Integrand size = 23, antiderivative size = 114 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^4(c+d x) \, dx=\frac {1}{8} \left (3 a^2-24 a b+8 b^2\right ) x-\frac {a (a-8 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac {\left (a^2-8 a b+4 b^2\right ) \tanh (c+d x)}{4 d}+\frac {a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \]
1/8*(3*a^2-24*a*b+8*b^2)*x-1/8*a*(a-8*b)*cosh(d*x+c)*sinh(d*x+c)/d-1/4*(a^ 2-8*a*b+4*b^2)*tanh(d*x+c)/d+1/4*a^2*sinh(d*x+c)^4*tanh(d*x+c)/d-1/3*b^2*t anh(d*x+c)^3/d
Time = 2.73 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.34 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^4(c+d x) \, dx=\frac {\left (b+a \cosh ^2(c+d x)\right )^2 \text {sech}^3(c+d x) \left (32 b^2 \text {sech}(c) \sinh (d x)+64 (3 a-2 b) b \cosh ^2(c+d x) \text {sech}(c) \sinh (d x)+3 \cosh ^3(c+d x) \left (4 \left (3 a^2-24 a b+8 b^2\right ) d x-8 a (a-2 b) \sinh (2 (c+d x))+a^2 \sinh (4 (c+d x))\right )+32 b^2 \cosh (c+d x) \tanh (c)\right )}{24 d (a+2 b+a \cosh (2 (c+d x)))^2} \]
((b + a*Cosh[c + d*x]^2)^2*Sech[c + d*x]^3*(32*b^2*Sech[c]*Sinh[d*x] + 64* (3*a - 2*b)*b*Cosh[c + d*x]^2*Sech[c]*Sinh[d*x] + 3*Cosh[c + d*x]^3*(4*(3* a^2 - 24*a*b + 8*b^2)*d*x - 8*a*(a - 2*b)*Sinh[2*(c + d*x)] + a^2*Sinh[4*( c + d*x)]) + 32*b^2*Cosh[c + d*x]*Tanh[c]))/(24*d*(a + 2*b + a*Cosh[2*(c + d*x)])^2)
Time = 0.37 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4620, 366, 360, 25, 1467, 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.
| \(\displaystyle \int \sinh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (i c+i d x)^4 \left (a+b \sec (i c+i d x)^2\right )^2dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (-b \tanh ^2(c+d x)+a+b\right )^2}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle \frac {\frac {a^2 \tanh ^5(c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {\tanh ^4(c+d x) \left (5 a^2-4 (a+b)^2+4 b^2 \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle \frac {\frac {1}{4} \left (-\frac {1}{2} \int -\frac {8 b^2 \tanh ^4(c+d x)+2 a (a-8 b) \tanh ^2(c+d x)+a (a-8 b)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {a (a-8 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a^2 \tanh ^5(c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {8 b^2 \tanh ^4(c+d x)+2 a (a-8 b) \tanh ^2(c+d x)+a (a-8 b)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {a (a-8 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a^2 \tanh ^5(c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 1467 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \left (-8 b^2 \tanh ^2(c+d x)-2 \left (a^2-8 b a+4 b^2\right )+\frac {3 a^2-24 b a+8 b^2}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)-\frac {a (a-8 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a^2 \tanh ^5(c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (\left (3 a^2-24 a b+8 b^2\right ) \text {arctanh}(\tanh (c+d x))-2 \left (a^2-8 a b+4 b^2\right ) \tanh (c+d x)-\frac {8}{3} b^2 \tanh ^3(c+d x)\right )-\frac {a (a-8 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a^2 \tanh ^5(c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
((a^2*Tanh[c + d*x]^5)/(4*(1 - Tanh[c + d*x]^2)^2) + (-1/2*(a*(a - 8*b)*Ta nh[c + d*x])/(1 - Tanh[c + d*x]^2) + ((3*a^2 - 24*a*b + 8*b^2)*ArcTanh[Tan h[c + d*x]] - 2*(a^2 - 8*a*b + 4*b^2)*Tanh[c + d*x] - (8*b^2*Tanh[c + d*x] ^3)/3)/2)/4)/d
3.1.9.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Time = 13.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (d x +c -\tanh \left (d x +c \right )-\frac {\tanh \left (d x +c \right )^{3}}{3}\right )}{d}\) | \(109\) |
| default | \(\frac {a^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (d x +c -\tanh \left (d x +c \right )-\frac {\tanh \left (d x +c \right )^{3}}{3}\right )}{d}\) | \(109\) |
| parts | \(\frac {a^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {b^{2} \left (d x +c -\tanh \left (d x +c \right )-\frac {\tanh \left (d x +c \right )^{3}}{3}\right )}{d}+\frac {2 a b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(114\) |
| risch | \(\frac {3 a^{2} x}{8}-3 a b x +b^{2} x +\frac {a^{2} {\mathrm e}^{4 d x +4 c}}{64 d}-\frac {a^{2} {\mathrm e}^{2 d x +2 c}}{8 d}+\frac {a \,{\mathrm e}^{2 d x +2 c} b}{4 d}+\frac {a^{2} {\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {a \,{\mathrm e}^{-2 d x -2 c} b}{4 d}-\frac {a^{2} {\mathrm e}^{-4 d x -4 c}}{64 d}-\frac {4 b \left (3 a \,{\mathrm e}^{4 d x +4 c}-3 b \,{\mathrm e}^{4 d x +4 c}+6 \,{\mathrm e}^{2 d x +2 c} a -3 b \,{\mathrm e}^{2 d x +2 c}+3 a -2 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}\) | \(192\) |
1/d*(a^2*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+2 *a*b*(1/2*sinh(d*x+c)^3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*tanh(d*x+c))+b^2*(d* x+c-tanh(d*x+c)-1/3*tanh(d*x+c)^3))
Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (104) = 208\).
Time = 0.25 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.00 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^4(c+d x) \, dx=\frac {3 \, a^{2} \sinh \left (d x + c\right )^{7} + 3 \, {\left (21 \, a^{2} \cosh \left (d x + c\right )^{2} - 5 \, a^{2} + 16 \, a b\right )} \sinh \left (d x + c\right )^{5} + 8 \, {\left (3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} d x - 48 \, a b + 32 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 24 \, {\left (3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} d x - 48 \, a b + 32 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (105 \, a^{2} \cosh \left (d x + c\right )^{4} - 30 \, {\left (5 \, a^{2} - 16 \, a b\right )} \cosh \left (d x + c\right )^{2} - 63 \, a^{2} + 528 \, a b - 256 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 24 \, {\left (3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} d x - 48 \, a b + 32 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (7 \, a^{2} \cosh \left (d x + c\right )^{6} - 5 \, {\left (5 \, a^{2} - 16 \, a b\right )} \cosh \left (d x + c\right )^{4} - {\left (63 \, a^{2} - 528 \, a b + 256 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 15 \, a^{2} + 160 \, a b\right )} \sinh \left (d x + c\right )}{192 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]
1/192*(3*a^2*sinh(d*x + c)^7 + 3*(21*a^2*cosh(d*x + c)^2 - 5*a^2 + 16*a*b) *sinh(d*x + c)^5 + 8*(3*(3*a^2 - 24*a*b + 8*b^2)*d*x - 48*a*b + 32*b^2)*co sh(d*x + c)^3 + 24*(3*(3*a^2 - 24*a*b + 8*b^2)*d*x - 48*a*b + 32*b^2)*cosh (d*x + c)*sinh(d*x + c)^2 + (105*a^2*cosh(d*x + c)^4 - 30*(5*a^2 - 16*a*b) *cosh(d*x + c)^2 - 63*a^2 + 528*a*b - 256*b^2)*sinh(d*x + c)^3 + 24*(3*(3* a^2 - 24*a*b + 8*b^2)*d*x - 48*a*b + 32*b^2)*cosh(d*x + c) + 3*(7*a^2*cosh (d*x + c)^6 - 5*(5*a^2 - 16*a*b)*cosh(d*x + c)^4 - (63*a^2 - 528*a*b + 256 *b^2)*cosh(d*x + c)^2 - 15*a^2 + 160*a*b)*sinh(d*x + c))/(d*cosh(d*x + c)^ 3 + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(d*x + c))
Timed out. \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^4(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (104) = 208\).
Time = 0.21 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.85 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^4(c+d x) \, dx=\frac {1}{64} \, a^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{3} \, b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} - \frac {1}{4} \, a b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]
1/64*a^2*(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) + 1/3*b^2*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e ^(-6*d*x - 6*c) + 1))) - 1/4*a*b*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (1 7*e^(-2*d*x - 2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c))))
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (104) = 208\).
Time = 0.30 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.03 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^4(c+d x) \, dx=\frac {3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 24 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 48 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} {\left (d x + c\right )} - 3 \, {\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 144 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 16 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {256 \, {\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b - 2 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{192 \, d} \]
1/192*(3*a^2*e^(4*d*x + 4*c) - 24*a^2*e^(2*d*x + 2*c) + 48*a*b*e^(2*d*x + 2*c) + 24*(3*a^2 - 24*a*b + 8*b^2)*(d*x + c) - 3*(18*a^2*e^(4*d*x + 4*c) - 144*a*b*e^(4*d*x + 4*c) + 48*b^2*e^(4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) + 16*a*b*e^(2*d*x + 2*c) + a^2)*e^(-4*d*x - 4*c) - 256*(3*a*b*e^(4*d*x + 4 *c) - 3*b^2*e^(4*d*x + 4*c) + 6*a*b*e^(2*d*x + 2*c) - 3*b^2*e^(2*d*x + 2*c ) + 3*a*b - 2*b^2)/(e^(2*d*x + 2*c) + 1)^3)/d
Time = 0.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.36 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^4(c+d x) \, dx=x\,\left (\frac {3\,a^2}{8}-3\,a\,b+b^2\right )-\frac {\frac {4\,\left (a\,b-b^2\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b-b^2\right )}{3\,d}+\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b-b^2\right )}{3\,d}+\frac {4\,a\,b}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {4\,\left (a\,b-b^2\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a\,b-a^2\right )}{8\,d}-\frac {a^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {a^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}+\frac {a\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a-2\,b\right )}{8\,d} \]
x*((3*a^2)/8 - 3*a*b + b^2) - ((4*(a*b - b^2))/(3*d) + (4*exp(4*c + 4*d*x) *(a*b - b^2))/(3*d) + (8*a*b*exp(2*c + 2*d*x))/(3*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - ((4*exp(2*c + 2*d*x)*(a*b - b^2))/(3*d) + (4*a*b)/(3*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - (4*(a*b - b^2))/(3*d*(exp(2*c + 2*d*x) + 1)) + (exp(2*c + 2*d*x)*(2*a*b - a^2))/(8*d) - (a^2*exp(- 4*c - 4*d*x))/(64*d) + (a^2*exp(4*c + 4*d*x))/( 64*d) + (a*exp(- 2*c - 2*d*x)*(a - 2*b))/(8*d)