Integrand size = 23, antiderivative size = 76 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {(a+b)^2 \log (\cosh (c+d x))}{d}-\frac {(a+b)^2 \tanh ^2(c+d x)}{2 d}-\frac {b (2 a+b) \tanh ^4(c+d x)}{4 d}-\frac {b^2 \tanh ^6(c+d x)}{6 d} \] Output:
(a+b)^2*ln(cosh(d*x+c))/d-1/2*(a+b)^2*tanh(d*x+c)^2/d-1/4*b*(2*a+b)*tanh(d *x+c)^4/d-1/6*b^2*tanh(d*x+c)^6/d
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {12 (a+b)^2 \log (\cosh (c+d x))+6 \left (a^2+4 a b+3 b^2\right ) \text {sech}^2(c+d x)-3 b (2 a+3 b) \text {sech}^4(c+d x)+2 b^2 \text {sech}^6(c+d x)}{12 d} \] Input:
Integrate[Tanh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(12*(a + b)^2*Log[Cosh[c + d*x]] + 6*(a^2 + 4*a*b + 3*b^2)*Sech[c + d*x]^2 - 3*b*(2*a + 3*b)*Sech[c + d*x]^4 + 2*b^2*Sech[c + d*x]^6)/(12*d)
Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4153, 26, 354, 86, 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.
\(\displaystyle \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int i \tan (i c+i d x)^3 \left (a-b \tan (i c+i d x)^2\right )^2dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \tan (i c+i d x)^3 \left (a-b \tan (i c+i d x)^2\right )^2dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {i \int -\frac {i \tanh ^3(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\int \frac {\tanh ^3(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\tanh ^2(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (-b^2 \tanh ^4(c+d x)-b (2 a+b) \tanh ^2(c+d x)-(a+b)^2-\frac {(a+b)^2}{\tanh ^2(c+d x)-1}\right )d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} b (2 a+b) \tanh ^4(c+d x)-(a+b)^2 \tanh ^2(c+d x)-(a+b)^2 \log \left (1-\tanh ^2(c+d x)\right )-\frac {1}{3} b^2 \tanh ^6(c+d x)}{2 d}\) |
Input:
Int[Tanh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(-((a + b)^2*Log[1 - Tanh[c + d*x]^2]) - (a + b)^2*Tanh[c + d*x]^2 - (b*(2 *a + b)*Tanh[c + d*x]^4)/2 - (b^2*Tanh[c + d*x]^6)/3)/(2*d)
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[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.71
method | result | size |
derivativedivides | \(\frac {-\frac {\tanh \left (d x +c \right )^{4} a b}{2}-\tanh \left (d x +c \right )^{2} a b +\frac {\left (-a^{2}-2 a b -b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )}{2}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {\tanh \left (d x +c \right )^{4} b^{2}}{4}-\frac {\tanh \left (d x +c \right )^{2} a^{2}}{2}-\frac {b^{2} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{6} b^{2}}{6}}{d}\) | \(130\) |
default | \(\frac {-\frac {\tanh \left (d x +c \right )^{4} a b}{2}-\tanh \left (d x +c \right )^{2} a b +\frac {\left (-a^{2}-2 a b -b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )}{2}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {\tanh \left (d x +c \right )^{4} b^{2}}{4}-\frac {\tanh \left (d x +c \right )^{2} a^{2}}{2}-\frac {b^{2} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{6} b^{2}}{6}}{d}\) | \(130\) |
parallelrisch | \(-\frac {2 \tanh \left (d x +c \right )^{6} b^{2}+6 \tanh \left (d x +c \right )^{4} a b +3 \tanh \left (d x +c \right )^{4} b^{2}+12 a^{2} d x +24 a b d x +12 b^{2} d x +6 \tanh \left (d x +c \right )^{2} a^{2}+12 \tanh \left (d x +c \right )^{2} a b +6 b^{2} \tanh \left (d x +c \right )^{2}+12 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2}+24 \ln \left (1-\tanh \left (d x +c \right )\right ) a b +12 \ln \left (1-\tanh \left (d x +c \right )\right ) b^{2}}{12 d}\) | \(150\) |
parts | \(\frac {a^{2} \left (-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}+\frac {b^{2} \left (-\frac {\tanh \left (d x +c \right )^{6}}{6}-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 a b \left (-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}\) | \(152\) |
risch | \(-a^{2} x -2 a b x -b^{2} x -\frac {2 a^{2} c}{d}-\frac {4 a b c}{d}-\frac {2 b^{2} c}{d}+\frac {2 \,{\mathrm e}^{2 d x +2 c} \left (3 a^{2} {\mathrm e}^{8 d x +8 c}+12 a b \,{\mathrm e}^{8 d x +8 c}+9 b^{2} {\mathrm e}^{8 d x +8 c}+12 a^{2} {\mathrm e}^{6 d x +6 c}+36 a b \,{\mathrm e}^{6 d x +6 c}+18 b^{2} {\mathrm e}^{6 d x +6 c}+18 a^{2} {\mathrm e}^{4 d x +4 c}+48 a b \,{\mathrm e}^{4 d x +4 c}+34 b^{2} {\mathrm e}^{4 d x +4 c}+12 a^{2} {\mathrm e}^{2 d x +2 c}+36 a b \,{\mathrm e}^{2 d x +2 c}+18 b^{2} {\mathrm e}^{2 d x +2 c}+3 a^{2}+12 a b +9 b^{2}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{6}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a b}{d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) b^{2}}{d}\) | \(308\) |
Input:
int(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2*tanh(d*x+c)^4*a*b-tanh(d*x+c)^2*a*b+1/2*(-a^2-2*a*b-b^2)*ln(1+ta nh(d*x+c))-1/2*(a^2+2*a*b+b^2)*ln(-1+tanh(d*x+c))-1/4*tanh(d*x+c)^4*b^2-1/ 2*tanh(d*x+c)^2*a^2-1/2*b^2*tanh(d*x+c)^2-1/6*tanh(d*x+c)^6*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 3441 vs. \(2 (70) = 140\).
Time = 0.13 (sec) , antiderivative size = 3441, normalized size of antiderivative = 45.28 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\text {Too large to display} \] Input:
integrate(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
Too large to include
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (65) = 130\).
Time = 0.23 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.24 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\begin {cases} a^{2} x - \frac {a^{2} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a^{2} \tanh ^{2}{\left (c + d x \right )}}{2 d} + 2 a b x - \frac {2 a b \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a b \tanh ^{4}{\left (c + d x \right )}}{2 d} - \frac {a b \tanh ^{2}{\left (c + d x \right )}}{d} + b^{2} x - \frac {b^{2} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {b^{2} \tanh ^{6}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \tanh ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{2} \tanh ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tanh(d*x+c)**3*(a+b*tanh(d*x+c)**2)**2,x)
Output:
Piecewise((a**2*x - a**2*log(tanh(c + d*x) + 1)/d - a**2*tanh(c + d*x)**2/ (2*d) + 2*a*b*x - 2*a*b*log(tanh(c + d*x) + 1)/d - a*b*tanh(c + d*x)**4/(2 *d) - a*b*tanh(c + d*x)**2/d + b**2*x - b**2*log(tanh(c + d*x) + 1)/d - b* *2*tanh(c + d*x)**6/(6*d) - b**2*tanh(c + d*x)**4/(4*d) - b**2*tanh(c + d* x)**2/(2*d), Ne(d, 0)), (x*(a + b*tanh(c)**2)**2*tanh(c)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (70) = 140\).
Time = 0.12 (sec) , antiderivative size = 333, normalized size of antiderivative = 4.38 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{3} \, b^{2} {\left (3 \, x + \frac {3 \, c}{d} + \frac {3 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, {\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} + 18 \, e^{\left (-4 \, d x - 4 \, c\right )} + 34 \, e^{\left (-6 \, d x - 6 \, c\right )} + 18 \, e^{\left (-8 \, d x - 8 \, c\right )} + 9 \, e^{\left (-10 \, d x - 10 \, c\right )}\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} + 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}}\right )} + 2 \, a b {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} \] Input:
integrate(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/3*b^2*(3*x + 3*c/d + 3*log(e^(-2*d*x - 2*c) + 1)/d + 2*(9*e^(-2*d*x - 2* c) + 18*e^(-4*d*x - 4*c) + 34*e^(-6*d*x - 6*c) + 18*e^(-8*d*x - 8*c) + 9*e ^(-10*d*x - 10*c))/(d*(6*e^(-2*d*x - 2*c) + 15*e^(-4*d*x - 4*c) + 20*e^(-6 *d*x - 6*c) + 15*e^(-8*d*x - 8*c) + 6*e^(-10*d*x - 10*c) + e^(-12*d*x - 12 *c) + 1))) + 2*a*b*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 4*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + e^(-6*d*x - 6*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))) + a^2*(x + c /d + log(e^(-2*d*x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2* c) + e^(-4*d*x - 4*c) + 1)))
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (70) = 140\).
Time = 0.20 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.51 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - \frac {2 \, {\left (3 \, {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} e^{\left (10 \, d x + 10 \, c\right )} + 6 \, {\left (2 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} e^{\left (8 \, d x + 8 \, c\right )} + 2 \, {\left (9 \, a^{2} + 24 \, a b + 17 \, b^{2}\right )} e^{\left (6 \, d x + 6 \, c\right )} + 6 \, {\left (2 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{6}}}{3 \, d} \] Input:
integrate(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
-1/3*(3*(a^2 + 2*a*b + b^2)*(d*x + c) - 3*(a^2 + 2*a*b + b^2)*log(e^(2*d*x + 2*c) + 1) - 2*(3*(a^2 + 4*a*b + 3*b^2)*e^(10*d*x + 10*c) + 6*(2*a^2 + 6 *a*b + 3*b^2)*e^(8*d*x + 8*c) + 2*(9*a^2 + 24*a*b + 17*b^2)*e^(6*d*x + 6*c ) + 6*(2*a^2 + 6*a*b + 3*b^2)*e^(4*d*x + 4*c) + 3*(a^2 + 4*a*b + 3*b^2)*e^ (2*d*x + 2*c))/(e^(2*d*x + 2*c) + 1)^6)/d
Time = 2.52 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.32 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=x\,\left (a^2+2\,a\,b+b^2\right )-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^4\,\left (b^2+2\,a\,b\right )}{4\,d}-\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )\,\left (a^2+2\,a\,b+b^2\right )}{d}-\frac {b^2\,{\mathrm {tanh}\left (c+d\,x\right )}^6}{6\,d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^2\,\left (a^2+2\,a\,b+b^2\right )}{2\,d} \] Input:
int(tanh(c + d*x)^3*(a + b*tanh(c + d*x)^2)^2,x)
Output:
x*(2*a*b + a^2 + b^2) - (tanh(c + d*x)^4*(2*a*b + b^2))/(4*d) - (log(tanh( c + d*x) + 1)*(2*a*b + a^2 + b^2))/d - (b^2*tanh(c + d*x)^6)/(6*d) - (tanh (c + d*x)^2*(2*a*b + a^2 + b^2))/(2*d)
Time = 0.24 (sec) , antiderivative size = 1141, normalized size of antiderivative = 15.01 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx =\text {Too large to display} \] Input:
int(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x)
Output:
(3*e**(12*c + 12*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2 + 6*e**(12*c + 12*d*x )*log(e**(2*c + 2*d*x) + 1)*a*b + 3*e**(12*c + 12*d*x)*log(e**(2*c + 2*d*x ) + 1)*b**2 - 3*e**(12*c + 12*d*x)*a**2*d*x - e**(12*c + 12*d*x)*a**2 - 6* e**(12*c + 12*d*x)*a*b*d*x - 4*e**(12*c + 12*d*x)*a*b - 3*e**(12*c + 12*d* x)*b**2*d*x - 3*e**(12*c + 12*d*x)*b**2 + 18*e**(10*c + 10*d*x)*log(e**(2* c + 2*d*x) + 1)*a**2 + 36*e**(10*c + 10*d*x)*log(e**(2*c + 2*d*x) + 1)*a*b + 18*e**(10*c + 10*d*x)*log(e**(2*c + 2*d*x) + 1)*b**2 - 18*e**(10*c + 10 *d*x)*a**2*d*x - 36*e**(10*c + 10*d*x)*a*b*d*x - 18*e**(10*c + 10*d*x)*b** 2*d*x + 45*e**(8*c + 8*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2 + 90*e**(8*c + 8*d*x)*log(e**(2*c + 2*d*x) + 1)*a*b + 45*e**(8*c + 8*d*x)*log(e**(2*c + 2 *d*x) + 1)*b**2 - 45*e**(8*c + 8*d*x)*a**2*d*x + 9*e**(8*c + 8*d*x)*a**2 - 90*e**(8*c + 8*d*x)*a*b*d*x + 12*e**(8*c + 8*d*x)*a*b - 45*e**(8*c + 8*d* x)*b**2*d*x - 9*e**(8*c + 8*d*x)*b**2 + 60*e**(6*c + 6*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2 + 120*e**(6*c + 6*d*x)*log(e**(2*c + 2*d*x) + 1)*a*b + 60 *e**(6*c + 6*d*x)*log(e**(2*c + 2*d*x) + 1)*b**2 - 60*e**(6*c + 6*d*x)*a** 2*d*x + 16*e**(6*c + 6*d*x)*a**2 - 120*e**(6*c + 6*d*x)*a*b*d*x + 16*e**(6 *c + 6*d*x)*a*b - 60*e**(6*c + 6*d*x)*b**2*d*x + 8*e**(6*c + 6*d*x)*b**2 + 45*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2 + 90*e**(4*c + 4*d*x)* log(e**(2*c + 2*d*x) + 1)*a*b + 45*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) + 1)*b**2 - 45*e**(4*c + 4*d*x)*a**2*d*x + 9*e**(4*c + 4*d*x)*a**2 - 90*...