Integrand size = 23, antiderivative size = 47 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-\frac {a^2 \coth (c+d x)}{d}+\frac {a b \tanh ^2(c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \] Output:
-a^2*coth(d*x+c)/d+a*b*tanh(d*x+c)^2/d+1/5*b^2*tanh(d*x+c)^5/d
Time = 0.58 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.00 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-\frac {a^2 \coth (c+d x)}{d}-\frac {a b \text {sech}^2(c+d x)}{d}+\frac {b^2 \tanh (c+d x)}{5 d}-\frac {2 b^2 \text {sech}^2(c+d x) \tanh (c+d x)}{5 d}+\frac {b^2 \text {sech}^4(c+d x) \tanh (c+d x)}{5 d} \] Input:
Integrate[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^2,x]
Output:
-((a^2*Coth[c + d*x])/d) - (a*b*Sech[c + d*x]^2)/d + (b^2*Tanh[c + d*x])/( 5*d) - (2*b^2*Sech[c + d*x]^2*Tanh[c + d*x])/(5*d) + (b^2*Sech[c + d*x]^4* Tanh[c + d*x])/(5*d)
Time = 0.47 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 25, 4146, 802, 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.
| \(\displaystyle \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a+i b \tan (i c+i d x)^3\right )^2}{\sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (i b \tan (i c+i d x)^3+a\right )^2}{\sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \coth ^2(c+d x) \left (b \tanh ^3(c+d x)+a\right )^2d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 802 |
| \(\displaystyle \frac {\int \left (b^2 \tanh ^4(c+d x)+2 a b \tanh (c+d x)+a^2 \coth ^2(c+d x)\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-a^2 \coth (c+d x)+a b \tanh ^2(c+d x)+\frac {1}{5} b^2 \tanh ^5(c+d x)}{d}\) |
Input:
Int[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^2,x]
Output:
(-(a^2*Coth[c + d*x]) + a*b*Tanh[c + d*x]^2 + (b^2*Tanh[c + d*x]^5)/5)/d
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Leaf count of result is larger than twice the leaf count of optimal. \(97\) vs. \(2(45)=90\).
Time = 8.51 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.09
| method | result | size |
| derivativedivides | \(\frac {-\coth \left (d x +c \right ) a^{2}-\frac {a b}{\cosh \left (d x +c \right )^{2}}+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(98\) |
| default | \(\frac {-\coth \left (d x +c \right ) a^{2}-\frac {a b}{\cosh \left (d x +c \right )^{2}}+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(98\) |
| risch | \(-\frac {2 \left (5 \,{\mathrm e}^{10 d x +10 c} a^{2}+10 \,{\mathrm e}^{10 d x +10 c} a b +5 \,{\mathrm e}^{10 d x +10 c} b^{2}+25 \,{\mathrm e}^{8 d x +8 c} a^{2}+20 \,{\mathrm e}^{8 d x +8 c} a b -5 \,{\mathrm e}^{8 d x +8 c} b^{2}+50 \,{\mathrm e}^{6 d x +6 c} a^{2}+10 \,{\mathrm e}^{6 d x +6 c} b^{2}+50 \,{\mathrm e}^{4 d x +4 c} a^{2}-20 \,{\mathrm e}^{4 d x +4 c} a b -10 \,{\mathrm e}^{4 d x +4 c} b^{2}+25 \,{\mathrm e}^{2 d x +2 c} a^{2}-10 \,{\mathrm e}^{2 d x +2 c} b a +b^{2} {\mathrm e}^{2 d x +2 c}+5 a^{2}-b^{2}\right )}{5 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5} \left ({\mathrm e}^{2 d x +2 c}-1\right )}\) | \(234\) |
Input:
int(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-coth(d*x+c)*a^2-a*b/cosh(d*x+c)^2+b^2*(-1/2*sinh(d*x+c)^3/cosh(d*x+c )^5-3/8*sinh(d*x+c)/cosh(d*x+c)^5+3/8*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d* x+c)^2)*tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (45) = 90\).
Time = 0.10 (sec) , antiderivative size = 518, normalized size of antiderivative = 11.02 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-\frac {4 \, {\left ({\left (5 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (5 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (5 \, a b + 3 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + {\left (25 \, a^{2} + 5 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (10 \, {\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 15 \, a b - 3 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + {\left (10 \, {\left (5 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (25 \, a^{2} + 5 \, 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} - a b\right )} \cosh \left (d x + c\right ) + {\left (5 \, {\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 9 \, {\left (5 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 10 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )\right )}}{5 \, {\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 3 \, d \cosh \left (d x + c\right )^{5} + {\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (7 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + d \cosh \left (d x + c\right )^{3} + {\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (7 \, d \cosh \left (d x + c\right )^{5} + 10 \, d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5 \, d \cosh \left (d x + c\right ) + {\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="fricas")
Output:
-4/5*((5*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c)^5 + 5*(5*a^2 + 5*a*b + 2*b^2)* cosh(d*x + c)*sinh(d*x + c)^4 + (5*a*b + 3*b^2)*sinh(d*x + c)^5 + (25*a^2 + 5*a*b - 2*b^2)*cosh(d*x + c)^3 + (10*(5*a*b + 3*b^2)*cosh(d*x + c)^2 + 1 5*a*b - 3*b^2)*sinh(d*x + c)^3 + (10*(5*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c) ^3 + 3*(25*a^2 + 5*a*b - 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^2 - a*b)*cosh(d*x + c) + (5*(5*a*b + 3*b^2)*cosh(d*x + c)^4 + 9*(5*a*b - b^ 2)*cosh(d*x + c)^2 + 10*a*b + 10*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 3*d*cosh(d*x + c)^ 5 + (21*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^4 + d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c)^4 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 + 3*(7*d*cosh(d*x + c )^5 + 10*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^2 - 5*d*cosh(d *x + c) + (7*d*cosh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*x + c) ^2 + 5*d)*sinh(d*x + c))
\[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right )^{2} \operatorname {csch}^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(csch(d*x+c)**2*(a+b*tanh(d*x+c)**3)**2,x)
Output:
Integral((a + b*tanh(c + d*x)**3)**2*csch(c + d*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (45) = 90\).
Time = 0.04 (sec) , antiderivative size = 256, normalized size of antiderivative = 5.45 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {2}{5} \, b^{2} {\left (\frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {5 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} - \frac {4 \, a b}{d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2}} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="maxima")
Output:
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))) + 2*a^2/(d*(e^(-2*d*x - 2*c) - 1)) - 4*a*b/( d*(e^(d*x + c) + e^(-d*x - c))^2)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (45) = 90\).
Time = 0.21 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.60 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (\frac {5 \, a^{2}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + \frac {10 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 5 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 30 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 30 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, a b e^{\left (2 \, d x + 2 \, c\right )} + b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}\right )}}{5 \, d} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="giac")
Output:
-2/5*(5*a^2/(e^(2*d*x + 2*c) - 1) + (10*a*b*e^(8*d*x + 8*c) + 5*b^2*e^(8*d *x + 8*c) + 30*a*b*e^(6*d*x + 6*c) + 30*a*b*e^(4*d*x + 4*c) + 10*b^2*e^(4* d*x + 4*c) + 10*a*b*e^(2*d*x + 2*c) + b^2)/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 2.92 (sec) , antiderivative size = 483, normalized size of antiderivative = 10.28 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-\frac {\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+a\,b\right )}{5\,d}-\frac {2\,\left (2\,a\,b-b^2\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a\,b-b^2\right )}{5\,d}+\frac {12\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,b^2}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b-b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,\left (a\,b-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,\left (b^2+a\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b-b^2\right )}{5\,d}+\frac {6\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:
int((a + b*tanh(c + d*x)^3)^2/sinh(c + d*x)^2,x)
Output:
- ((2*exp(8*c + 8*d*x)*(2*a*b + b^2))/(5*d) - (8*exp(2*c + 2*d*x)*(a*b + b ^2))/(5*d) - (2*(2*a*b - b^2))/(5*d) + (8*exp(6*c + 6*d*x)*(a*b - b^2))/(5 *d) + (12*b^2*exp(4*c + 4*d*x))/(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)/(5*d) + (2*exp(4*c + 4*d*x)*(2*a*b + b^2))/(5*d) + (4*exp(2*c + 2*d*x)*(a*b - b^2))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + e xp(6*c + 6*d*x) + 1) - ((2*(a*b - b^2))/(5*d) + (2*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*exp(6*c + 6*d*x)*(2*a*b + b^2))/(5*d) - (2*(a*b + b^2))/(5*d) + (6*exp(4*c + 4*d*x )*(a*b - b^2))/(5*d) + (6*b^2*exp(2*c + 2*d*x))/(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*a^ 2)/(d*(exp(2*c + 2*d*x) - 1)) - (2*(2*a*b + b^2))/(5*d*(exp(2*c + 2*d*x) + 1))
Time = 0.26 (sec) , antiderivative size = 274, normalized size of antiderivative = 5.83 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {5 e^{12 d x +12 c} a^{2}+10 e^{12 d x +12 c} a b +5 e^{12 d x +12 c} b^{2}-75 e^{8 d x +8 c} a^{2}-30 e^{8 d x +8 c} a b +45 e^{8 d x +8 c} b^{2}-200 e^{6 d x +6 c} a^{2}-40 e^{6 d x +6 c} b^{2}-225 e^{4 d x +4 c} a^{2}+30 e^{4 d x +4 c} a b +15 e^{4 d x +4 c} b^{2}-120 e^{2 d x +2 c} a^{2}-24 e^{2 d x +2 c} b^{2}-25 a^{2}-10 a b -b^{2}}{10 d \left (e^{12 d x +12 c}+4 e^{10 d x +10 c}+5 e^{8 d x +8 c}-5 e^{4 d x +4 c}-4 e^{2 d x +2 c}-1\right )} \] Input:
int(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x)
Output:
(5*e**(12*c + 12*d*x)*a**2 + 10*e**(12*c + 12*d*x)*a*b + 5*e**(12*c + 12*d *x)*b**2 - 75*e**(8*c + 8*d*x)*a**2 - 30*e**(8*c + 8*d*x)*a*b + 45*e**(8*c + 8*d*x)*b**2 - 200*e**(6*c + 6*d*x)*a**2 - 40*e**(6*c + 6*d*x)*b**2 - 22 5*e**(4*c + 4*d*x)*a**2 + 30*e**(4*c + 4*d*x)*a*b + 15*e**(4*c + 4*d*x)*b* *2 - 120*e**(2*c + 2*d*x)*a**2 - 24*e**(2*c + 2*d*x)*b**2 - 25*a**2 - 10*a *b - b**2)/(10*d*(e**(12*c + 12*d*x) + 4*e**(10*c + 10*d*x) + 5*e**(8*c + 8*d*x) - 5*e**(4*c + 4*d*x) - 4*e**(2*c + 2*d*x) - 1))