\(\int \coth ^7(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\) [167]

Optimal result

Integrand size = 23, antiderivative size = 103 \[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {a \left (a^2+3 a b+3 b^2\right ) \coth ^2(c+d x)}{2 d}-\frac {a^2 (a+3 b) \coth ^4(c+d x)}{4 d}-\frac {a^3 \coth ^6(c+d x)}{6 d}+\frac {(a+b)^3 \log (\cosh (c+d x))}{d}+\frac {(a+b)^3 \log (\tanh (c+d x))}{d} \] Output:

-1/2*a*(a^2+3*a*b+3*b^2)*coth(d*x+c)^2/d-1/4*a^2*(a+3*b)*coth(d*x+c)^4/d-1 
/6*a^3*coth(d*x+c)^6/d+(a+b)^3*ln(cosh(d*x+c))/d+(a+b)^3*ln(tanh(d*x+c))/d
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {a (a+b)^2 \coth ^2(c+d x)+\frac {1}{2} (a+b) \left (b+a \coth ^2(c+d x)\right )^2+\frac {1}{3} \left (b+a \coth ^2(c+d x)\right )^3-2 (a+b)^3 \log (\sinh (c+d x))}{2 d} \] Input:

Integrate[Coth[c + d*x]^7*(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, 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.

\(\displaystyle \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \left (a-b \tan (i c+i d x)^2\right )^3}{\tan (i c+i d x)^7}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\tan (i c+i d x)^7}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

Int[Coth[c + d*x]^7*(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

(-(a*(a^2 + 3*a*b + 3*b^2)*Coth[c + d*x]) - (a^2*(a + 3*b)*Coth[c + d*x]^2 
)/2 - (a^3*Coth[c + d*x]^3)/3 + (a + b)^3*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 = 0.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {-12 \left (a +b \right )^{3} \ln \left (1-\tanh \left (d x +c \right )\right )+12 \left (a +b \right )^{3} \ln \left (\tanh \left (d x +c \right )\right )-2 \coth \left (d x +c \right )^{6} a^{3}-3 a^{2} \coth \left (d x +c \right )^{4} \left (a +3 b \right )+\left (-6 a^{3}-18 a^{2} b -18 a \,b^{2}\right ) \coth \left (d x +c \right )^{2}-12 d x \left (a +b \right )^{3}}{12 d}\)	\(106\)
derivativedivides	\(-\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\frac {a^{3}}{6 \tanh \left (d x +c \right )^{6}}+\frac {a \left (a^{2}+3 a b +3 b^{2}\right )}{2 \tanh \left (d x +c \right )^{2}}+\frac {a^{2} \left (a +3 b \right )}{4 \tanh \left (d x +c \right )^{4}}+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\)	\(159\)
default	\(-\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\frac {a^{3}}{6 \tanh \left (d x +c \right )^{6}}+\frac {a \left (a^{2}+3 a b +3 b^{2}\right )}{2 \tanh \left (d x +c \right )^{2}}+\frac {a^{2} \left (a +3 b \right )}{4 \tanh \left (d x +c \right )^{4}}+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\)	\(159\)
risch	\(-a^{3} x -3 a^{2} b x -3 a \,b^{2} x -b^{3} x -\frac {2 a^{3} c}{d}-\frac {6 a^{2} b c}{d}-\frac {6 a \,b^{2} c}{d}-\frac {2 b^{3} c}{d}-\frac {2 a \,{\mathrm e}^{2 d x +2 c} \left (9 a^{2} {\mathrm e}^{8 d x +8 c}+18 a b \,{\mathrm e}^{8 d x +8 c}+9 b^{2} {\mathrm e}^{8 d x +8 c}-18 a^{2} {\mathrm e}^{6 d x +6 c}-54 a b \,{\mathrm e}^{6 d x +6 c}-36 b^{2} {\mathrm e}^{6 d x +6 c}+34 a^{2} {\mathrm e}^{4 d x +4 c}+72 a b \,{\mathrm e}^{4 d x +4 c}+54 b^{2} {\mathrm e}^{4 d x +4 c}-18 a^{2} {\mathrm e}^{2 d x +2 c}-54 a b \,{\mathrm e}^{2 d x +2 c}-36 b^{2} {\mathrm e}^{2 d x +2 c}+9 a^{2}+18 a b +9 b^{2}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{6}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b}{d}+\frac {3 a \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b^{3}}{d}\)	\(353\)

Input:

int(coth(d*x+c)^7*(a+b*tanh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/12*(-12*(a+b)^3*ln(1-tanh(d*x+c))+12*(a+b)^3*ln(tanh(d*x+c))-2*coth(d*x+ 
c)^6*a^3-3*a^2*coth(d*x+c)^4*(a+3*b)+(-6*a^3-18*a^2*b-18*a*b^2)*coth(d*x+c 
)^2-12*d*x*(a+b)^3)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4305 vs. \(2 (97) = 194\).

Time = 0.17 (sec) , antiderivative size = 4305, normalized size of antiderivative = 41.80 \[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \] Input:

integrate(coth(d*x+c)^7*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:

integrate(coth(d*x+c)**7*(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. 420 vs. \(2 (97) = 194\).

Time = 0.05 (sec) , antiderivative size = 420, normalized size of antiderivative = 4.08 \[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {1}{3} \, a^{3} {\left (3 \, x + \frac {3 \, c}{d} + \frac {3 \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {3 \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, {\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 18 \, e^{\left (-4 \, d x - 4 \, c\right )} + 34 \, e^{\left (-6 \, d x - 6 \, c\right )} - 18 \, e^{\left (-8 \, d x - 8 \, c\right )} + 9 \, e^{\left (-10 \, d x - 10 \, c\right )}\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} - 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} - 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} - e^{\left (-12 \, d x - 12 \, c\right )} - 1\right )}}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 3 \, a b^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + \frac {b^{3} \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}{d} \] Input:

integrate(coth(d*x+c)^7*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/3*a^3*(3*x + 3*c/d + 3*log(e^(-d*x - c) + 1)/d + 3*log(e^(-d*x - c) - 1) 
/d + 2*(9*e^(-2*d*x - 2*c) - 18*e^(-4*d*x - 4*c) + 34*e^(-6*d*x - 6*c) - 1 
8*e^(-8*d*x - 8*c) + 9*e^(-10*d*x - 10*c))/(d*(6*e^(-2*d*x - 2*c) - 15*e^( 
-4*d*x - 4*c) + 20*e^(-6*d*x - 6*c) - 15*e^(-8*d*x - 8*c) + 6*e^(-10*d*x - 
 10*c) - e^(-12*d*x - 12*c) - 1))) + 3*a^2*b*(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*b^2*(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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (97) = 194\).

Time = 0.43 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.11 \[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac {2 \, {\left (9 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} e^{\left (10 \, d x + 10 \, c\right )} - 18 \, {\left (a^{3} + 3 \, a^{2} b + 2 \, a b^{2}\right )} e^{\left (8 \, d x + 8 \, c\right )} + 2 \, {\left (17 \, a^{3} + 36 \, a^{2} b + 27 \, a b^{2}\right )} e^{\left (6 \, d x + 6 \, c\right )} - 18 \, {\left (a^{3} + 3 \, a^{2} b + 2 \, a b^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{6}}}{3 \, d} \] Input:

integrate(coth(d*x+c)^7*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

-1/3*(3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c) - 3*(a^3 + 3*a^2*b + 3*a 
*b^2 + b^3)*log(abs(e^(2*d*x + 2*c) - 1)) + 2*(9*(a^3 + 2*a^2*b + a*b^2)*e 
^(10*d*x + 10*c) - 18*(a^3 + 3*a^2*b + 2*a*b^2)*e^(8*d*x + 8*c) + 2*(17*a^ 
3 + 36*a^2*b + 27*a*b^2)*e^(6*d*x + 6*c) - 18*(a^3 + 3*a^2*b + 2*a*b^2)*e^ 
(4*d*x + 4*c) + 9*(a^3 + 2*a^2*b + a*b^2)*e^(2*d*x + 2*c))/(e^(2*d*x + 2*c 
) - 1)^6)/d
 

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.69 \[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{d}-\frac {4\,\left (11\,a^3+3\,b\,a^2\right )}{d\,\left (6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {32\,a^3}{3\,d\,\left (15\,{\mathrm {e}}^{4\,c+4\,d\,x}-6\,{\mathrm {e}}^{2\,c+2\,d\,x}-20\,{\mathrm {e}}^{6\,c+6\,d\,x}+15\,{\mathrm {e}}^{8\,c+8\,d\,x}-6\,{\mathrm {e}}^{10\,c+10\,d\,x}+{\mathrm {e}}^{12\,c+12\,d\,x}+1\right )}-\frac {6\,\left (3\,a^3+4\,a^2\,b+a\,b^2\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {6\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {8\,\left (13\,a^3+9\,b\,a^2\right )}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {32\,a^3}{d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1\right )}-x\,{\left (a+b\right )}^3 \] Input:

int(coth(c + d*x)^7*(a + b*tanh(c + d*x)^2)^3,x)
 

Output:

(log(exp(2*c)*exp(2*d*x) - 1)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/d - (4*(3*a 
^2*b + 11*a^3))/(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)) - (32*a^3)/(3*d*(15*exp(4*c + 4*d*x) - 6*e 
xp(2*c + 2*d*x) - 20*exp(6*c + 6*d*x) + 15*exp(8*c + 8*d*x) - 6*exp(10*c + 
 10*d*x) + exp(12*c + 12*d*x) + 1)) - (6*(a*b^2 + 4*a^2*b + 3*a^3))/(d*(ex 
p(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - (6*(a*b^2 + 2*a^2*b + a^3))/(d 
*(exp(2*c + 2*d*x) - 1)) - (8*(9*a^2*b + 13*a^3))/(3*d*(3*exp(2*c + 2*d*x) 
 - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1)) - (32*a^3)/(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)) - x*(a + b)^3
 

Reduce [F]

\[ \int \coth ^7(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\int \coth \left (d x +c \right )^{7} \left (\tanh \left (d x +c \right )^{2} b +a \right )^{3}d x \] Input:

int(coth(d*x+c)^7*(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

int(coth(d*x+c)^7*(a+b*tanh(d*x+c)^2)^3,x)