/sinh(d*x+c)^5*cosh(d*x+c)+3/8*(-8/15-1/ 5*csch(d*x+c)^4+4/15*csch(d*x+c)^2)*coth(d*x+c))+b^2*(-1/4/sinh(d*x+c)^5*c osh(d*x+c)-1/4*(-8/15-1/5*csch(d*x+c)^4+4/15*csch(d*x+c)^2)*coth(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (60) = 120\).
Time = 0.24 (sec) , antiderivative size = 425, normalized size of antiderivative = 6.64 \[ \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {{\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{2} - 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (15 \, a^{2} d x - 2 \, {\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{2} - 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{2} + 6 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 30 \, a^{2} d x - 3 \, {\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 46 \, a^{2} + 12 \, a b - 4 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]
-1/15*((23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^5 + 5*(23*a^2 + 6*a*b - 2*b^ 2)*cosh(d*x + c)*sinh(d*x + c)^4 - (15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*s inh(d*x + c)^5 - 5*(5*a^2 - 6*a*b - 2*b^2)*cosh(d*x + c)^3 + 5*(15*a^2*d*x - 2*(15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^2 + 23*a^2 + 6*a* b - 2*b^2)*sinh(d*x + c)^3 + 5*(2*(23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^3 - 3*(5*a^2 - 6*a*b - 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^2 + 6*a*b + 4*b^2)*cosh(d*x + c) - 5*((15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*co sh(d*x + c)^4 + 30*a^2*d*x - 3*(15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*cosh( d*x + c)^2 + 46*a^2 + 12*a*b - 4*b^2)*sinh(d*x + c))/(d*sinh(d*x + c)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5*(d*cosh(d*x + c)^4 - 3*d*c osh(d*x + c)^2 + 2*d)*sinh(d*x + c))
Timed out. \[ \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (60) = 120\).
Time = 0.21 (sec) , antiderivative size = 613, normalized size of antiderivative = 9.58 \[ \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {1}{15} \, a^{2} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\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 )}}\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 )} + \frac {4}{5} \, a b {\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 )} \]
1/15*a^2*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) - 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c) - 45*e^(-8*d*x - 8*c) - 23)/(d*(5*e^(-2*d*x - 2*c) - 1 0*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))) + 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))) + 4/5*a*b*(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. 170 vs. \(2 (60) = 120\).
Time = 0.36 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.66 \[ \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {15 \, {\left (d x + c\right )} a^{2} - \frac {2 \, {\left (45 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 30 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 30 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \]
1/15*(15*(d*x + c)*a^2 - 2*(45*a^2*e^(8*d*x + 8*c) + 30*a*b*e^(8*d*x + 8*c ) - 90*a^2*e^(6*d*x + 6*c) + 30*b^2*e^(6*d*x + 6*c) + 140*a^2*e^(4*d*x + 4 *c) + 60*a*b*e^(4*d*x + 4*c) + 10*b^2*e^(4*d*x + 4*c) - 70*a^2*e^(2*d*x + 2*c) + 10*b^2*e^(2*d*x + 2*c) + 23*a^2 + 6*a*b - 2*b^2)/(e^(2*d*x + 2*c) - 1)^5)/d
Time = 2.18 (sec) , antiderivative size = 511, normalized size of antiderivative = 7.98 \[ \int \coth ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=a^2\,x-\frac {\frac {2\,\left (5\,a^2+6\,a\,b+4\,b^2\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^2+6\,a\,b+4\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2+2\,b\,a\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^2+6\,a\,b+4\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {2\,\left (3\,a^2+2\,b\,a\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]
a^2*x - ((2*(6*a*b + 5*a^2 + 4*b^2))/(15*d) + (4*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d) + (2*exp(4*c + 4*d*x)*(2*a*b + 3*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*(2*a*b + b^2))/(5* d) + (2*exp(2*c + 2*d*x)*(2*a*b + 3*a^2))/(5*d))/(exp(4*c + 4*d*x) - 2*exp (2*c + 2*d*x) + 1) - ((2*(2*a*b + b^2))/(5*d) + (6*exp(4*c + 4*d*x)*(2*a*b + b^2))/(5*d) + (2*exp(6*c + 6*d*x)*(2*a*b + 3*a^2))/(5*d) + (2*exp(2*c + 2*d*x)*(6*a*b + 5*a^2 + 4*b^2))/(5*d))/(6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((2*(2*a*b + 3*a^2)) /(5*d) + (8*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d) + (8*exp(6*c + 6*d*x)*(2 *a*b + b^2))/(5*d) + (2*exp(8*c + 8*d*x)*(2*a*b + 3*a^2))/(5*d) + (4*exp(4 *c + 4*d*x)*(6*a*b + 5*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))/(5*d*(exp(2*c + 2*d*x) - 1))