\(\int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx\) [107]

Optimal result

Integrand size = 18, antiderivative size = 163 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=-\frac {4 e^{2 a+3 d+5 b x}}{b \left (1-e^{2 d+2 b x}\right )^4}+\frac {10 e^{2 a+d+3 b x}}{3 b \left (1-e^{2 d+2 b x}\right )^3}-\frac {5 e^{2 a-d+b x}}{2 b \left (1-e^{2 d+2 b x}\right )^2}+\frac {5 e^{2 a-d+b x}}{4 b \left (1-e^{2 d+2 b x}\right )}+\frac {5 e^{2 a-2 d} \text {arctanh}\left (e^{d+b x}\right )}{4 b} \] Output:

-4*exp(5*b*x+2*a+3*d)/b/(1-exp(2*b*x+2*d))^4+10/3*exp(3*b*x+2*a+d)/b/(1-ex 
p(2*b*x+2*d))^3-5/2*exp(b*x+2*a-d)/b/(1-exp(2*b*x+2*d))^2+5/4*exp(b*x+2*a- 
d)/b/(1-exp(2*b*x+2*d))+5/4*exp(2*a-2*d)*arctanh(exp(b*x+d))/b
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.97 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=\frac {e^{2 a} \left (15 \cosh (2 d) \log \left (\left (1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (-1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )-15 \cosh (2 d) \log \left (\left (-1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )-\frac {96 e^{b x} (\cosh (d)-\sinh (d))^5}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^4}-\frac {272 e^{b x} (\cosh (d)-\sinh (d))^4}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^3}-\frac {236 e^{b x} (\cosh (d)-\sinh (d))^3}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^2}-\frac {30 e^{b x} (\cosh (d)-\sinh (d))^2}{\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)}-15 \log \left (\left (1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (-1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right ) \sinh (2 d)+15 \log \left (\left (-1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right ) \sinh (2 d)\right )}{24 b} \] Input:

Integrate[E^(2*(a + b*x))*Csch[d + b*x]^5,x]
 

Output:

(E^(2*a)*(15*Cosh[2*d]*Log[(1 + E^(b*x))*Cosh[d/2] + (-1 + E^(b*x))*Sinh[d 
/2]] - 15*Cosh[2*d]*Log[(-1 + E^(b*x))*Cosh[d/2] + (1 + E^(b*x))*Sinh[d/2] 
] - (96*E^(b*x)*(Cosh[d] - Sinh[d])^5)/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^ 
(2*b*x))*Sinh[d])^4 - (272*E^(b*x)*(Cosh[d] - Sinh[d])^4)/((-1 + E^(2*b*x) 
)*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^3 - (236*E^(b*x)*(Cosh[d] - Sinh[d])^ 
3)/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^2 - (30*E^(b*x)*(C 
osh[d] - Sinh[d])^2)/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d]) 
- 15*Log[(1 + E^(b*x))*Cosh[d/2] + (-1 + E^(b*x))*Sinh[d/2]]*Sinh[2*d] + 1 
5*Log[(-1 + E^(b*x))*Cosh[d/2] + (1 + E^(b*x))*Sinh[d/2]]*Sinh[2*d]))/(24* 
b)
 

Rubi [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2720, 27, 252, 252, 252, 215, 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.

\(\displaystyle \int e^{2 (a+b x)} \text {csch}^5(b x+d) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int -\frac {32 e^{2 a+6 b x}}{\left (1-e^{2 b x}\right )^5}de^{b x}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {32 e^{2 a} \int \frac {e^{6 b x}}{\left (1-e^{2 b x}\right )^5}de^{b x}}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {32 e^{2 a} \left (\frac {e^{5 b x}}{8 \left (1-e^{2 b x}\right )^4}-\frac {5}{8} \int \frac {e^{4 b x}}{\left (1-e^{2 b x}\right )^4}de^{b x}\right )}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {32 e^{2 a} \left (\frac {e^{5 b x}}{8 \left (1-e^{2 b x}\right )^4}-\frac {5}{8} \left (\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}-\frac {1}{2} \int \frac {e^{2 b x}}{\left (1-e^{2 b x}\right )^3}de^{b x}\right )\right )}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {32 e^{2 a} \left (\frac {e^{5 b x}}{8 \left (1-e^{2 b x}\right )^4}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{\left (1-e^{2 b x}\right )^2}de^{b x}-\frac {e^{b x}}{4 \left (1-e^{2 b x}\right )^2}\right )+\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}\right )\right )}{b}\)

\(\Big \downarrow \) 215

\(\displaystyle -\frac {32 e^{2 a} \left (\frac {e^{5 b x}}{8 \left (1-e^{2 b x}\right )^4}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1-e^{2 b x}}de^{b x}+\frac {e^{b x}}{2 \left (1-e^{2 b x}\right )}\right )-\frac {e^{b x}}{4 \left (1-e^{2 b x}\right )^2}\right )+\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}\right )\right )}{b}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {32 e^{2 a} \left (\frac {e^{5 b x}}{8 \left (1-e^{2 b x}\right )^4}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \text {arctanh}\left (e^{b x}\right )+\frac {e^{b x}}{2 \left (1-e^{2 b x}\right )}\right )-\frac {e^{b x}}{4 \left (1-e^{2 b x}\right )^2}\right )+\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}\right )\right )}{b}\)

Input:

Int[E^(2*(a + b*x))*Csch[d + b*x]^5,x]
 

Output:

(-32*E^(2*a)*(E^(5*b*x)/(8*(1 - E^(2*b*x))^4) - (5*(E^(3*b*x)/(6*(1 - E^(2 
*b*x))^3) + (-1/4*E^(b*x)/(1 - E^(2*b*x))^2 + (E^(b*x)/(2*(1 - E^(2*b*x))) 
 + ArcTanh[E^(b*x)]/2)/4)/2))/8))/b
 

Defintions of rubi rules used

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 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\left (15 \,{\mathrm e}^{6 b x +6 a +6 d}+73 \,{\mathrm e}^{4 b x +6 a +4 d}-55 \,{\mathrm e}^{2 b x +6 a +2 d}+15 \,{\mathrm e}^{6 a}\right ) {\mathrm e}^{b x +4 a -d}}{12 \left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{4} b}-\frac {5 \ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{8 b}+\frac {5 \ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{8 b}\)	\(144\)

Input:

int(exp(2*b*x+2*a)*csch(b*x+d)^5,x,method=_RETURNVERBOSE)
 

Output:

-1/12/(-exp(2*b*x+2*a+2*d)+exp(2*a))^4/b*(15*exp(6*b*x+6*a+6*d)+73*exp(4*b 
*x+6*a+4*d)-55*exp(2*b*x+6*a+2*d)+15*exp(6*a))*exp(b*x+4*a-d)-5/8*ln(exp(b 
*x+a)-exp(a-d))/b*exp(2*a-2*d)+5/8*ln(exp(b*x+a)+exp(a-d))/b*exp(2*a-2*d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2509 vs. \(2 (137) = 274\).

Time = 0.11 (sec) , antiderivative size = 2509, normalized size of antiderivative = 15.39 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=\text {Too large to display} \] Input:

integrate(exp(2*b*x+2*a)*csch(b*x+d)^5,x, algorithm="fricas")
 

Output:

-1/24*(30*cosh(b*x + d)^7*cosh(-2*a + 2*d) + 30*(cosh(-2*a + 2*d) - sinh(- 
2*a + 2*d))*sinh(b*x + d)^7 + 210*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b 
*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^6 + 146*cosh(b*x + d)^5*cosh(-2*a 
+ 2*d) + 2*(315*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (315*cosh(b*x + d)^2 + 
73)*sinh(-2*a + 2*d) + 73*cosh(-2*a + 2*d))*sinh(b*x + d)^5 + 10*(105*cosh 
(b*x + d)^3*cosh(-2*a + 2*d) + 73*cosh(b*x + d)*cosh(-2*a + 2*d) - (105*co 
sh(b*x + d)^3 + 73*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^4 - 110* 
cosh(b*x + d)^3*cosh(-2*a + 2*d) + 10*(105*cosh(b*x + d)^4*cosh(-2*a + 2*d 
) + 146*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (105*cosh(b*x + d)^4 + 146*cosh 
(b*x + d)^2 - 11)*sinh(-2*a + 2*d) - 11*cosh(-2*a + 2*d))*sinh(b*x + d)^3 
+ 10*(63*cosh(b*x + d)^5*cosh(-2*a + 2*d) + 146*cosh(b*x + d)^3*cosh(-2*a 
+ 2*d) - 33*cosh(b*x + d)*cosh(-2*a + 2*d) - (63*cosh(b*x + d)^5 + 146*cos 
h(b*x + d)^3 - 33*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^2 + 30*co 
sh(b*x + d)*cosh(-2*a + 2*d) - 15*(cosh(b*x + d)^8*cosh(-2*a + 2*d) + (cos 
h(-2*a + 2*d) - sinh(-2*a + 2*d))*sinh(b*x + d)^8 + 8*(cosh(b*x + d)*cosh( 
-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^7 - 4*cosh(b*x 
 + d)^6*cosh(-2*a + 2*d) + 4*(7*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (7*cosh 
(b*x + d)^2 - 1)*sinh(-2*a + 2*d) - cosh(-2*a + 2*d))*sinh(b*x + d)^6 + 8* 
(7*cosh(b*x + d)^3*cosh(-2*a + 2*d) - 3*cosh(b*x + d)*cosh(-2*a + 2*d) - ( 
7*cosh(b*x + d)^3 - 3*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^5 ...
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=e^{2 a} \int e^{2 b x} \operatorname {csch}^{5}{\left (b x + d \right )}\, dx \] Input:

integrate(exp(2*b*x+2*a)*csch(b*x+d)**5,x)
 

Output:

exp(2*a)*Integral(exp(2*b*x)*csch(b*x + d)**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.96 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=\frac {5 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} + 1\right )}{8 \, b} - \frac {5 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} - 1\right )}{8 \, b} + \frac {{\left (15 \, e^{\left (-b x - d\right )} + 73 \, e^{\left (-3 \, b x - 3 \, d\right )} - 55 \, e^{\left (-5 \, b x - 5 \, d\right )} + 15 \, e^{\left (-7 \, b x - 7 \, d\right )}\right )} e^{\left (2 \, a - 2 \, d\right )}}{12 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, d\right )} - 6 \, e^{\left (-4 \, b x - 4 \, d\right )} + 4 \, e^{\left (-6 \, b x - 6 \, d\right )} - e^{\left (-8 \, b x - 8 \, d\right )} - 1\right )}} \] Input:

integrate(exp(2*b*x+2*a)*csch(b*x+d)^5,x, algorithm="maxima")
 

Output:

5/8*e^(2*a - 2*d)*log(e^(-b*x - d) + 1)/b - 5/8*e^(2*a - 2*d)*log(e^(-b*x 
- d) - 1)/b + 1/12*(15*e^(-b*x - d) + 73*e^(-3*b*x - 3*d) - 55*e^(-5*b*x - 
 5*d) + 15*e^(-7*b*x - 7*d))*e^(2*a - 2*d)/(b*(4*e^(-2*b*x - 2*d) - 6*e^(- 
4*b*x - 4*d) + 4*e^(-6*b*x - 6*d) - e^(-8*b*x - 8*d) - 1))
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.63 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=\frac {{\left (15 \, e^{\left (-2 \, d\right )} \log \left (e^{\left (b x + d\right )} + 1\right ) - 15 \, e^{\left (-2 \, d\right )} \log \left ({\left | e^{\left (b x + d\right )} - 1 \right |}\right ) - \frac {2 \, {\left (15 \, e^{\left (7 \, b x + 7 \, d\right )} + 73 \, e^{\left (5 \, b x + 5 \, d\right )} - 55 \, e^{\left (3 \, b x + 3 \, d\right )} + 15 \, e^{\left (b x + d\right )}\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{4}}\right )} e^{\left (2 \, a\right )}}{24 \, b} \] Input:

integrate(exp(2*b*x+2*a)*csch(b*x+d)^5,x, algorithm="giac")
 

Output:

1/24*(15*e^(-2*d)*log(e^(b*x + d) + 1) - 15*e^(-2*d)*log(abs(e^(b*x + d) - 
 1)) - 2*(15*e^(7*b*x + 7*d) + 73*e^(5*b*x + 5*d) - 55*e^(3*b*x + 3*d) + 1 
5*e^(b*x + d))*e^(-2*d)/(e^(2*b*x + 2*d) - 1)^4)*e^(2*a)/b
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.45 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=\frac {5\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {-b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )}{4\,\sqrt {-b^2}}-\frac {5\,{\mathrm {e}}^{2\,a-d+b\,x}}{4\,b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}-1\right )}-\frac {5\,{\mathrm {e}}^{2\,a-d+b\,x}}{2\,b\,\left ({\mathrm {e}}^{4\,d+4\,b\,x}-2\,{\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}-\frac {10\,{\mathrm {e}}^{2\,a+d+3\,b\,x}}{3\,b\,\left (3\,{\mathrm {e}}^{2\,d+2\,b\,x}-3\,{\mathrm {e}}^{4\,d+4\,b\,x}+{\mathrm {e}}^{6\,d+6\,b\,x}-1\right )}-\frac {4\,{\mathrm {e}}^{2\,a+3\,d+5\,b\,x}}{b\,\left (6\,{\mathrm {e}}^{4\,d+4\,b\,x}-4\,{\mathrm {e}}^{2\,d+2\,b\,x}-4\,{\mathrm {e}}^{6\,d+6\,b\,x}+{\mathrm {e}}^{8\,d+8\,b\,x}+1\right )} \] Input:

int(exp(2*a + 2*b*x)/sinh(d + b*x)^5,x)
 

Output:

(5*exp(4*a - 4*d)^(1/2)*atan((exp(2*a)*exp(-d)*exp(b*x)*(-b^2)^(1/2))/(b*( 
exp(4*a)*exp(-4*d))^(1/2))))/(4*(-b^2)^(1/2)) - (5*exp(2*a - d + b*x))/(4* 
b*(exp(2*d + 2*b*x) - 1)) - (5*exp(2*a - d + b*x))/(2*b*(exp(4*d + 4*b*x) 
- 2*exp(2*d + 2*b*x) + 1)) - (10*exp(2*a + d + 3*b*x))/(3*b*(3*exp(2*d + 2 
*b*x) - 3*exp(4*d + 4*b*x) + exp(6*d + 6*b*x) - 1)) - (4*exp(2*a + 3*d + 5 
*b*x))/(b*(6*exp(4*d + 4*b*x) - 4*exp(2*d + 2*b*x) - 4*exp(6*d + 6*b*x) + 
exp(8*d + 8*b*x) + 1))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.92 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \, dx=\frac {e^{2 a} \left (-15 e^{8 b x +8 d} \mathrm {log}\left (e^{b x +d}-1\right )+15 e^{8 b x +8 d} \mathrm {log}\left (e^{b x +d}+1\right )-30 e^{7 b x +7 d}+60 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}-1\right )-60 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}+1\right )-146 e^{5 b x +5 d}-90 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )+90 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )+110 e^{3 b x +3 d}+60 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )-60 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )-30 e^{b x +d}-15 \,\mathrm {log}\left (e^{b x +d}-1\right )+15 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{24 e^{2 d} b \left (e^{8 b x +8 d}-4 e^{6 b x +6 d}+6 e^{4 b x +4 d}-4 e^{2 b x +2 d}+1\right )} \] Input:

int(exp(2*b*x+2*a)*csch(b*x+d)^5,x)
 

Output:

(e**(2*a)*( - 15*e**(8*b*x + 8*d)*log(e**(b*x + d) - 1) + 15*e**(8*b*x + 8 
*d)*log(e**(b*x + d) + 1) - 30*e**(7*b*x + 7*d) + 60*e**(6*b*x + 6*d)*log( 
e**(b*x + d) - 1) - 60*e**(6*b*x + 6*d)*log(e**(b*x + d) + 1) - 146*e**(5* 
b*x + 5*d) - 90*e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) + 90*e**(4*b*x + 4* 
d)*log(e**(b*x + d) + 1) + 110*e**(3*b*x + 3*d) + 60*e**(2*b*x + 2*d)*log( 
e**(b*x + d) - 1) - 60*e**(2*b*x + 2*d)*log(e**(b*x + d) + 1) - 30*e**(b*x 
 + d) - 15*log(e**(b*x + d) - 1) + 15*log(e**(b*x + d) + 1)))/(24*e**(2*d) 
*b*(e**(8*b*x + 8*d) - 4*e**(6*b*x + 6*d) + 6*e**(4*b*x + 4*d) - 4*e**(2*b 
*x + 2*d) + 1))