\(\int \coth ^6(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\) [166]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=(a+b)^3 x-\frac {a \left (a^2+3 a b+3 b^2\right ) \coth (c+d x)}{d}-\frac {a^2 (a+3 b) \coth ^3(c+d x)}{3 d}-\frac {a^3 \coth ^5(c+d x)}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {a \coth (c+d x) \left (15 \left (a^2+3 a b+3 b^2\right )+5 a (a+3 b) \coth ^2(c+d x)+3 a^2 \coth ^4(c+d x)\right )}{15 d}+\frac {(a+b)^3 \text {arctanh}\left (\sqrt {\tanh ^2(c+d x)}\right ) \tanh (c+d x)}{d \sqrt {\tanh ^2(c+d x)}} \] Input:

Integrate[Coth[c + d*x]^6*(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

-1/15*(a*Coth[c + d*x]*(15*(a^2 + 3*a*b + 3*b^2) + 5*a*(a + 3*b)*Coth[c + 
d*x]^2 + 3*a^2*Coth[c + d*x]^4))/d + ((a + b)^3*ArcTanh[Sqrt[Tanh[c + d*x] 
^2]]*Tanh[c + d*x])/(d*Sqrt[Tanh[c + d*x]^2])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 25, 4153, 25, 364, 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.

\(\displaystyle \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\tan (i c+i d x)^6}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\tan (i c+i d x)^6}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle -\frac {\int -\frac {\coth ^6(c+d x) \left (b \tanh ^2(c+d x)+a\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\coth ^6(c+d x) \left (b \tanh ^2(c+d x)+a\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 364

\(\displaystyle \frac {\int \left (a^3 \coth ^6(c+d x)+a^2 (a+3 b) \coth ^4(c+d x)+a \left (a^2+3 b a+3 b^2\right ) \coth ^2(c+d x)-\frac {(a+b)^3}{\tanh ^2(c+d x)-1}\right )d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 2009

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

Input:

Int[Coth[c + d*x]^6*(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

-((-((a + b)^3*ArcTanh[Tanh[c + d*x]]) + a*(a^2 + 3*a*b + 3*b^2)*Coth[c + 
d*x] + (a^2*(a + 3*b)*Coth[c + d*x]^3)/3 + (a^3*Coth[c + d*x]^5)/5)/d)
 

Defintions of rubi rules used

rule 25

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

rule 364

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), 
x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x 
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In 
tegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])
 

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 = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97

method	result	size
parallelrisch	\(\frac {-3 \coth \left (d x +c \right )^{5} a^{3}-5 a^{2} \coth \left (d x +c \right )^{3} \left (a +3 b \right )+\left (-15 a^{3}-45 a^{2} b -45 a \,b^{2}\right ) \coth \left (d x +c \right )+15 d x \left (a +b \right )^{3}}{15 d}\)	\(72\)
derivativedivides	\(-\frac {\frac {a^{3}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a^{2}+3 a b +3 b^{2}\right )}{\tanh \left (d x +c \right )}+\frac {a^{2} \left (a +3 b \right )}{3 \tanh \left (d x +c \right )^{3}}+\left (-\frac {1}{2} a^{3}-\frac {3}{2} a^{2} b -\frac {3}{2} a \,b^{2}-\frac {1}{2} b^{3}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\)	\(127\)
default	\(-\frac {\frac {a^{3}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a^{2}+3 a b +3 b^{2}\right )}{\tanh \left (d x +c \right )}+\frac {a^{2} \left (a +3 b \right )}{3 \tanh \left (d x +c \right )^{3}}+\left (-\frac {1}{2} a^{3}-\frac {3}{2} a^{2} b -\frac {3}{2} a \,b^{2}-\frac {1}{2} b^{3}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\)	\(127\)
risch	\(a^{3} x +3 a^{2} b x +3 a \,b^{2} x +b^{3} x -\frac {2 a \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+90 a b \,{\mathrm e}^{8 d x +8 c}+45 b^{2} {\mathrm e}^{8 d x +8 c}-90 a^{2} {\mathrm e}^{6 d x +6 c}-270 a b \,{\mathrm e}^{6 d x +6 c}-180 b^{2} {\mathrm e}^{6 d x +6 c}+140 a^{2} {\mathrm e}^{4 d x +4 c}+330 a b \,{\mathrm e}^{4 d x +4 c}+270 b^{2} {\mathrm e}^{4 d x +4 c}-70 a^{2} {\mathrm e}^{2 d x +2 c}-210 a b \,{\mathrm e}^{2 d x +2 c}-180 b^{2} {\mathrm e}^{2 d x +2 c}+23 a^{2}+60 a b +45 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\)	\(224\)

Input:

int(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/15*(-3*coth(d*x+c)^5*a^3-5*a^2*coth(d*x+c)^3*(a+3*b)+(-15*a^3-45*a^2*b-4 
5*a*b^2)*coth(d*x+c)+15*d*x*(a+b)^3)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (70) = 140\).

Time = 0.09 (sec) , antiderivative size = 557, normalized size of antiderivative = 7.53 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {{\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{3} + 24 \, a^{2} b + 27 \, a b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 2 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{3} + 24 \, a^{2} b + 27 \, a b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{4} + 46 \, a^{3} + 120 \, a^{2} b + 90 \, a b^{2} + 30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 3 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \] Input:

integrate(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

-1/15*((23*a^3 + 60*a^2*b + 45*a*b^2)*cosh(d*x + c)^5 + 5*(23*a^3 + 60*a^2 
*b + 45*a*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 - (23*a^3 + 60*a^2*b + 45*a*b 
^2 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*sinh(d*x + c)^5 - 5*(5*a^3 + 
24*a^2*b + 27*a*b^2)*cosh(d*x + c)^3 + 5*(23*a^3 + 60*a^2*b + 45*a*b^2 + 1 
5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x - 2*(23*a^3 + 60*a^2*b + 45*a*b^2 + 
15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 
 5*(2*(23*a^3 + 60*a^2*b + 45*a*b^2)*cosh(d*x + c)^3 - 3*(5*a^3 + 24*a^2*b 
 + 27*a*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^3 + 6*a^2*b + 9*a*b^ 
2)*cosh(d*x + c) - 5*((23*a^3 + 60*a^2*b + 45*a*b^2 + 15*(a^3 + 3*a^2*b + 
3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^4 + 46*a^3 + 120*a^2*b + 90*a*b^2 + 30*( 
a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x - 3*(23*a^3 + 60*a^2*b + 45*a*b^2 + 15* 
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*si 
nh(d*x + c)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5*(d*cosh(d* 
x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:

integrate(coth(d*x+c)**6*(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. 239 vs. \(2 (70) = 140\).

Time = 0.05 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.23 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {1}{15} \, a^{3} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + a^{2} b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 3 \, a b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + b^{3} x \] Input:

integrate(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

1/15*a^3*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) - 140*e^(-4*d*x - 4*c) + 
90*e^(-6*d*x - 6*c) - 45*e^(-8*d*x - 8*c) - 23)/(d*(5*e^(-2*d*x - 2*c) - 1 
0*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x 
 - 10*c) - 1))) + a^2*b*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x 
 - 4*c) - 2)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c 
) - 1))) + 3*a*b^2*(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1))) + b^3*x
                                                                                    
                                                                                    
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (70) = 140\).

Time = 0.35 (sec) , antiderivative size = 241, normalized size of antiderivative = 3.26 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (45 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 45 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} - 270 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 180 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 330 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 270 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 210 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 180 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \] Input:

integrate(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c) - 2*(45*a^3*e^(8*d*x + 
8*c) + 90*a^2*b*e^(8*d*x + 8*c) + 45*a*b^2*e^(8*d*x + 8*c) - 90*a^3*e^(6*d 
*x + 6*c) - 270*a^2*b*e^(6*d*x + 6*c) - 180*a*b^2*e^(6*d*x + 6*c) + 140*a^ 
3*e^(4*d*x + 4*c) + 330*a^2*b*e^(4*d*x + 4*c) + 270*a*b^2*e^(4*d*x + 4*c) 
- 70*a^3*e^(2*d*x + 2*c) - 210*a^2*b*e^(2*d*x + 2*c) - 180*a*b^2*e^(2*d*x 
+ 2*c) + 23*a^3 + 60*a^2*b + 45*a*b^2)/(e^(2*d*x + 2*c) - 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 2.90 (sec) , antiderivative size = 568, normalized size of antiderivative = 7.68 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=x\,{\left (a+b\right )}^3-\frac {\frac {6\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {\frac {2\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{15\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}-\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}+\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {6\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \] Input:

int(coth(c + d*x)^6*(a + b*tanh(c + d*x)^2)^3,x)
 

Output:

x*(a + b)^3 - ((6*(a*b^2 + 2*a^2*b + a^3))/(5*d) + (6*exp(8*c + 8*d*x)*(a* 
b^2 + 2*a^2*b + a^3))/(5*d) - (24*exp(2*c + 2*d*x)*(a*b^2 + a^2*b))/(5*d) 
- (24*exp(6*c + 6*d*x)*(a*b^2 + a^2*b))/(5*d) + (4*exp(4*c + 4*d*x)*(9*a*b 
^2 + 6*a^2*b + 5*a^3))/(5*d))/(5*exp(2*c + 2*d*x) - 10*exp(4*c + 4*d*x) + 
10*exp(6*c + 6*d*x) - 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) - 1) - ((2*( 
9*a*b^2 + 6*a^2*b + 5*a^3))/(15*d) + (6*exp(4*c + 4*d*x)*(a*b^2 + 2*a^2*b 
+ a^3))/(5*d) - (12*exp(2*c + 2*d*x)*(a*b^2 + a^2*b))/(5*d))/(3*exp(2*c + 
2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) + ((6*(a*b^2 + a^2*b)) 
/(5*d) - (6*exp(6*c + 6*d*x)*(a*b^2 + 2*a^2*b + a^3))/(5*d) + (18*exp(4*c 
+ 4*d*x)*(a*b^2 + a^2*b))/(5*d) - (2*exp(2*c + 2*d*x)*(9*a*b^2 + 6*a^2*b + 
 5*a^3))/(5*d))/(6*exp(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) + ((6*(a*b^2 + a^2*b))/(5*d) - (6*exp(2*c + 2* 
d*x)*(a*b^2 + 2*a^2*b + a^3))/(5*d))/(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x 
) + 1) - (6*(a*b^2 + 2*a^2*b + a^3))/(5*d*(exp(2*c + 2*d*x) - 1))
 

Reduce [F]

\[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\int \coth \left (d x +c \right )^{6} \left (\tanh \left (d x +c \right )^{2} b +a \right )^{3}d x \] Input:

int(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

int(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^3,x)