\(\int \coth ^5(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\) [165]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {a^2 (a+3 b) \coth ^2(c+d x)}{2 d}-\frac {a^3 \coth ^4(c+d x)}{4 d}+\frac {(a+b)^3 \log (\cosh (c+d x))}{d}+\frac {a \left (a^2+3 a b+3 b^2\right ) \log (\tanh (c+d x))}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {-a^2 (a+3 b) \coth ^2(c+d x)-\frac {1}{2} a^3 \coth ^4(c+d x)+2 (a+b)^3 \log (\sinh (c+d x))-2 b^3 \log (\tanh (c+d x))}{2 d} \] Input:

Integrate[Coth[c + d*x]^5*(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 83, 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, 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.

\(\displaystyle \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \left (a-b \tan (i c+i d x)^2\right )^3}{\tan (i c+i d x)^5}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\tan (i c+i d x)^5}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

Int[Coth[c + d*x]^5*(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

(-(a^2*(a + 3*b)*Coth[c + d*x]) - (a^3*Coth[c + d*x]^2)/2 + a*(a^2 + 3*a*b 
 + 3*b^2)*Log[Tanh[c + d*x]^2] - (a + b)^3*Log[1 - 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 = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(\frac {-4 \left (a +b \right )^{3} \ln \left (1-\tanh \left (d x +c \right )\right )+4 \left (a^{3}+3 a^{2} b +3 a \,b^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )-\coth \left (d x +c \right )^{4} a^{3}-2 a^{2} \coth \left (d x +c \right )^{2} \left (a +3 b \right )-4 d x \left (a +b \right )^{3}}{4 d}\)	\(90\)
derivativedivides	\(-\frac {\frac {a^{3}}{4 \tanh \left (d x +c \right )^{4}}-a \left (a^{2}+3 a b +3 b^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\frac {a^{2} \left (a +3 b \right )}{2 \tanh \left (d x +c \right )^{2}}+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\)	\(127\)
default	\(-\frac {\frac {a^{3}}{4 \tanh \left (d x +c \right )^{4}}-a \left (a^{2}+3 a b +3 b^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\frac {a^{2} \left (a +3 b \right )}{2 \tanh \left (d x +c \right )^{2}}+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\)	\(127\)
risch	\(-a^{3} x -3 a^{2} b x -3 a \,b^{2} x -b^{3} x -\frac {2 a^{3} c}{d}-\frac {6 a^{2} b c}{d}-\frac {6 a \,b^{2} c}{d}-\frac {2 b^{3} c}{d}-\frac {2 a^{2} {\mathrm e}^{2 d x +2 c} \left (2 a \,{\mathrm e}^{4 d x +4 c}+3 b \,{\mathrm e}^{4 d x +4 c}-2 a \,{\mathrm e}^{2 d x +2 c}-6 b \,{\mathrm e}^{2 d x +2 c}+2 a +3 b \right )}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b}{d}+\frac {3 a \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) b^{3}}{d}\)	\(231\)

Input:

int(coth(d*x+c)^5*(a+b*tanh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2393 vs. \(2 (79) = 158\).

Time = 0.11 (sec) , antiderivative size = 2393, normalized size of antiderivative = 28.83 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \] Input:

integrate(coth(d*x+c)^5*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

-((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^8 + 8*(a^3 + 3*a^2*b + 
 3*a*b^2 + b^3)*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + (a^3 + 3*a^2*b + 3*a*b 
^2 + b^3)*d*x*sinh(d*x + c)^8 + 2*(2*a^3 + 3*a^2*b - 2*(a^3 + 3*a^2*b + 3* 
a*b^2 + b^3)*d*x)*cosh(d*x + c)^6 + 2*(14*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)* 
d*x*cosh(d*x + c)^2 + 2*a^3 + 3*a^2*b - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)* 
d*x)*sinh(d*x + c)^6 + 4*(14*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x 
+ c)^3 + 3*(2*a^3 + 3*a^2*b - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh( 
d*x + c))*sinh(d*x + c)^5 - 2*(2*a^3 + 6*a^2*b - 3*(a^3 + 3*a^2*b + 3*a*b^ 
2 + b^3)*d*x)*cosh(d*x + c)^4 + 2*(35*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x* 
cosh(d*x + c)^4 - 2*a^3 - 6*a^2*b + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x 
+ 15*(2*a^3 + 3*a^2*b - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + 
c)^2)*sinh(d*x + c)^4 + 8*(7*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x 
+ c)^5 + 5*(2*a^3 + 3*a^2*b - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh( 
d*x + c)^3 - (2*a^3 + 6*a^2*b - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cos 
h(d*x + c))*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x + 2*(2*a 
^3 + 3*a^2*b - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2 + 2* 
(14*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^6 + 15*(2*a^3 + 3*a^ 
2*b - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^4 + 2*a^3 + 3*a 
^2*b - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x - 6*(2*a^3 + 6*a^2*b - 3*(a^3 
 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - (b^...
 

Sympy [F(-1)]

Timed out. \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:

integrate(coth(d*x+c)**5*(a+b*tanh(d*x+c)**2)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (79) = 158\).

Time = 0.06 (sec) , antiderivative size = 264, normalized size of antiderivative = 3.18 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=a^{3} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + \frac {b^{3} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}{d} + \frac {3 \, a b^{2} \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}{d} \] Input:

integrate(coth(d*x+c)^5*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (79) = 158\).

Time = 0.34 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.22 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {2 \, b^{3} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right ) + 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right ) - \frac {3 \, a^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 9 \, a^{2} b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 9 \, a b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 4 \, a^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 12 \, a^{2} b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 36 \, a b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 4 \, a^{3} - 12 \, a^{2} b + 36 \, a b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right )}^{2}}}{4 \, d} \] Input:

integrate(coth(d*x+c)^5*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

1/4*(2*b^3*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2) + 2*(a^3 + 3*a^2*b 
+ 3*a*b^2)*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2) - (3*a^3*(e^(2*d*x 
+ 2*c) + e^(-2*d*x - 2*c))^2 + 9*a^2*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) 
)^2 + 9*a*b^2*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 4*a^3*(e^(2*d*x + 2 
*c) + e^(-2*d*x - 2*c)) - 12*a^2*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - 
36*a*b^2*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - 4*a^3 - 12*a^2*b + 36*a*b^ 
2)/(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2)^2)/d
 

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 381, normalized size of antiderivative = 4.59 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {\ln \left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )\,\left (b^3\,d+d\,\left (a^3+3\,a^2\,b+3\,a\,b^2\right )\right )}{2\,d^2}-x\,{\left (a+b\right )}^3-\frac {2\,\left (2\,a^3+3\,b\,a^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {8\,a^3}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (a^3\,\sqrt {-d^2}-b^3\,\sqrt {-d^2}+3\,a\,b^2\,\sqrt {-d^2}+3\,a^2\,b\,\sqrt {-d^2}\right )}{d\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+16\,a^3\,b^3+3\,a^2\,b^4-6\,a\,b^5+b^6}}\right )\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+16\,a^3\,b^3+3\,a^2\,b^4-6\,a\,b^5+b^6}}{\sqrt {-d^2}}-\frac {2\,\left (4\,a^3+3\,b\,a^2\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {4\,a^3}{d\,\left (6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )} \] Input:

int(coth(c + d*x)^5*(a + b*tanh(c + d*x)^2)^3,x)
 

Output:

(log(exp(4*c + 4*d*x) - 1)*(b^3*d + d*(3*a*b^2 + 3*a^2*b + a^3)))/(2*d^2) 
- x*(a + b)^3 - (2*(3*a^2*b + 2*a^3))/(d*(exp(2*c + 2*d*x) - 1)) - (8*a^3) 
/(d*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1)) - (a 
tan((exp(2*c)*exp(2*d*x)*(a^3*(-d^2)^(1/2) - b^3*(-d^2)^(1/2) + 3*a*b^2*(- 
d^2)^(1/2) + 3*a^2*b*(-d^2)^(1/2)))/(d*(6*a^5*b - 6*a*b^5 + a^6 + b^6 + 3* 
a^2*b^4 + 16*a^3*b^3 + 15*a^4*b^2)^(1/2)))*(6*a^5*b - 6*a*b^5 + a^6 + b^6 
+ 3*a^2*b^4 + 16*a^3*b^3 + 15*a^4*b^2)^(1/2))/(-d^2)^(1/2) - (2*(3*a^2*b + 
 4*a^3))/(d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - (4*a^3)/(d*(6*e 
xp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8*c + 8*d* 
x) + 1))
 

Reduce [F]

\[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\int \coth \left (d x +c \right )^{5} \left (\tanh \left (d x +c \right )^{2} b +a \right )^{3}d x \] Input:

int(coth(d*x+c)^5*(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

int(coth(d*x+c)^5*(a+b*tanh(d*x+c)^2)^3,x)