Integrand size = 23, antiderivative size = 87 \[ \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {b^2 (6 a+5 b) \arctan (\sinh (c+d x))}{2 d}+\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}-\frac {b^3 \text {sech}(c+d x) \tanh (c+d x)}{2 d} \]
1/2*b^2*(6*a+5*b)*arctan(sinh(d*x+c))/d+(a-2*b)*(a+b)^2*sinh(d*x+c)/d+1/3* (a+b)^3*sinh(d*x+c)^3/d-1/2*b^3*sech(d*x+c)*tanh(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 7.06 (sec) , antiderivative size = 494, normalized size of antiderivative = 5.68 \[ \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {\text {csch}^5(c+d x) \left (-256 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};-\sinh ^2(c+d x)\right ) \sinh ^8(c+d x) \left (a+a \sinh ^2(c+d x)+b \sinh ^2(c+d x)\right )^3-\frac {315 \text {arctanh}\left (\sqrt {-\sinh ^2(c+d x)}\right ) \left (b^3 \sinh ^6(c+d x) \left (2161+1875 \sinh ^2(c+d x)+243 \sinh ^4(c+d x)+\sinh ^6(c+d x)\right )+a^3 \cosh ^6(c+d x) \left (2401+1875 \sinh ^2(c+d x)+243 \sinh ^4(c+d x)+\sinh ^6(c+d x)\right )+3 a^2 b \left (\sinh (c+d x)+\sinh ^3(c+d x)\right )^2 \left (2401+1875 \sinh ^2(c+d x)+243 \sinh ^4(c+d x)+\sinh ^6(c+d x)\right )+3 a b^2 \sinh ^4(c+d x) \left (2401+4180 \sinh ^2(c+d x)+2118 \sinh ^4(c+d x)+244 \sinh ^6(c+d x)+\sinh ^8(c+d x)\right )\right )}{\sqrt {-\sinh ^2(c+d x)}}+21 \left (b^3 \sinh ^6(c+d x) \left (32415+17320 \sinh ^2(c+d x)+753 \sinh ^4(c+d x)\right )+3 a b^2 \sinh ^4(c+d x) \left (36015+50695 \sinh ^2(c+d x)+18073 \sinh ^4(c+d x)+753 \sinh ^6(c+d x)\right )+3 a^2 b \sinh ^2(c+d x) \left (36015+88150 \sinh ^2(c+d x)+69728 \sinh ^4(c+d x)+18826 \sinh ^6(c+d x)+753 \sinh ^8(c+d x)\right )+a^3 \left (36015+124165 \sinh ^2(c+d x)+157878 \sinh ^4(c+d x)+89514 \sinh ^6(c+d x)+19579 \sinh ^8(c+d x)+753 \sinh ^{10}(c+d x)\right )\right )\right )}{30240 d} \]
(Csch[c + d*x]^5*(-256*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2 }, -Sinh[c + d*x]^2]*Sinh[c + d*x]^8*(a + a*Sinh[c + d*x]^2 + b*Sinh[c + d *x]^2)^3 - (315*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^3*Sinh[c + d*x]^6*(2161 + 1875*Sinh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + a^3*Cos h[c + d*x]^6*(2401 + 1875*Sinh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + 3*a^2*b*(Sinh[c + d*x] + Sinh[c + d*x]^3)^2*(2401 + 1875*Sinh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + 3*a*b^2*Sinh[c + d*x] ^4*(2401 + 4180*Sinh[c + d*x]^2 + 2118*Sinh[c + d*x]^4 + 244*Sinh[c + d*x] ^6 + Sinh[c + d*x]^8)))/Sqrt[-Sinh[c + d*x]^2] + 21*(b^3*Sinh[c + d*x]^6*( 32415 + 17320*Sinh[c + d*x]^2 + 753*Sinh[c + d*x]^4) + 3*a*b^2*Sinh[c + d* x]^4*(36015 + 50695*Sinh[c + d*x]^2 + 18073*Sinh[c + d*x]^4 + 753*Sinh[c + d*x]^6) + 3*a^2*b*Sinh[c + d*x]^2*(36015 + 88150*Sinh[c + d*x]^2 + 69728* Sinh[c + d*x]^4 + 18826*Sinh[c + d*x]^6 + 753*Sinh[c + d*x]^8) + a^3*(3601 5 + 124165*Sinh[c + d*x]^2 + 157878*Sinh[c + d*x]^4 + 89514*Sinh[c + d*x]^ 6 + 19579*Sinh[c + d*x]^8 + 753*Sinh[c + d*x]^10))))/(30240*d)
Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4159, 300, 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.
| \(\displaystyle \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\sec (i c+i d x)^3}dx\) |
| \(\Big \downarrow \) 4159 |
| \(\displaystyle \frac {\int \frac {\left ((a+b) \sinh ^2(c+d x)+a\right )^3}{\left (\sinh ^2(c+d x)+1\right )^2}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (\sinh ^2(c+d x) (a+b)^3+(a-2 b) (a+b)^2+\frac {3 (a+b) \sinh ^2(c+d x) b^2+(3 a+2 b) b^2}{\left (\sinh ^2(c+d x)+1\right )^2}\right )d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} b^2 (6 a+5 b) \arctan (\sinh (c+d x))+\frac {1}{3} (a+b)^3 \sinh ^3(c+d x)+(a-2 b) (a+b)^2 \sinh (c+d x)-\frac {b^3 \sinh (c+d x)}{2 \left (\sinh ^2(c+d x)+1\right )}}{d}\) |
((b^2*(6*a + 5*b)*ArcTan[Sinh[c + d*x]])/2 + (a - 2*b)*(a + b)^2*Sinh[c + d*x] + ((a + b)^3*Sinh[c + d*x]^3)/3 - (b^3*Sinh[c + d*x])/(2*(1 + Sinh[c + d*x]^2)))/d
3.1.98.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 22.77 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.78
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {2}{3}+\frac {\cosh \left (d x +c \right )^{2}}{3}\right ) \sinh \left (d x +c \right )+a^{2} b \sinh \left (d x +c \right )^{3}+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right )^{3}}{3}-\sinh \left (d x +c \right )+2 \arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {5 \,\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+5 \arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}\) | \(155\) |
| default | \(\frac {a^{3} \left (\frac {2}{3}+\frac {\cosh \left (d x +c \right )^{2}}{3}\right ) \sinh \left (d x +c \right )+a^{2} b \sinh \left (d x +c \right )^{3}+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right )^{3}}{3}-\sinh \left (d x +c \right )+2 \arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {5 \,\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+5 \arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}\) | \(155\) |
| risch | \(\frac {{\mathrm e}^{3 d x +3 c} a^{3}}{24 d}+\frac {{\mathrm e}^{3 d x +3 c} a^{2} b}{8 d}+\frac {{\mathrm e}^{3 d x +3 c} a \,b^{2}}{8 d}+\frac {{\mathrm e}^{3 d x +3 c} b^{3}}{24 d}+\frac {3 \,{\mathrm e}^{d x +c} a^{3}}{8 d}-\frac {3 \,{\mathrm e}^{d x +c} a^{2} b}{8 d}-\frac {15 \,{\mathrm e}^{d x +c} a \,b^{2}}{8 d}-\frac {9 b^{3} {\mathrm e}^{d x +c}}{8 d}-\frac {3 \,{\mathrm e}^{-d x -c} a^{3}}{8 d}+\frac {3 \,{\mathrm e}^{-d x -c} a^{2} b}{8 d}+\frac {15 \,{\mathrm e}^{-d x -c} a \,b^{2}}{8 d}+\frac {9 \,{\mathrm e}^{-d x -c} b^{3}}{8 d}-\frac {{\mathrm e}^{-3 d x -3 c} a^{3}}{24 d}-\frac {{\mathrm e}^{-3 d x -3 c} a^{2} b}{8 d}-\frac {{\mathrm e}^{-3 d x -3 c} a \,b^{2}}{8 d}-\frac {{\mathrm e}^{-3 d x -3 c} b^{3}}{24 d}-\frac {b^{3} {\mathrm e}^{d x +c} \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}+\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{d}+\frac {5 i b^{3} \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 d}-\frac {5 i b^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 d}-\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{d}\) | \(386\) |
1/d*(a^3*(2/3+1/3*cosh(d*x+c)^2)*sinh(d*x+c)+a^2*b*sinh(d*x+c)^3+3*a*b^2*( 1/3*sinh(d*x+c)^3-sinh(d*x+c)+2*arctan(exp(d*x+c)))+b^3*(1/3*sinh(d*x+c)^5 /cosh(d*x+c)^2-5/3*sinh(d*x+c)^3/cosh(d*x+c)^2-5*sinh(d*x+c)/cosh(d*x+c)^2 +5/2*sech(d*x+c)*tanh(d*x+c)+5*arctan(exp(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 1840 vs. \(2 (81) = 162\).
Time = 0.28 (sec) , antiderivative size = 1840, normalized size of antiderivative = 21.15 \[ \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \]
1/24*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^10 + 10*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^9 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^10 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^8 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3 + 45*(a^3 + 3*a^2*b + 3*a* b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^8 + 8*(15*(a^3 + 3*a^2*b + 3*a*b ^2 + b^3)*cosh(d*x + c)^3 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d* x + c))*sinh(d*x + c)^7 + 2*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^6 + 2*(105*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3 + 14*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*c osh(d*x + c)^2)*sinh(d*x + c)^6 + 4*(63*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*co sh(d*x + c)^5 + 14*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^3 + 3*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 - 2*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^4 + 2*(105*(a^3 + 3 *a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^6 + 35*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^4 - 5*a^3 + 3*a^2*b + 21*a*b^2 + 25*b^3 + 15*(5*a^ 3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(15* (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^7 + 7*(11*a^3 - 3*a^2*b - 39 *a*b^2 - 25*b^3)*cosh(d*x + c)^5 + 5*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3) *cosh(d*x + c)^3 - (5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c))*si nh(d*x + c)^3 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 - (11*a^3 - 3*a^2*b - 39*...
\[ \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \cosh ^{3}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (81) = 162\).
Time = 0.28 (sec) , antiderivative size = 284, normalized size of antiderivative = 3.26 \[ \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3}}{8 \, d} - \frac {1}{8} \, a b^{2} {\left (\frac {{\left (15 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )} e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {15 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {48 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} + \frac {1}{24} \, b^{3} {\left (\frac {27 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} - \frac {120 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {25 \, e^{\left (-2 \, d x - 2 \, c\right )} + 77 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d {\left (e^{\left (-3 \, d x - 3 \, c\right )} + 2 \, e^{\left (-5 \, d x - 5 \, c\right )} + e^{\left (-7 \, d x - 7 \, c\right )}\right )}}\right )} + \frac {1}{24} \, a^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} - \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]
1/8*a^2*b*(e^(d*x + c) - e^(-d*x - c))^3/d - 1/8*a*b^2*((15*e^(-2*d*x - 2* c) - 1)*e^(3*d*x + 3*c)/d - (15*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + 48*ar ctan(e^(-d*x - c))/d) + 1/24*b^3*((27*e^(-d*x - c) - e^(-3*d*x - 3*c))/d - 120*arctan(e^(-d*x - c))/d - (25*e^(-2*d*x - 2*c) + 77*e^(-4*d*x - 4*c) + 3*e^(-6*d*x - 6*c) - 1)/(d*(e^(-3*d*x - 3*c) + 2*e^(-5*d*x - 5*c) + e^(-7 *d*x - 7*c)))) + 1/24*a^3*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d - e^(-3*d*x - 3*c)/d)
Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (81) = 162\).
Time = 0.47 (sec) , antiderivative size = 264, normalized size of antiderivative = 3.03 \[ \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 12 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 36 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 24 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - \frac {24 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4} + 6 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (6 \, a b^{2} + 5 \, b^{3}\right )}}{24 \, d} \]
1/24*(a^3*(e^(d*x + c) - e^(-d*x - c))^3 + 3*a^2*b*(e^(d*x + c) - e^(-d*x - c))^3 + 3*a*b^2*(e^(d*x + c) - e^(-d*x - c))^3 + b^3*(e^(d*x + c) - e^(- d*x - c))^3 + 12*a^3*(e^(d*x + c) - e^(-d*x - c)) - 36*a*b^2*(e^(d*x + c) - e^(-d*x - c)) - 24*b^3*(e^(d*x + c) - e^(-d*x - c)) - 24*b^3*(e^(d*x + c ) - e^(-d*x - c))/((e^(d*x + c) - e^(-d*x - c))^2 + 4) + 6*(pi + 2*arctan( 1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(6*a*b^2 + 5*b^3))/d
Time = 0.33 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.67 \[ \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (5\,b^3\,\sqrt {d^2}+6\,a\,b^2\,\sqrt {d^2}\right )}{d\,\sqrt {36\,a^2\,b^4+60\,a\,b^5+25\,b^6}}\right )\,\sqrt {36\,a^2\,b^4+60\,a\,b^5+25\,b^6}}{\sqrt {d^2}}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,{\left (a+b\right )}^3}{24\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,{\left (a+b\right )}^3}{24\,d}+\frac {3\,{\mathrm {e}}^{c+d\,x}\,{\left (a+b\right )}^2\,\left (a-3\,b\right )}{8\,d}-\frac {b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{-c-d\,x}\,{\left (a+b\right )}^2\,\left (a-3\,b\right )}{8\,d}+\frac {2\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]
(atan((exp(d*x)*exp(c)*(5*b^3*(d^2)^(1/2) + 6*a*b^2*(d^2)^(1/2)))/(d*(60*a *b^5 + 25*b^6 + 36*a^2*b^4)^(1/2)))*(60*a*b^5 + 25*b^6 + 36*a^2*b^4)^(1/2) )/(d^2)^(1/2) - (exp(- 3*c - 3*d*x)*(a + b)^3)/(24*d) + (exp(3*c + 3*d*x)* (a + b)^3)/(24*d) + (3*exp(c + d*x)*(a + b)^2*(a - 3*b))/(8*d) - (b^3*exp( c + d*x))/(d*(exp(2*c + 2*d*x) + 1)) - (3*exp(- c - d*x)*(a + b)^2*(a - 3* b))/(8*d) + (2*b^3*exp(c + d*x))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))