Integrand size = 23, antiderivative size = 109 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^3 d}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^3 d}-\frac {a^2}{4 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{2 b^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )} \]
ln(cosh(d*x+c))/(a+b)^3/d+1/2*ln(a+b*tanh(d*x+c)^2)/(a+b)^3/d-1/4*a^2/b^2/ (a+b)/d/(a+b*tanh(d*x+c)^2)^2+1/2*a*(a+2*b)/b^2/(a+b)^2/d/(a+b*tanh(d*x+c) ^2)
Time = 1.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {-4 \log (\cosh (c+d x))-2 \log \left (a+b \tanh ^2(c+d x)\right )+\frac {a^2 (a+b)^2}{b^2 \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {2 a (a+b) (a+2 b)}{b^2 \left (a+b \tanh ^2(c+d x)\right )}}{4 (a+b)^3 d} \]
-1/4*(-4*Log[Cosh[c + d*x]] - 2*Log[a + b*Tanh[c + d*x]^2] + (a^2*(a + b)^ 2)/(b^2*(a + b*Tanh[c + d*x]^2)^2) - (2*a*(a + b)*(a + 2*b))/(b^2*(a + b*T anh[c + d*x]^2)))/((a + b)^3*d)
Time = 0.35 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.96, 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, 99, 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.
| \(\displaystyle \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \tan (i c+i d x)^5}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\tan (i c+i d x)^5}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {i \int \frac {i \tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\int \frac {\tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\tanh ^4(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {a^2}{b (a+b) \left (b \tanh ^2(c+d x)+a\right )^3}-\frac {(a+2 b) a}{b (a+b)^2 \left (b \tanh ^2(c+d x)+a\right )^2}-\frac {1}{(a+b)^3 \left (\tanh ^2(c+d x)-1\right )}+\frac {b}{(a+b)^3 \left (b \tanh ^2(c+d x)+a\right )}\right )d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^2}{2 b^2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{b^2 (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{(a+b)^3}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{(a+b)^3}}{2 d}\) |
(-(Log[1 - Tanh[c + d*x]^2]/(a + b)^3) + Log[a + b*Tanh[c + d*x]^2]/(a + b )^3 - a^2/(2*b^2*(a + b)*(a + b*Tanh[c + d*x]^2)^2) + (a*(a + 2*b))/(b^2*( a + b)^2*(a + b*Tanh[c + d*x]^2)))/(2*d)
3.2.91.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.17
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {-\frac {a \left (a^{2}+3 a b +2 b^{2}\right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}-\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(127\) |
| default | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {-\frac {a \left (a^{2}+3 a b +2 b^{2}\right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}-\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(127\) |
| risch | \(-\frac {x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {2 c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {4 \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right ) a \,{\mathrm e}^{2 d x +2 c}}{\left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )^{2} d \left (a +b \right )^{3}}+\frac {\ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) | \(231\) |
| parallelrisch | \(-\frac {4 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2} b^{2}+4 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{4} b^{4}-2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{4} b^{4}-4 a \,b^{3} \tanh \left (d x +c \right )^{2}-2 \tanh \left (d x +c \right )^{2} a^{3} b -2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) a^{2} b^{2}+8 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2} a \,b^{3}+4 b^{4} \tanh \left (d x +c \right )^{4} x d -a^{4}+4 a^{2} b^{2} d x -3 a^{2} b^{2}-4 a^{3} b -6 a^{2} b^{2} \tanh \left (d x +c \right )^{2}+8 a \,b^{3} \tanh \left (d x +c \right )^{2} x d -4 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a \,b^{3}}{4 \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) b^{2} d}\) | \(291\) |
1/d*(-1/2/(a+b)^3*ln(tanh(d*x+c)-1)-1/2/(a+b)^3*(-a*(a^2+3*a*b+2*b^2)/b^2/ (a+b*tanh(d*x+c)^2)+1/2*a^2*(a^2+2*a*b+b^2)/b^2/(a+b*tanh(d*x+c)^2)^2-ln(a +b*tanh(d*x+c)^2))-1/2/(a+b)^3*ln(tanh(d*x+c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 2584 vs. \(2 (103) = 206\).
Time = 0.30 (sec) , antiderivative size = 2584, normalized size of antiderivative = 23.71 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
-1/2*(2*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^8 + 16*(a^2 + 2*a*b + b^2)*d *x*cosh(d*x + c)*sinh(d*x + c)^7 + 2*(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c) ^8 + 8*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 + (a^2 - b^2)*d*x - a^2 - a*b)*sinh(d*x + c)^6 + 16*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^3 + 3*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 4*((3*a^2 - 2*a*b + 3*b^2)*d*x - 2 *a^2 + 4*a*b)*cosh(d*x + c)^4 + 4*(35*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c )^4 + (3*a^2 - 2*a*b + 3*b^2)*d*x + 30*((a^2 - b^2)*d*x - a^2 - a*b)*cosh( d*x + c)^2 - 2*a^2 + 4*a*b)*sinh(d*x + c)^4 + 16*(7*(a^2 + 2*a*b + b^2)*d* x*cosh(d*x + c)^5 + 10*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^3 + ((3 *a^2 - 2*a*b + 3*b^2)*d*x - 2*a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(a^2 + 2*a*b + b^2)*d*x + 8*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c )^2 + 8*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 + 15*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^4 + (a^2 - b^2)*d*x + 3*((3*a^2 - 2*a*b + 3*b^2) *d*x - 2*a^2 + 4*a*b)*cosh(d*x + c)^2 - a^2 - a*b)*sinh(d*x + c)^2 - ((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh (d*x + c)^7 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^8 + 4*(a^2 - b^2)*cosh(d*x + c)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + 3*(a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*a^2 - 2*a*b + 3*b^2)*cosh(d*x + c)^4 + 2*(...
Timed out. \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (103) = 206\).
Time = 0.24 (sec) , antiderivative size = 376, normalized size of antiderivative = 3.45 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {d x + c}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac {4 \, {\left ({\left (a^{2} + a b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} - 2 \, a b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{2} + a b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5} + 4 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} + 7 \, a^{4} b + 6 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 7 \, a b^{4} + 3 \, b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {\log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} \]
(d*x + c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) + 4*((a^2 + a*b)*e^(-2*d*x - 2*c) + (a^2 - 2*a*b)*e^(-4*d*x - 4*c) + (a^2 + a*b)*e^(-6*d*x - 6*c))/((a ^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5 + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*e^(-2*d*x - 2*c) + 2*(3*a^5 + 7*a ^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*e^(-4*d*x - 4*c) + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*e^(-6*d*x - 6*c) + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*e^(-8*d*x - 8*c))*d) + 1/2*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/ ((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d)
Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (103) = 206\).
Time = 0.47 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.25 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {2 \, \log \left ({\left | a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {3 \, a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 3 \, b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} - 4 \, a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 12 \, b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 4 \, a + 12 \, b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )}^{2}}}{4 \, d} \]
1/4*(2*log(abs(a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a - 2*b))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (3*a* (e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 3*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 - 4*a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - 12*b*(e^(2*d*x + 2* c) + e^(-2*d*x - 2*c)) - 4*a + 12*b)/((a^2 + 2*a*b + b^2)*(a*(e^(2*d*x + 2 *c) + e^(-2*d*x - 2*c)) + b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a - 2 *b)^2))/d
Time = 0.74 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.82 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {a^4+a^3\,b\,\left (2\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\right )-a\,b^3\,\left (-4\,{\mathrm {tanh}\left (c+d\,x\right )}^2+{\mathrm {tanh}\left (c+d\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,8{}\mathrm {i}\right )+a^2\,b^2\,\left (6\,{\mathrm {tanh}\left (c+d\,x\right )}^2+3-\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,4{}\mathrm {i}\right )-b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^4\,\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,4{}\mathrm {i}}{4\,d\,a^5\,b^2+8\,d\,a^4\,b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^2+12\,d\,a^4\,b^3+4\,d\,a^3\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^4+24\,d\,a^3\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^2+12\,d\,a^3\,b^4+12\,d\,a^2\,b^5\,{\mathrm {tanh}\left (c+d\,x\right )}^4+24\,d\,a^2\,b^5\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\,d\,a^2\,b^5+12\,d\,a\,b^6\,{\mathrm {tanh}\left (c+d\,x\right )}^4+8\,d\,a\,b^6\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\,d\,b^7\,{\mathrm {tanh}\left (c+d\,x\right )}^4} \]
(a^4 + a^3*b*(2*tanh(c + d*x)^2 + 4) - a*b^3*(tanh(c + d*x)^2*atan((a*tanh (c + d*x)^2*1i + b*tanh(c + d*x)^2*1i)/(2*a - a*tanh(c + d*x)^2 + b*tanh(c + d*x)^2))*8i - 4*tanh(c + d*x)^2) + a^2*b^2*(6*tanh(c + d*x)^2 - atan((a *tanh(c + d*x)^2*1i + b*tanh(c + d*x)^2*1i)/(2*a - a*tanh(c + d*x)^2 + b*t anh(c + d*x)^2))*4i + 3) - b^4*tanh(c + d*x)^4*atan((a*tanh(c + d*x)^2*1i + b*tanh(c + d*x)^2*1i)/(2*a - a*tanh(c + d*x)^2 + b*tanh(c + d*x)^2))*4i) /(4*a^2*b^5*d + 12*a^3*b^4*d + 12*a^4*b^3*d + 4*a^5*b^2*d + 4*b^7*d*tanh(c + d*x)^4 + 24*a^2*b^5*d*tanh(c + d*x)^2 + 24*a^3*b^4*d*tanh(c + d*x)^2 + 8*a^4*b^3*d*tanh(c + d*x)^2 + 12*a^2*b^5*d*tanh(c + d*x)^4 + 4*a^3*b^4*d*t anh(c + d*x)^4 + 8*a*b^6*d*tanh(c + d*x)^2 + 12*a*b^6*d*tanh(c + d*x)^4)