Integrand size = 23, antiderivative size = 148 \[ \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx=\frac {3 b^3 x}{8}-\frac {3 a^3 \text {arctanh}(\cosh (c+d x))}{8 d}+\frac {3 a b^2 \cosh (c+d x)}{d}-\frac {3 a^2 b \coth (c+d x)}{d}+\frac {3 a^3 \coth (c+d x) \text {csch}(c+d x)}{8 d}-\frac {a^3 \coth (c+d x) \text {csch}^3(c+d x)}{4 d}-\frac {3 b^3 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {b^3 \cosh (c+d x) \sinh ^3(c+d x)}{4 d} \]
3/8*b^3*x-3/8*a^3*arctanh(cosh(d*x+c))/d+3*a*b^2*cosh(d*x+c)/d-3*a^2*b*cot h(d*x+c)/d+3/8*a^3*coth(d*x+c)*csch(d*x+c)/d-1/4*a^3*coth(d*x+c)*csch(d*x+ c)^3/d-3/8*b^3*cosh(d*x+c)*sinh(d*x+c)/d+1/4*b^3*cosh(d*x+c)*sinh(d*x+c)^3 /d
Time = 7.21 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.61 \[ \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx=\frac {3 b^3 (c+d x)}{8 d}+\frac {3 a b^2 \cosh (c+d x)}{d}-\frac {3 a^2 b \coth \left (\frac {1}{2} (c+d x)\right )}{2 d}+\frac {3 a^3 \text {csch}^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {a^3 \text {csch}^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {3 a^3 \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {3 a^3 \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {3 a^3 \text {sech}^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {a^3 \text {sech}^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {b^3 \sinh (2 (c+d x))}{4 d}+\frac {b^3 \sinh (4 (c+d x))}{32 d}-\frac {3 a^2 b \tanh \left (\frac {1}{2} (c+d x)\right )}{2 d} \]
(3*b^3*(c + d*x))/(8*d) + (3*a*b^2*Cosh[c + d*x])/d - (3*a^2*b*Coth[(c + d *x)/2])/(2*d) + (3*a^3*Csch[(c + d*x)/2]^2)/(32*d) - (a^3*Csch[(c + d*x)/2 ]^4)/(64*d) - (3*a^3*Log[Cosh[(c + d*x)/2]])/(8*d) + (3*a^3*Log[Sinh[(c + d*x)/2]])/(8*d) + (3*a^3*Sech[(c + d*x)/2]^2)/(32*d) + (a^3*Sech[(c + d*x) /2]^4)/(64*d) - (b^3*Sinh[2*(c + d*x)])/(4*d) + (b^3*Sinh[4*(c + d*x)])/(3 2*d) - (3*a^2*b*Tanh[(c + d*x)/2])/(2*d)
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 26, 3699, 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.
| \(\displaystyle \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i \left (a+i b \sin (i c+i d x)^3\right )^3}{\sin (i c+i d x)^5}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {\left (i b \sin (i c+i d x)^3+a\right )^3}{\sin (i c+i d x)^5}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle i \int \left (-i a^3 \text {csch}^5(c+d x)-3 i a^2 b \text {csch}^2(c+d x)-i b^3 \sinh ^4(c+d x)-3 i a b^2 \sinh (c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle i \left (\frac {3 i a^3 \text {arctanh}(\cosh (c+d x))}{8 d}+\frac {i a^3 \coth (c+d x) \text {csch}^3(c+d x)}{4 d}-\frac {3 i a^3 \coth (c+d x) \text {csch}(c+d x)}{8 d}+\frac {3 i a^2 b \coth (c+d x)}{d}-\frac {3 i a b^2 \cosh (c+d x)}{d}-\frac {i b^3 \sinh ^3(c+d x) \cosh (c+d x)}{4 d}+\frac {3 i b^3 \sinh (c+d x) \cosh (c+d x)}{8 d}-\frac {3}{8} i b^3 x\right )\) |
I*(((-3*I)/8)*b^3*x + (((3*I)/8)*a^3*ArcTanh[Cosh[c + d*x]])/d - ((3*I)*a* b^2*Cosh[c + d*x])/d + ((3*I)*a^2*b*Coth[c + d*x])/d - (((3*I)/8)*a^3*Coth [c + d*x]*Csch[c + d*x])/d + ((I/4)*a^3*Coth[c + d*x]*Csch[c + d*x]^3)/d + (((3*I)/8)*b^3*Cosh[c + d*x]*Sinh[c + d*x])/d - ((I/4)*b^3*Cosh[c + d*x]* Sinh[c + d*x]^3)/d)
3.2.68.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n) ^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4] || Gt Q[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Time = 0.73 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\left (-\frac {\operatorname {csch}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {csch}\left (d x +c \right )}{8}\right ) \coth \left (d x +c \right )-\frac {3 \,\operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )}{4}\right )-3 a^{2} b \coth \left (d x +c \right )+3 a \,b^{2} \cosh \left (d x +c \right )+b^{3} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(108\) |
| default | \(\frac {a^{3} \left (\left (-\frac {\operatorname {csch}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {csch}\left (d x +c \right )}{8}\right ) \coth \left (d x +c \right )-\frac {3 \,\operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )}{4}\right )-3 a^{2} b \coth \left (d x +c \right )+3 a \,b^{2} \cosh \left (d x +c \right )+b^{3} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(108\) |
| parallelrisch | \(\frac {512 a^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{\frac {3}{8}}\right )-11 \operatorname {sech}\left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{3} \left (\cosh \left (d x +c \right )-\frac {19 \cosh \left (2 d x +2 c \right )}{22}-\frac {3 \cosh \left (3 d x +3 c \right )}{11}+\frac {19 \cosh \left (4 d x +4 c \right )}{88}+\frac {57}{88}\right ) \operatorname {csch}\left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+768 \,\operatorname {sech}\left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {csch}\left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b -1536 \left (a^{2} \coth \left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\frac {b x d}{8}+\cosh \left (d x +c \right ) a -\frac {b \sinh \left (2 d x +2 c \right )}{12}+\frac {b \sinh \left (4 d x +4 c \right )}{96}+a \right ) b \right ) b}{512 d}\) | \(175\) |
| risch | \(\frac {3 b^{3} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} b^{3}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} b^{3}}{8 d}+\frac {3 \,{\mathrm e}^{d x +c} a \,b^{2}}{2 d}+\frac {3 \,{\mathrm e}^{-d x -c} a \,b^{2}}{2 d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{3}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{3}}{64 d}+\frac {a^{2} \left (3 a \,{\mathrm e}^{7 d x +7 c}-24 b \,{\mathrm e}^{6 d x +6 c}-11 a \,{\mathrm e}^{5 d x +5 c}+72 b \,{\mathrm e}^{4 d x +4 c}-11 a \,{\mathrm e}^{3 d x +3 c}-72 b \,{\mathrm e}^{2 d x +2 c}+3 \,{\mathrm e}^{d x +c} a +24 b \right )}{4 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}+\frac {3 a^{3} \ln \left ({\mathrm e}^{d x +c}-1\right )}{8 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{d x +c}+1\right )}{8 d}\) | \(249\) |
1/d*(a^3*((-1/4*csch(d*x+c)^3+3/8*csch(d*x+c))*coth(d*x+c)-3/4*arctanh(exp (d*x+c)))-3*a^2*b*coth(d*x+c)+3*a*b^2*cosh(d*x+c)+b^3*((1/4*sinh(d*x+c)^3- 3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c))
Leaf count of result is larger than twice the leaf count of optimal. 4541 vs. \(2 (136) = 272\).
Time = 0.32 (sec) , antiderivative size = 4541, normalized size of antiderivative = 30.68 \[ \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx=\text {Too large to display} \]
1/64*(b^3*cosh(d*x + c)^16 + 16*b^3*cosh(d*x + c)*sinh(d*x + c)^15 + b^3*s inh(d*x + c)^16 - 12*b^3*cosh(d*x + c)^14 + 96*a*b^2*cosh(d*x + c)^13 + 12 *(10*b^3*cosh(d*x + c)^2 - b^3)*sinh(d*x + c)^14 + 8*(70*b^3*cosh(d*x + c) ^3 - 21*b^3*cosh(d*x + c) + 12*a*b^2)*sinh(d*x + c)^13 + 2*(12*b^3*d*x + 1 9*b^3)*cosh(d*x + c)^12 + 2*(910*b^3*cosh(d*x + c)^4 + 12*b^3*d*x - 546*b^ 3*cosh(d*x + c)^2 + 624*a*b^2*cosh(d*x + c) + 19*b^3)*sinh(d*x + c)^12 + 4 8*(a^3 - 6*a*b^2)*cosh(d*x + c)^11 + 24*(182*b^3*cosh(d*x + c)^5 - 182*b^3 *cosh(d*x + c)^3 + 312*a*b^2*cosh(d*x + c)^2 + 2*a^3 - 12*a*b^2 + (12*b^3* d*x + 19*b^3)*cosh(d*x + c))*sinh(d*x + c)^11 - 4*(24*b^3*d*x + 96*a^2*b + 11*b^3)*cosh(d*x + c)^10 + 4*(2002*b^3*cosh(d*x + c)^6 - 3003*b^3*cosh(d* x + c)^4 + 6864*a*b^2*cosh(d*x + c)^3 - 24*b^3*d*x - 96*a^2*b - 11*b^3 + 3 3*(12*b^3*d*x + 19*b^3)*cosh(d*x + c)^2 + 132*(a^3 - 6*a*b^2)*cosh(d*x + c ))*sinh(d*x + c)^10 - 16*(11*a^3 - 12*a*b^2)*cosh(d*x + c)^9 + 8*(1430*b^3 *cosh(d*x + c)^7 - 3003*b^3*cosh(d*x + c)^5 + 8580*a*b^2*cosh(d*x + c)^4 + 55*(12*b^3*d*x + 19*b^3)*cosh(d*x + c)^3 - 22*a^3 + 24*a*b^2 + 330*(a^3 - 6*a*b^2)*cosh(d*x + c)^2 - 5*(24*b^3*d*x + 96*a^2*b + 11*b^3)*cosh(d*x + c))*sinh(d*x + c)^9 + 144*(b^3*d*x + 8*a^2*b)*cosh(d*x + c)^8 + 18*(715*b^ 3*cosh(d*x + c)^8 - 2002*b^3*cosh(d*x + c)^6 + 6864*a*b^2*cosh(d*x + c)^5 + 8*b^3*d*x + 55*(12*b^3*d*x + 19*b^3)*cosh(d*x + c)^4 + 440*(a^3 - 6*a*b^ 2)*cosh(d*x + c)^3 + 64*a^2*b - 10*(24*b^3*d*x + 96*a^2*b + 11*b^3)*cos...
Timed out. \[ \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.72 \[ \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx=\frac {1}{64} \, b^{3} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {3}{2} \, a b^{2} {\left (\frac {e^{\left (d x + c\right )}}{d} + \frac {e^{\left (-d x - c\right )}}{d}\right )} - \frac {1}{8} \, a^{3} {\left (\frac {3 \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac {3 \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, {\left (3 \, e^{\left (-d x - c\right )} - 11 \, e^{\left (-3 \, d x - 3 \, c\right )} - 11 \, e^{\left (-5 \, d x - 5 \, c\right )} + 3 \, e^{\left (-7 \, d x - 7 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + \frac {6 \, a^{2} b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]
1/64*b^3*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2 *c)/d - e^(-4*d*x - 4*c)/d) + 3/2*a*b^2*(e^(d*x + c)/d + e^(-d*x - c)/d) - 1/8*a^3*(3*log(e^(-d*x - c) + 1)/d - 3*log(e^(-d*x - c) - 1)/d + 2*(3*e^( -d*x - c) - 11*e^(-3*d*x - 3*c) - 11*e^(-5*d*x - 5*c) + 3*e^(-7*d*x - 7*c) )/(d*(4*e^(-2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) - e^(-8 *d*x - 8*c) - 1))) + 6*a^2*b/(d*(e^(-2*d*x - 2*c) - 1))
Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (136) = 272\).
Time = 0.46 (sec) , antiderivative size = 329, normalized size of antiderivative = 2.22 \[ \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx=\frac {24 \, {\left (d x + c\right )} b^{3} + b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 96 \, a b^{2} e^{\left (d x + c\right )} - 24 \, a^{3} \log \left (e^{\left (d x + c\right )} + 1\right ) + 24 \, a^{3} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right ) + \frac {{\left (96 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - b^{3} + 48 \, {\left (a^{3} + 2 \, a b^{2}\right )} e^{\left (11 \, d x + 11 \, c\right )} - 8 \, {\left (48 \, a^{2} b - b^{3}\right )} e^{\left (10 \, d x + 10 \, c\right )} - 16 \, {\left (11 \, a^{3} + 24 \, a b^{2}\right )} e^{\left (9 \, d x + 9 \, c\right )} + 3 \, {\left (384 \, a^{2} b - 11 \, b^{3}\right )} e^{\left (8 \, d x + 8 \, c\right )} - 16 \, {\left (11 \, a^{3} - 36 \, a b^{2}\right )} e^{\left (7 \, d x + 7 \, c\right )} - 4 \, {\left (288 \, a^{2} b - 13 \, b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )} + 48 \, {\left (a^{3} - 8 \, a b^{2}\right )} e^{\left (5 \, d x + 5 \, c\right )} + 2 \, {\left (192 \, a^{2} b - 19 \, b^{3}\right )} e^{\left (4 \, d x + 4 \, c\right )}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{{\left (e^{\left (d x + c\right )} + 1\right )}^{4} {\left (e^{\left (d x + c\right )} - 1\right )}^{4}}}{64 \, d} \]
1/64*(24*(d*x + c)*b^3 + b^3*e^(4*d*x + 4*c) - 8*b^3*e^(2*d*x + 2*c) + 96* a*b^2*e^(d*x + c) - 24*a^3*log(e^(d*x + c) + 1) + 24*a^3*log(abs(e^(d*x + c) - 1)) + (96*a*b^2*e^(3*d*x + 3*c) + 12*b^3*e^(2*d*x + 2*c) - b^3 + 48*( a^3 + 2*a*b^2)*e^(11*d*x + 11*c) - 8*(48*a^2*b - b^3)*e^(10*d*x + 10*c) - 16*(11*a^3 + 24*a*b^2)*e^(9*d*x + 9*c) + 3*(384*a^2*b - 11*b^3)*e^(8*d*x + 8*c) - 16*(11*a^3 - 36*a*b^2)*e^(7*d*x + 7*c) - 4*(288*a^2*b - 13*b^3)*e^ (6*d*x + 6*c) + 48*(a^3 - 8*a*b^2)*e^(5*d*x + 5*c) + 2*(192*a^2*b - 19*b^3 )*e^(4*d*x + 4*c))*e^(-4*d*x - 4*c)/((e^(d*x + c) + 1)^4*(e^(d*x + c) - 1) ^4))/d
Time = 1.64 (sec) , antiderivative size = 451, normalized size of antiderivative = 3.05 \[ \int \text {csch}^5(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx=\frac {3\,b^3\,x}{8}-\frac {\frac {3\,a^2\,b}{2\,d}+\frac {2\,a^3\,{\mathrm {e}}^{c+d\,x}}{d}-\frac {3\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{d}+\frac {3\,a^2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{2\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {\frac {4\,a^3\,{\mathrm {e}}^{3\,c+3\,d\,x}}{d}-\frac {3\,a^2\,b}{2\,d}+\frac {9\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{2\,d}-\frac {9\,a^2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{2\,d}+\frac {3\,a^2\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{2\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {3\,a^2\,b}{d}-\frac {3\,a^3\,{\mathrm {e}}^{c+d\,x}}{4\,d}}{{\mathrm {e}}^{2\,c+2\,d\,x}-1}-\frac {3\,\mathrm {atan}\left (\frac {a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-d^2}}{d\,\sqrt {a^6}}\right )\,\sqrt {a^6}}{4\,\sqrt {-d^2}}+\frac {b^3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {b^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {b^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {b^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}+\frac {3\,a\,b^2\,{\mathrm {e}}^{-c-d\,x}}{2\,d}+\frac {3\,a\,b^2\,{\mathrm {e}}^{c+d\,x}}{2\,d}-\frac {a^3\,{\mathrm {e}}^{c+d\,x}}{2\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]
(3*b^3*x)/8 - ((3*a^2*b)/(2*d) + (2*a^3*exp(c + d*x))/d - (3*a^2*b*exp(2*c + 2*d*x))/d + (3*a^2*b*exp(4*c + 4*d*x))/(2*d))/(3*exp(2*c + 2*d*x) - 3*e xp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - ((4*a^3*exp(3*c + 3*d*x))/d - (3 *a^2*b)/(2*d) + (9*a^2*b*exp(2*c + 2*d*x))/(2*d) - (9*a^2*b*exp(4*c + 4*d* x))/(2*d) + (3*a^2*b*exp(6*c + 6*d*x))/(2*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) - ((3*a^2*b)/d - (3*a^3*exp(c + d*x))/(4*d))/(exp(2*c + 2*d*x) - 1) - (3*atan((a^3*exp(d*x )*exp(c)*(-d^2)^(1/2))/(d*(a^6)^(1/2)))*(a^6)^(1/2))/(4*(-d^2)^(1/2)) + (b ^3*exp(- 2*c - 2*d*x))/(8*d) - (b^3*exp(2*c + 2*d*x))/(8*d) - (b^3*exp(- 4 *c - 4*d*x))/(64*d) + (b^3*exp(4*c + 4*d*x))/(64*d) + (3*a*b^2*exp(- c - d *x))/(2*d) + (3*a*b^2*exp(c + d*x))/(2*d) - (a^3*exp(c + d*x))/(2*d*(exp(4 *c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1))