\(\int \frac {\tanh ^5(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [180]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^2 d}-\frac {a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 b^2 (a+b)^2 d}-\frac {a^2}{2 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \] Output:

ln(cosh(d*x+c))/(a+b)^2/d-1/2*a*(a+2*b)*ln(a+b*tanh(d*x+c)^2)/b^2/(a+b)^2/ 
d-1/2*a^2/b^2/(a+b)/d/(a+b*tanh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {-2 \log (\cosh (c+d x))+\frac {a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{b^2}+\frac {a^2 (a+b)}{b^2 \left (a+b \tanh ^2(c+d x)\right )}}{2 (a+b)^2 d} \] Input:

Integrate[Tanh[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

-1/2*(-2*Log[Cosh[c + d*x]] + (a*(a + 2*b)*Log[a + b*Tanh[c + d*x]^2])/b^2 
 + (a^2*(a + b))/(b^2*(a + b*Tanh[c + d*x]^2)))/((a + b)^2*d)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01, 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.

\(\displaystyle \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \tan (i c+i d x)^5}{\left (a-b \tan (i c+i d x)^2\right )^2}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\tan (i c+i d x)^5}{\left (a-b \tan (i c+i d x)^2\right )^2}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle -\frac {i \int \frac {i \tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\int \frac {\tanh ^5(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {a^2}{b (a+b) \left (b \tanh ^2(c+d x)+a\right )^2}-\frac {(a+2 b) a}{b (a+b)^2 \left (b \tanh ^2(c+d x)+a\right )}-\frac {1}{(a+b)^2 \left (\tanh ^2(c+d x)-1\right )}\right )d\tanh ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {a^2}{b^2 (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{b^2 (a+b)^2}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{(a+b)^2}}{2 d}\)

Input:

Int[Tanh[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

(-(Log[1 - Tanh[c + d*x]^2]/(a + b)^2) - (a*(a + 2*b)*Log[a + b*Tanh[c + d 
*x]^2])/(b^2*(a + b)^2) - a^2/(b^2*(a + b)*(a + b*Tanh[c + d*x]^2)))/(2*d)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 99

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]))
 

rule 354

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4153

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]))
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {a \left (\frac {\left (a +2 b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b^{2}}+\frac {a \left (a +b \right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(91\)
default	\(\frac {-\frac {a \left (\frac {\left (a +2 b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b^{2}}+\frac {a \left (a +b \right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(91\)
parallelrisch	\(-\frac {a^{3}+a^{2} b +\ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) a^{3}+2 a \,b^{2} d x +2 \ln \left (1-\tanh \left (d x +c \right )\right ) a \,b^{2}+2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) a^{2} b +2 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2} b^{3}+2 b^{3} \tanh \left (d x +c \right )^{2} x d +\ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a^{2} b +2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a \,b^{2}}{2 \left (a +b \tanh \left (d x +c \right )^{2}\right ) d \left (a +b \right )^{2} b^{2}}\)	\(190\)
risch	\(\frac {x}{a^{2}+2 a b +b^{2}}-\frac {2 x}{b^{2}}-\frac {2 c}{b^{2} d}+\frac {2 a^{2} x}{b^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 a^{2} c}{b^{2} d \left (a^{2}+2 a b +b^{2}\right )}+\frac {4 a x}{b \left (a^{2}+2 a b +b^{2}\right )}+\frac {4 a c}{b d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 a^{2} {\mathrm e}^{2 d x +2 c}}{b d \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{b^{2} d}-\frac {a^{2} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 b^{2} d \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{b d \left (a^{2}+2 a b +b^{2}\right )}\)	\(329\)

Input:

int(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/2*a/(a+b)^2*((a+2*b)/b^2*ln(a+b*tanh(d*x+c)^2)+a*(a+b)/b^2/(a+b*ta 
nh(d*x+c)^2))-1/2/(a+b)^2*ln(1+tanh(d*x+c))-1/2/(a+b)^2*ln(-1+tanh(d*x+c)) 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1141 vs. \(2 (79) = 158\).

Time = 0.18 (sec) , antiderivative size = 1141, normalized size of antiderivative = 13.75 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

-1/2*(2*(a*b^2 + b^3)*d*x*cosh(d*x + c)^4 + 8*(a*b^2 + b^3)*d*x*cosh(d*x + 
 c)*sinh(d*x + c)^3 + 2*(a*b^2 + b^3)*d*x*sinh(d*x + c)^4 + 2*(a*b^2 + b^3 
)*d*x + 4*(a^2*b + (a*b^2 - b^3)*d*x)*cosh(d*x + c)^2 + 4*(3*(a*b^2 + b^3) 
*d*x*cosh(d*x + c)^2 + a^2*b + (a*b^2 - b^3)*d*x)*sinh(d*x + c)^2 + ((a^3 
+ 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^4 + 4*(a^3 + 3*a^2*b + 2*a*b^2)*cosh(d* 
x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 2*a*b^2)*sinh(d*x + c)^4 + a^3 + 
 3*a^2*b + 2*a*b^2 + 2*(a^3 + a^2*b - 2*a*b^2)*cosh(d*x + c)^2 + 2*(a^3 + 
a^2*b - 2*a*b^2 + 3*(a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + 
c)^2 + 4*((a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^3 + (a^3 + a^2*b - 2*a*b 
^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*((a + b)*cosh(d*x + c)^2 + (a + b) 
*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) 
 + sinh(d*x + c)^2)) - 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 
+ 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 
 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 
+ 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b - a*b^2 - 
 b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 
+ 4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^ 
2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) 
- sinh(d*x + c))) + 8*((a*b^2 + b^3)*d*x*cosh(d*x + c)^3 + (a^2*b + (a*b^2 
 - b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^3*b^2 + 3*a^2*b^3 + 3*a*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:

integrate(tanh(d*x+c)**5/(a+b*tanh(d*x+c)**2)**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (79) = 158\).

Time = 0.13 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.61 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + a^{2} b^{2} - a b^{3} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {{\left (a^{2} + 2 \, a b\right )} \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{2} d} \] Input:

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-2*a^2*e^(-2*d*x - 2*c)/((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4 + 2*(a^3*b + a 
^2*b^2 - a*b^3 - b^4)*e^(-2*d*x - 2*c) + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^ 
4)*e^(-4*d*x - 4*c))*d) - 1/2*(a^2 + 2*a*b)*log(2*(a - b)*e^(-2*d*x - 2*c) 
 + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^2*b^2 + 2*a*b^3 + b^4)*d) + (d*x 
+ c)/((a^2 + 2*a*b + b^2)*d) + log(e^(-2*d*x - 2*c) + 1)/(b^2*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (79) = 158\).

Time = 0.24 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.34 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (a^{2} + 2 \, a b\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{a^{2} b^{2} + 2 \, a b^{3} + b^{4}} + \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} + \frac {4 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} {\left (a + b\right )}^{2} b} - \frac {2 \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{b^{2}}}{2 \, d} \] Input:

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

-1/2*((a^2 + 2*a*b)*log(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d 
*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)/(a^2*b^2 + 2*a*b^3 + b^4) + 2*(d* 
x + c)/(a^2 + 2*a*b + b^2) + 4*a^2*e^(2*d*x + 2*c)/((a*e^(4*d*x + 4*c) + b 
*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)*(a + 
 b)^2*b) - 2*log(e^(2*d*x + 2*c) + 1)/b^2)/d
 

Mupad [B] (verification not implemented)

Time = 2.93 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.05 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {a^2}{2\,\left (d\,a^2\,b^2+d\,a\,b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^2+d\,a\,b^3+d\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left ({\mathrm {tanh}\left (c+d\,x\right )}^2-1\right )}{2\,\left (d\,a^2+2\,d\,a\,b+d\,b^2\right )}-\frac {a^2\,\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{2\,\left (d\,a^2\,b^2+2\,d\,a\,b^3+d\,b^4\right )}-\frac {a\,b\,\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{d\,a^2\,b^2+2\,d\,a\,b^3+d\,b^4} \] Input:

int(tanh(c + d*x)^5/(a + b*tanh(c + d*x)^2)^2,x)
 

Output:

- a^2/(2*(a^2*b^2*d + b^4*d*tanh(c + d*x)^2 + a*b^3*d + a*b^3*d*tanh(c + d 
*x)^2)) - log(tanh(c + d*x)^2 - 1)/(2*(a^2*d + b^2*d + 2*a*b*d)) - (a^2*lo 
g(a + b*tanh(c + d*x)^2))/(2*(b^4*d + a^2*b^2*d + 2*a*b^3*d)) - (a*b*log(a 
 + b*tanh(c + d*x)^2))/(b^4*d + a^2*b^2*d + 2*a*b^3*d)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 3130, normalized size of antiderivative = 37.71 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x)
 

Output:

( - e**(4*c + 4*d*x)*log(tanh(c + d*x)**2*b + a)*tanh(c + d*x)**2*a**5*b - 
 2*e**(4*c + 4*d*x)*log(tanh(c + d*x)**2*b + a)*tanh(c + d*x)**2*a**4*b**2 
 + 2*e**(4*c + 4*d*x)*log(tanh(c + d*x)**2*b + a)*tanh(c + d*x)**2*a**2*b* 
*4 + e**(4*c + 4*d*x)*log(tanh(c + d*x)**2*b + a)*tanh(c + d*x)**2*a*b**5 
- e**(4*c + 4*d*x)*log(tanh(c + d*x)**2*b + a)*a**6 - 2*e**(4*c + 4*d*x)*l 
og(tanh(c + d*x)**2*b + a)*a**5*b + 2*e**(4*c + 4*d*x)*log(tanh(c + d*x)** 
2*b + a)*a**3*b**3 + e**(4*c + 4*d*x)*log(tanh(c + d*x)**2*b + a)*a**2*b** 
4 + e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e* 
*(c + d*x)*sqrt(b))*tanh(c + d*x)**2*a**3*b**3 - e**(4*c + 4*d*x)*log(e**( 
2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*tanh(c + 
d*x)**2*a*b**5 + e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt( 
a + b) - 2*e**(c + d*x)*sqrt(b))*a**4*b**2 - e**(4*c + 4*d*x)*log(e**(2*c 
+ 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a**2*b**4 + e 
**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + 
 d*x)*sqrt(b))*tanh(c + d*x)**2*a**3*b**3 - e**(4*c + 4*d*x)*log(e**(2*c + 
 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*tanh(c + d*x)* 
*2*a*b**5 + e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b 
) + 2*e**(c + d*x)*sqrt(b))*a**4*b**2 - e**(4*c + 4*d*x)*log(e**(2*c + 2*d 
*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*a**2*b**4 + e**(4* 
c + 4*d*x)*tanh(c + d*x)**2*a**5*b + e**(4*c + 4*d*x)*tanh(c + d*x)**2*...