\(\int \frac {\coth ^4(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [189]

Optimal result

Integrand size = 23, antiderivative size = 159 \[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {x}{(a+b)^2}+\frac {b^{5/2} (7 a+5 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{7/2} (a+b)^2 d}-\frac {\left (2 a^2-2 a b-5 b^2\right ) \coth (c+d x)}{2 a^3 (a+b) d}-\frac {(2 a+5 b) \coth ^3(c+d x)}{6 a^2 (a+b) d}+\frac {b \coth ^3(c+d x)}{2 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \] Output:

x/(a+b)^2+1/2*b^(5/2)*(7*a+5*b)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/a^(7/2 
)/(a+b)^2/d-1/2*(2*a^2-2*a*b-5*b^2)*coth(d*x+c)/a^3/(a+b)/d-1/6*(2*a+5*b)* 
coth(d*x+c)^3/a^2/(a+b)/d+1/2*b*coth(d*x+c)^3/a/(a+b)/d/(a+b*tanh(d*x+c)^2 
)
 

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87 \[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {6 (c+d x)}{(a+b)^2}+\frac {3 b^{5/2} (7 a+5 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{7/2} (a+b)^2}+\frac {4 (-2 a+3 b) \coth (c+d x)}{a^3}-\frac {2 \coth (c+d x) \text {csch}^2(c+d x)}{a^2}+\frac {3 b^3 \sinh (2 (c+d x))}{a^3 (a+b) (a-b+(a+b) \cosh (2 (c+d x)))}}{6 d} \] Input:

Integrate[Coth[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

((6*(c + d*x))/(a + b)^2 + (3*b^(5/2)*(7*a + 5*b)*ArcTan[(Sqrt[b]*Tanh[c + 
 d*x])/Sqrt[a]])/(a^(7/2)*(a + b)^2) + (4*(-2*a + 3*b)*Coth[c + d*x])/a^3 
- (2*Coth[c + d*x]*Csch[c + d*x]^2)/a^2 + (3*b^3*Sinh[2*(c + d*x)])/(a^3*( 
a + b)*(a - b + (a + b)*Cosh[2*(c + d*x)])))/(6*d)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4153, 374, 25, 445, 27, 445, 25, 397, 218, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 374

\(\displaystyle \frac {\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {\int -\frac {\coth ^4(c+d x) \left (-5 b \tanh ^2(c+d x)+2 a+5 b\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\coth ^4(c+d x) \left (-5 b \tanh ^2(c+d x)+2 a+5 b\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {\int -\frac {3 \coth ^2(c+d x) \left (2 a^2-2 b a-5 b^2+b (2 a+5 b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{3 a}-\frac {(2 a+5 b) \coth ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {\coth ^2(c+d x) \left (2 a^2-2 b a-5 b^2+b (2 a+5 b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{a}-\frac {(2 a+5 b) \coth ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {\frac {-\frac {\int -\frac {2 a^3-2 b a^2+2 b^2 a+5 b^3+b \left (2 a^2-2 b a-5 b^2\right ) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{a}-\frac {\left (2 a^2-2 a b-5 b^2\right ) \coth (c+d x)}{a}}{a}-\frac {(2 a+5 b) \coth ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {2 a^3-2 b a^2+2 b^2 a+5 b^3+b \left (2 a^2-2 b a-5 b^2\right ) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{a}-\frac {\left (2 a^2-2 a b-5 b^2\right ) \coth (c+d x)}{a}}{a}-\frac {(2 a+5 b) \coth ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {\frac {2 a^3 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {b^3 (7 a+5 b) \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}}{a}-\frac {\left (2 a^2-2 a b-5 b^2\right ) \coth (c+d x)}{a}}{a}-\frac {(2 a+5 b) \coth ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {\frac {2 a^3 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {b^{5/2} (7 a+5 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{a}-\frac {\left (2 a^2-2 a b-5 b^2\right ) \coth (c+d x)}{a}}{a}-\frac {(2 a+5 b) \coth ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\frac {\frac {2 a^3 \text {arctanh}(\tanh (c+d x))}{a+b}+\frac {b^{5/2} (7 a+5 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{a}-\frac {\left (2 a^2-2 a b-5 b^2\right ) \coth (c+d x)}{a}}{a}-\frac {(2 a+5 b) \coth ^3(c+d x)}{3 a}}{2 a (a+b)}+\frac {b \coth ^3(c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\)

Input:

Int[Coth[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

((-1/3*((2*a + 5*b)*Coth[c + d*x]^3)/a + (((b^(5/2)*(7*a + 5*b)*ArcTan[(Sq 
rt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a + b)) + (2*a^3*ArcTanh[Tanh[c + 
 d*x]])/(a + b))/a - ((2*a^2 - 2*a*b - 5*b^2)*Coth[c + d*x])/a)/a)/(2*a*(a 
 + b)) + (b*Coth[c + d*x]^3)/(2*a*(a + b)*(a + b*Tanh[c + d*x]^2)))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 374

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
 + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - 
a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, 
c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, 
 c, d, e, m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 445

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ 
.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p 
+ 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) 
 Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c 
+ a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
 

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 = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(-\frac {-\frac {b^{3} \left (\frac {\left (\frac {b}{2}+\frac {a}{2}\right ) \tanh \left (d x +c \right )}{a +b \tanh \left (d x +c \right )^{2}}+\frac {\left (7 a +5 b \right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a +b \right )^{2} a^{3}}-\frac {2 b -a}{a^{3} \tanh \left (d x +c \right )}+\frac {1}{3 a^{2} \tanh \left (d x +c \right )^{3}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(142\)
default	\(-\frac {-\frac {b^{3} \left (\frac {\left (\frac {b}{2}+\frac {a}{2}\right ) \tanh \left (d x +c \right )}{a +b \tanh \left (d x +c \right )^{2}}+\frac {\left (7 a +5 b \right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a +b \right )^{2} a^{3}}-\frac {2 b -a}{a^{3} \tanh \left (d x +c \right )}+\frac {1}{3 a^{2} \tanh \left (d x +c \right )^{3}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(142\)
risch	\(\frac {x}{a^{2}+2 a b +b^{2}}-\frac {12 a^{4} {\mathrm e}^{8 d x +8 c}+24 a^{3} b \,{\mathrm e}^{8 d x +8 c}-21 a \,b^{3} {\mathrm e}^{8 d x +8 c}-15 b^{4} {\mathrm e}^{8 d x +8 c}+12 a^{4} {\mathrm e}^{6 d x +6 c}-12 a^{3} b \,{\mathrm e}^{6 d x +6 c}-12 a^{2} b^{2} {\mathrm e}^{6 d x +6 c}+54 a \,b^{3} {\mathrm e}^{6 d x +6 c}+60 b^{4} {\mathrm e}^{6 d x +6 c}-4 a^{4} {\mathrm e}^{4 d x +4 c}+60 a^{3} b \,{\mathrm e}^{4 d x +4 c}+60 a^{2} b^{2} {\mathrm e}^{4 d x +4 c}-76 a \,b^{3} {\mathrm e}^{4 d x +4 c}-90 b^{4} {\mathrm e}^{4 d x +4 c}+4 a^{4} {\mathrm e}^{2 d x +2 c}-20 a^{3} b \,{\mathrm e}^{2 d x +2 c}-4 a^{2} b^{2} {\mathrm e}^{2 d x +2 c}+74 a \,b^{3} {\mathrm e}^{2 d x +2 c}+60 b^{4} {\mathrm e}^{2 d x +2 c}+8 a^{4}+12 a^{3} b -12 a^{2} b^{2}-31 a \,b^{3}-15 b^{4}}{3 a^{3} d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3} \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {7 \sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {-a b}-b}{a +b}\right )}{4 a^{3} \left (a +b \right )^{2} d}+\frac {5 \sqrt {-a b}\, b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {-a b}-b}{a +b}\right )}{4 a^{4} \left (a +b \right )^{2} d}-\frac {7 \sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {-a b}+b}{a +b}\right )}{4 a^{3} \left (a +b \right )^{2} d}-\frac {5 \sqrt {-a b}\, b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {-a b}+b}{a +b}\right )}{4 a^{4} \left (a +b \right )^{2} d}\)	\(618\)

Input:

int(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/d*(-b^3/(a+b)^2/a^3*((1/2*b+1/2*a)*tanh(d*x+c)/(a+b*tanh(d*x+c)^2)+1/2* 
(7*a+5*b)/(a*b)^(1/2)*arctan(b*tanh(d*x+c)/(a*b)^(1/2)))-(2*b-a)/a^3/tanh( 
d*x+c)+1/3/a^2/tanh(d*x+c)^3-1/2/(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. 4080 vs. \(2 (143) = 286\).

Time = 0.23 (sec) , antiderivative size = 8482, normalized size of antiderivative = 53.35 \[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {\coth ^{4}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(coth(d*x+c)**4/(a+b*tanh(d*x+c)**2)**2,x)
 

Output:

Integral(coth(c + d*x)**4/(a + b*tanh(c + d*x)**2)**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2345 vs. \(2 (143) = 286\).

Time = 0.50 (sec) , antiderivative size = 2345, normalized size of antiderivative = 14.75 \[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
 

Output:

-1/4*(a^2*b - a*b^2 - b^3)*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d* 
x + 2*c) + a + b)/((a^5 + 2*a^4*b + a^3*b^2)*d) + 1/4*(a^2*b - 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^5 
 + 2*a^4*b + a^3*b^2)*d) + 1/32*(3*a^3*b - 29*a^2*b^2 - 11*a*b^3 + 5*b^4)* 
arctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^5 + 2*a^4*b + 
a^3*b^2)*sqrt(a*b)*d) - 1/32*(3*a^3*b - 29*a^2*b^2 - 11*a*b^3 + 5*b^4)*arc 
tan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/((a^5 + 2*a^4*b + a^ 
3*b^2)*sqrt(a*b)*d) + 1/48*(44*a^4 + 117*a^3*b + 111*a^2*b^2 + 23*a*b^3 - 
15*b^4 + 3*(24*a^4 + 69*a^3*b + 45*a^2*b^2 + 27*a*b^3 - 5*b^4)*e^(8*d*x + 
8*c) + 6*(6*a^4 - 31*a^3*b - 50*a^2*b^2 - 51*a*b^3 + 10*b^4)*e^(6*d*x + 6* 
c) - 2*(50*a^4 - 78*a^3*b - 225*a^2*b^2 - 196*a*b^3 + 45*b^4)*e^(4*d*x + 4 
*c) - 2*(10*a^4 + 115*a^3*b + 182*a^2*b^2 + 95*a*b^3 - 30*b^4)*e^(2*d*x + 
2*c))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3 - (a^6 + 3*a^5*b + 3*a^4*b^2 + 
 a^3*b^3)*e^(10*d*x + 10*c) + (a^6 + 7*a^5*b + 11*a^4*b^2 + 5*a^3*b^3)*e^( 
8*d*x + 8*c) + 2*(a^6 - 3*a^5*b - 9*a^4*b^2 - 5*a^3*b^3)*e^(6*d*x + 6*c) - 
 2*(a^6 - 3*a^5*b - 9*a^4*b^2 - 5*a^3*b^3)*e^(4*d*x + 4*c) - (a^6 + 7*a^5* 
b + 11*a^4*b^2 + 5*a^3*b^3)*e^(2*d*x + 2*c))*d) - 1/48*(44*a^4 + 117*a^3*b 
 + 111*a^2*b^2 + 23*a*b^3 - 15*b^4 - 2*(10*a^4 + 115*a^3*b + 182*a^2*b^2 + 
 95*a*b^3 - 30*b^4)*e^(-2*d*x - 2*c) - 2*(50*a^4 - 78*a^3*b - 225*a^2*b^2 
- 196*a*b^3 + 45*b^4)*e^(-4*d*x - 4*c) + 6*(6*a^4 - 31*a^3*b - 50*a^2*b...
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.77 \[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {3 \, {\left (7 \, a b^{3} + 5 \, b^{4}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {a b}} + \frac {6 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} - \frac {6 \, {\left (a b^{3} e^{\left (2 \, d x + 2 \, c\right )} - b^{4} e^{\left (2 \, d x + 2 \, c\right )} + a b^{3} + b^{4}\right )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}} - \frac {8 \, {\left (3 \, a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, a e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - 3 \, b\right )}}{a^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{6 \, d} \] Input:

integrate(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

1/6*(3*(7*a*b^3 + 5*b^4)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) 
 + a - b)/sqrt(a*b))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt(a*b)) + 6*(d*x + c)/( 
a^2 + 2*a*b + b^2) - 6*(a*b^3*e^(2*d*x + 2*c) - b^4*e^(2*d*x + 2*c) + a*b^ 
3 + b^4)/((a^5 + 2*a^4*b + a^3*b^2)*(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)) - 8*(3*a*e^(4*d*x + 
 4*c) - 3*b*e^(4*d*x + 4*c) - 3*a*e^(2*d*x + 2*c) + 6*b*e^(2*d*x + 2*c) + 
2*a - 3*b)/(a^3*(e^(2*d*x + 2*c) - 1)^3))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {coth}\left (c+d\,x\right )}^4}{{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \] Input:

int(coth(c + d*x)^4/(a + b*tanh(c + d*x)^2)^2,x)
 

Output:

int(coth(c + d*x)^4/(a + b*tanh(c + d*x)^2)^2, x)
 

Reduce [F]

\[ \int \frac {\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {\coth \left (d x +c \right )^{4}}{\left (\tanh \left (d x +c \right )^{2} b +a \right )^{2}}d x \] Input:

int(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x)
 

Output:

int(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x)