\(\int \frac {\coth (c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [197]

Optimal result

Integrand size = 21, antiderivative size = 138 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^3 d}+\frac {\log (\tanh (c+d x))}{a^3 d}-\frac {b \left (3 a^2+3 a b+b^2\right ) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^3 (a+b)^3 d}+\frac {b}{4 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {b (2 a+b)}{2 a^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )} \] Output:

ln(cosh(d*x+c))/(a+b)^3/d+ln(tanh(d*x+c))/a^3/d-1/2*b*(3*a^2+3*a*b+b^2)*ln 
(a+b*tanh(d*x+c)^2)/a^3/(a+b)^3/d+1/4*b/a/(a+b)/d/(a+b*tanh(d*x+c)^2)^2+1/ 
2*b*(2*a+b)/a^2/(a+b)^2/d/(a+b*tanh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {4 \log (\cosh (c+d x))}{(a+b)^3}+\frac {4 \log (\tanh (c+d x))+\frac {b \left (-2 \left (3 a^2+3 a b+b^2\right ) \log \left (a+b \tanh ^2(c+d x)\right )+\frac {a (a+b) \left (a (5 a+3 b)+2 b (2 a+b) \tanh ^2(c+d x)\right )}{\left (a+b \tanh ^2(c+d x)\right )^2}\right )}{(a+b)^3}}{a^3}}{4 d} \] Input:

Integrate[Coth[c + d*x]/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

((4*Log[Cosh[c + d*x]])/(a + b)^3 + (4*Log[Tanh[c + d*x]] + (b*(-2*(3*a^2 
+ 3*a*b + b^2)*Log[a + b*Tanh[c + d*x]^2] + (a*(a + b)*(a*(5*a + 3*b) + 2* 
b*(2*a + b)*Tanh[c + d*x]^2))/(a + b*Tanh[c + d*x]^2)^2))/(a + b)^3)/a^3)/ 
(4*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 4153, 26, 354, 93, 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.

\(\displaystyle \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i}{\tan (i c+i d x) \left (a-b \tan (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {1}{\tan (i c+i d x) \left (a-b \tan (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \frac {i \int -\frac {i \coth (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\int \frac {\coth (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {\int \frac {\coth (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 93

\(\displaystyle \frac {\int \left (-\frac {\left (3 a^2+3 b a+b^2\right ) b^2}{a^3 (a+b)^3 \left (b \tanh ^2(c+d x)+a\right )}-\frac {(2 a+b) b^2}{a^2 (a+b)^2 \left (b \tanh ^2(c+d x)+a\right )^2}-\frac {b^2}{a (a+b) \left (b \tanh ^2(c+d x)+a\right )^3}+\frac {\coth (c+d x)}{a^3}-\frac {1}{(a+b)^3 \left (\tanh ^2(c+d x)-1\right )}\right )d\tanh ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\log \left (\tanh ^2(c+d x)\right )}{a^3}+\frac {b (2 a+b)}{a^2 (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}-\frac {b \left (3 a^2+3 a b+b^2\right ) \log \left (a+b \tanh ^2(c+d x)\right )}{a^3 (a+b)^3}+\frac {b}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{(a+b)^3}}{2 d}\)

Input:

Int[Coth[c + d*x]/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

(Log[Tanh[c + d*x]^2]/a^3 - Log[1 - Tanh[c + d*x]^2]/(a + b)^3 - (b*(3*a^2 
 + 3*a*b + b^2)*Log[a + b*Tanh[c + d*x]^2])/(a^3*(a + b)^3) + b/(2*a*(a + 
b)*(a + b*Tanh[c + d*x]^2)^2) + (b*(2*a + b))/(a^2*(a + b)^2*(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 93

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre 
eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
 

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 = 0.65 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(-\frac {\frac {b^{2} \left (-\frac {a \left (2 a^{2}+3 a b +b^{2}\right )}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\left (3 a^{2}+3 a b +b^{2}\right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}-\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}\right )}{2 \left (a +b \right )^{3} a^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )\right )}{a^{3}}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{3}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{3}}}{d}\)	\(161\)
default	\(-\frac {\frac {b^{2} \left (-\frac {a \left (2 a^{2}+3 a b +b^{2}\right )}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\left (3 a^{2}+3 a b +b^{2}\right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}-\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}\right )}{2 \left (a +b \right )^{3} a^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )\right )}{a^{3}}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{3}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{3}}}{d}\)	\(161\)
parallelrisch	\(\frac {-24 \left (\frac {\left (a +b \right )^{2} \cosh \left (4 d x +4 c \right )}{4}+\left (a^{2}-b^{2}\right ) \cosh \left (2 d x +2 c \right )+\frac {3 a^{2}}{4}-\frac {a b}{2}+\frac {3 b^{2}}{4}\right ) \left (a^{2}+a b +\frac {1}{3} b^{2}\right ) b \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )-16 \left (\frac {\left (a +b \right )^{2} \cosh \left (4 d x +4 c \right )}{4}+\left (a^{2}-b^{2}\right ) \cosh \left (2 d x +2 c \right )+\frac {3 a^{2}}{4}-\frac {a b}{2}+\frac {3 b^{2}}{4}\right ) a^{3} \ln \left (1-\tanh \left (d x +c \right )\right )-4 \left (a +b \right )^{2} \left (-\left (a +b \right )^{3} \ln \left (\tanh \left (d x +c \right )\right )+a^{3} d x +\frac {3 a \,b^{2}}{2}+\frac {3 b^{3}}{4}\right ) \cosh \left (4 d x +4 c \right )-16 \left (a +b \right ) \left (-\left (a -b \right ) \left (a +b \right )^{3} \ln \left (\tanh \left (d x +c \right )\right )+a^{4} d x -a^{3} b d x -\frac {5 a \,b^{3}}{4}-\frac {3 b^{4}}{4}\right ) \cosh \left (2 d x +2 c \right )+12 \left (a +b \right )^{3} \left (a^{2}-\frac {2}{3} a b +b^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )-12 a^{5} d x +8 a^{4} b d x +\left (-12 d x +6\right ) b^{2} a^{3}-5 a^{2} b^{3}-20 a \,b^{4}-9 b^{5}}{16 \left (\frac {\left (a +b \right )^{2} \cosh \left (4 d x +4 c \right )}{4}+\left (a^{2}-b^{2}\right ) \cosh \left (2 d x +2 c \right )+\frac {3 a^{2}}{4}-\frac {a b}{2}+\frac {3 b^{2}}{4}\right ) \left (a +b \right )^{3} a^{3} d}\)	\(390\)
risch	\(\frac {x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {2 x}{a^{3}}-\frac {2 c}{a^{3} d}+\frac {6 b x}{a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 b c}{a d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 b^{2} x}{a^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 b^{2} c}{a^{2} d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 b^{3} x}{a^{3} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 b^{3} c}{a^{3} d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (3 a^{2} {\mathrm e}^{4 d x +4 c}+4 a b \,{\mathrm e}^{4 d x +4 c}+b^{2} {\mathrm e}^{4 d x +4 c}+6 a^{2} {\mathrm e}^{2 d x +2 c}-2 a b \,{\mathrm e}^{2 d x +2 c}-2 b^{2} {\mathrm e}^{2 d x +2 c}+3 a^{2}+4 a b +b^{2}\right ) b^{2} {\mathrm e}^{2 d x +2 c}}{\left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )^{2} \left (a +b \right )^{3} d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{a^{3} d}-\frac {3 b \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 a d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 b^{2} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 a^{2} d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {b^{3} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 a^{3} d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\)	\(607\)

Input:

int(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/d*(1/2*b^2/(a+b)^3/a^3*(-a*(2*a^2+3*a*b+b^2)/b/(a+b*tanh(d*x+c)^2)+(3*a 
^2+3*a*b+b^2)/b*ln(a+b*tanh(d*x+c)^2)-1/2*a^2*(a^2+2*a*b+b^2)/b/(a+b*tanh( 
d*x+c)^2)^2)-1/a^3*ln(tanh(d*x+c))+1/2/(a+b)^3*ln(1+tanh(d*x+c))+1/2/(a+b) 
^3*ln(-1+tanh(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4800 vs. \(2 (132) = 264\).

Time = 0.36 (sec) , antiderivative size = 4800, normalized size of antiderivative = 34.78 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\coth {\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)**2)**3,x)
 

Output:

Integral(coth(c + d*x)/(a + b*tanh(c + d*x)**2)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (132) = 264\).

Time = 0.08 (sec) , antiderivative size = 498, normalized size of antiderivative = 3.61 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {{\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} d} + \frac {d x + c}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac {2 \, {\left ({\left (3 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{2} b^{2} - a b^{3} - b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (3 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5} + 4 \, {\left (a^{7} + 3 \, a^{6} b + 2 \, a^{5} b^{2} - 2 \, a^{4} b^{3} - 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{7} + 7 \, a^{6} b + 6 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 7 \, a^{3} b^{4} + 3 \, a^{2} b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{7} + 3 \, a^{6} b + 2 \, a^{5} b^{2} - 2 \, a^{4} b^{3} - 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{3} d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{3} d} \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/2*(3*a^2*b + 3*a*b^2 + b^3)*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^ 
(-4*d*x - 4*c) + a + b)/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d) + (d*x + 
 c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) + 2*((3*a^2*b^2 + 4*a*b^3 + b^4)*e 
^(-2*d*x - 2*c) + 2*(3*a^2*b^2 - a*b^3 - b^4)*e^(-4*d*x - 4*c) + (3*a^2*b^ 
2 + 4*a*b^3 + b^4)*e^(-6*d*x - 6*c))/((a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4 
*b^3 + 5*a^3*b^4 + a^2*b^5 + 4*(a^7 + 3*a^6*b + 2*a^5*b^2 - 2*a^4*b^3 - 3* 
a^3*b^4 - a^2*b^5)*e^(-2*d*x - 2*c) + 2*(3*a^7 + 7*a^6*b + 6*a^5*b^2 + 6*a 
^4*b^3 + 7*a^3*b^4 + 3*a^2*b^5)*e^(-4*d*x - 4*c) + 4*(a^7 + 3*a^6*b + 2*a^ 
5*b^2 - 2*a^4*b^3 - 3*a^3*b^4 - a^2*b^5)*e^(-6*d*x - 6*c) + (a^7 + 5*a^6*b 
 + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*e^(-8*d*x - 8*c))*d) + l 
og(e^(-d*x - c) + 1)/(a^3*d) + log(e^(-d*x - c) - 1)/(a^3*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (132) = 264\).

Time = 0.27 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.14 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}} + \frac {2 \, {\left (d x + c\right )}}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left ({\left (3 \, a^{2} b^{2} + a b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )} + {\left (3 \, a^{2} b^{2} + a b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )} + \frac {2 \, {\left (3 \, a^{3} b^{2} - a^{2} b^{3} - a b^{4}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{a + b}\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}^{2} {\left (a + b\right )}^{2} a^{3}}}{2 \, d} \] Input:

integrate(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

-1/2*((3*a^2*b + 3*a*b^2 + b^3)*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^6 + 3*a^5*b + 3*a^ 
4*b^2 + a^3*b^3) + 2*(d*x + c)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*log(abs 
(e^(2*d*x + 2*c) - 1))/a^3 - 4*((3*a^2*b^2 + a*b^3)*e^(6*d*x + 6*c) + (3*a 
^2*b^2 + a*b^3)*e^(2*d*x + 2*c) + 2*(3*a^3*b^2 - a^2*b^3 - a*b^4)*e^(4*d*x 
 + 4*c)/(a + b))/((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)^2*(a + b)^2*a^3))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\mathrm {coth}\left (c+d\,x\right )}{{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \] Input:

int(coth(c + d*x)/(a + b*tanh(c + d*x)^2)^3,x)
 

Output:

int(coth(c + d*x)/(a + b*tanh(c + d*x)^2)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 5158, normalized size of antiderivative = 37.38 \[ \int \frac {\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(coth(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

(2*e**(8*c + 8*d*x)*log(e**(c + d*x) - 1)*a**6 + 8*e**(8*c + 8*d*x)*log(e* 
*(c + d*x) - 1)*a**5*b + 10*e**(8*c + 8*d*x)*log(e**(c + d*x) - 1)*a**4*b* 
*2 - 10*e**(8*c + 8*d*x)*log(e**(c + d*x) - 1)*a**2*b**4 - 8*e**(8*c + 8*d 
*x)*log(e**(c + d*x) - 1)*a*b**5 - 2*e**(8*c + 8*d*x)*log(e**(c + d*x) - 1 
)*b**6 + 2*e**(8*c + 8*d*x)*log(e**(c + d*x) + 1)*a**6 + 8*e**(8*c + 8*d*x 
)*log(e**(c + d*x) + 1)*a**5*b + 10*e**(8*c + 8*d*x)*log(e**(c + d*x) + 1) 
*a**4*b**2 - 10*e**(8*c + 8*d*x)*log(e**(c + d*x) + 1)*a**2*b**4 - 8*e**(8 
*c + 8*d*x)*log(e**(c + d*x) + 1)*a*b**5 - 2*e**(8*c + 8*d*x)*log(e**(c + 
d*x) + 1)*b**6 - 3*e**(8*c + 8*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqr 
t(a + b) - 2*e**(c + d*x)*sqrt(b))*a**5*b - 6*e**(8*c + 8*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**(8*c + 8*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c 
+ d*x)*sqrt(b))*a**3*b**3 + 5*e**(8*c + 8*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 + 4*e**(8*c + 8*d* 
x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b) 
)*a*b**5 + e**(8*c + 8*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) 
 - 2*e**(c + d*x)*sqrt(b))*b**6 - 3*e**(8*c + 8*d*x)*log(e**(2*c + 2*d*x)* 
sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*a**5*b - 6*e**(8*c + 8 
*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**(8*c + 8*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sq...