\(\int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx\) [110]

Optimal result

Integrand size = 20, antiderivative size = 207 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=-\frac {2 e^{\frac {5 (a-d)}{3}+\frac {8}{3} (d+b x)}}{b \left (1-e^{2 (d+b x)}\right )^2}+\frac {8 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{3 b \left (1-e^{2 (d+b x)}\right )}-\frac {8 e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1+2 e^{\frac {2}{3} (d+b x)}}{\sqrt {3}}\right )}{3 \sqrt {3} b}+\frac {8 e^{\frac {5 (a-d)}{3}} \log \left (1-e^{\frac {2}{3} (d+b x)}\right )}{9 b}-\frac {4 e^{\frac {5 (a-d)}{3}} \log \left (1+e^{\frac {2}{3} (d+b x)}+e^{\frac {4}{3} (d+b x)}\right )}{9 b} \] Output:

-2*exp(5/3*a+d+8/3*b*x)/b/(1-exp(2*b*x+2*d))^2+8/3*exp(5/3*a-d+2/3*b*x)/b/ 
(1-exp(2*b*x+2*d))-8/9*3^(1/2)*exp(5/3*a-5/3*d)*arctan(1/3*(1+2*exp(2/3*b* 
x+2/3*d))*3^(1/2))/b+8/9*exp(5/3*a-5/3*d)*ln(1-exp(2/3*b*x+2/3*d))/b-4/9*e 
xp(5/3*a-5/3*d)*ln(1+exp(2/3*b*x+2/3*d)+exp(4/3*b*x+4/3*d))/b
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.14 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=\frac {2 e^{5 a/3} (\cosh (d)-\sinh (d))^2 \left (4 \text {RootSum}\left [\cosh \left (\frac {d}{2}\right )-\sinh \left (\frac {d}{2}\right )+\cosh \left (\frac {d}{2}\right ) \text {$\#$1}^3+\sinh \left (\frac {d}{2}\right ) \text {$\#$1}^3\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ]-4 \text {RootSum}\left [-\cosh \left (\frac {d}{2}\right )+\sinh \left (\frac {d}{2}\right )+\cosh \left (\frac {d}{2}\right ) \text {$\#$1}^3+\sinh \left (\frac {d}{2}\right ) \text {$\#$1}^3\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ]-\frac {27 e^{\frac {2 b x}{3}} (\cosh (d)-\sinh (d))}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^2}-\frac {63 e^{\frac {2 b x}{3}}}{\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)}\right )}{27 b} \] Input:

Integrate[E^((5*(a + b*x))/3)*Csch[d + b*x]^3,x]
 

Output:

(2*E^((5*a)/3)*(Cosh[d] - Sinh[d])^2*(4*RootSum[Cosh[d/2] - Sinh[d/2] + Co 
sh[d/2]*#1^3 + Sinh[d/2]*#1^3 & , (b*x - 3*Log[E^((b*x)/3) - #1])/#1 & ] - 
 4*RootSum[-Cosh[d/2] + Sinh[d/2] + Cosh[d/2]*#1^3 + Sinh[d/2]*#1^3 & , (b 
*x - 3*Log[E^((b*x)/3) - #1])/#1 & ] - (27*E^((2*b*x)/3)*(Cosh[d] - Sinh[d 
]))/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^2 - (63*E^((2*b*x 
)/3))/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])))/(27*b)
 

Rubi [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.65, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {2720, 27, 807, 817, 817, 750, 16, 1142, 1083, 217, 1103}

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.

\(\displaystyle \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(b x+d) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {3 \int -\frac {8 e^{\frac {5 a}{3}+\frac {13 b x}{3}}}{\left (1-e^{2 b x}\right )^3}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {24 e^{5 a/3} \int \frac {e^{\frac {13 b x}{3}}}{\left (1-e^{2 b x}\right )^3}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {12 e^{5 a/3} \int \frac {e^{2 b x}}{\left (1-e^{b x}\right )^3}de^{\frac {2 b x}{3}}}{b}\)

\(\Big \downarrow \) 817

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \int \frac {e^{b x}}{\left (1-e^{b x}\right )^2}de^{\frac {2 b x}{3}}\right )}{b}\)

\(\Big \downarrow \) 817

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \left (\frac {e^{\frac {2 b x}{3}}}{3 \left (1-e^{b x}\right )}-\frac {1}{3} \int \frac {1}{1-e^{b x}}de^{\frac {2 b x}{3}}\right )\right )}{b}\)

\(\Big \downarrow \) 750

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \left (\frac {1}{3} \left (-\frac {1}{3} \int \frac {1}{1-e^{\frac {2 b x}{3}}}de^{\frac {2 b x}{3}}-\frac {1}{3} \int \frac {2+e^{\frac {2 b x}{3}}}{1+2 e^{\frac {2 b x}{3}}}de^{\frac {2 b x}{3}}\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (1-e^{b x}\right )}\right )\right )}{b}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \log \left (1-e^{\frac {2 b x}{3}}\right )-\frac {1}{3} \int \frac {2+e^{\frac {2 b x}{3}}}{1+2 e^{\frac {2 b x}{3}}}de^{\frac {2 b x}{3}}\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (1-e^{b x}\right )}\right )\right )}{b}\)

\(\Big \downarrow \) 1142

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (-\frac {1}{2} \int 1de^{\frac {2 b x}{3}}-\frac {3}{2} \int \frac {1}{1+2 e^{\frac {2 b x}{3}}}de^{\frac {2 b x}{3}}\right )+\frac {1}{3} \log \left (1-e^{\frac {2 b x}{3}}\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (1-e^{b x}\right )}\right )\right )}{b}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (3 \int \frac {1}{-4-2 e^{\frac {2 b x}{3}}}d\left (1+2 e^{\frac {2 b x}{3}}\right )-\frac {1}{2} \int 1de^{\frac {2 b x}{3}}\right )+\frac {1}{3} \log \left (1-e^{\frac {2 b x}{3}}\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (1-e^{b x}\right )}\right )\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (-\frac {1}{2} \int 1de^{\frac {2 b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {2 b x}{3}}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (1-e^{\frac {2 b x}{3}}\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (1-e^{b x}\right )}\right )\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {12 e^{5 a/3} \left (\frac {e^{\frac {4 b x}{3}}}{6 \left (1-e^{b x}\right )^2}-\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (-\sqrt {3} \arctan \left (\frac {2 e^{\frac {2 b x}{3}}+1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (2 e^{\frac {2 b x}{3}}+1\right )\right )+\frac {1}{3} \log \left (1-e^{\frac {2 b x}{3}}\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (1-e^{b x}\right )}\right )\right )}{b}\)

Input:

Int[E^((5*(a + b*x))/3)*Csch[d + b*x]^3,x]
 

Output:

(-12*E^((5*a)/3)*(E^((4*b*x)/3)/(6*(1 - E^(b*x))^2) - (2*(E^((2*b*x)/3)/(3 
*(1 - E^(b*x))) + (Log[1 - E^((2*b*x)/3)]/3 + (-(Sqrt[3]*ArcTan[(1 + 2*E^( 
(2*b*x)/3))/Sqrt[3]]) - Log[1 + 2*E^((2*b*x)/3)]/2)/3)/3))/3))/b
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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 217

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

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*Rt[a, 3] - 
 Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; 
 FreeQ[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 817

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

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

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

Result contains complex when optimal does not.

Time = 0.54 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {2 \left (7 \,{\mathrm e}^{2 b x +2 d}-4\right ) {\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}}}{3 \left ({\mathrm e}^{2 b x +2 d}-1\right )^{2} b}+\frac {8 \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-1\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{9 b}-\frac {4 \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{9 b}+\frac {4 i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{9 b}-\frac {4 \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{9 b}-\frac {4 i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{9 b}\)	\(202\)

Input:

int(exp(5/3*b*x+5/3*a)*csch(b*x+d)^3,x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-2/3/(exp(2*b*x+2*d)-1)^2/b*(7*exp(2*b*x+2*d)-4)*exp(5/3*a-d+2/3*b*x)+8/9* 
ln(exp(2/3*b*x+2/3*d)-1)/b*exp(5/3*a-5/3*d)-4/9*ln(exp(2/3*b*x+2/3*d)+1/2- 
1/2*I*3^(1/2))/b*exp(5/3*a-5/3*d)+4/9*I*ln(exp(2/3*b*x+2/3*d)+1/2-1/2*I*3^ 
(1/2))/b*exp(5/3*a-5/3*d)*3^(1/2)-4/9*ln(exp(2/3*b*x+2/3*d)+1/2+1/2*I*3^(1 
/2))/b*exp(5/3*a-5/3*d)-4/9*I*ln(exp(2/3*b*x+2/3*d)+1/2+1/2*I*3^(1/2))/b*e 
xp(5/3*a-5/3*d)*3^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3987 vs. \(2 (155) = 310\).

Time = 0.13 (sec) , antiderivative size = 3987, normalized size of antiderivative = 19.26 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=\text {Too large to display} \] Input:

integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \operatorname {csch}^{3}{\left (b x + d \right )}\, dx \] Input:

integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)**3,x)
 

Output:

exp(5*a/3)*Integral(exp(5*b*x/3)*csch(b*x + d)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.22 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=-\frac {8 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{9 \, b} + \frac {8 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{9 \, b} - \frac {4 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{9 \, b} + \frac {8 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}{9 \, b} + \frac {8 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}{9 \, b} - \frac {4 \, e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{9 \, b} + \frac {2 \, {\left (7 \, e^{\left (-\frac {4}{3} \, b x - \frac {4}{3} \, d\right )} - 4 \, e^{\left (-\frac {10}{3} \, b x - \frac {10}{3} \, d\right )}\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{3 \, b {\left (2 \, e^{\left (-2 \, b x - 2 \, d\right )} - e^{\left (-4 \, b x - 4 \, d\right )} - 1\right )}} \] Input:

integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)^3,x, algorithm="maxima")
 

Output:

-8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(-1/3*b*x - 1/3*d) + 1))*e^(5/3*a - 5 
/3*d)/b + 8/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(-1/3*b*x - 1/3*d) - 1))*e^( 
5/3*a - 5/3*d)/b - 4/9*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) + e^(-2/ 
3*b*x - 2/3*d) + 1)/b + 8/9*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) + 1 
)/b + 8/9*e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) - 1)/b - 4/9*e^(5/3*a 
 - 5/3*d)*log(-e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x - 2/3*d) + 1)/b + 2/3*(7 
*e^(-4/3*b*x - 4/3*d) - 4*e^(-10/3*b*x - 10/3*d))*e^(5/3*a - 5/3*d)/(b*(2* 
e^(-2*b*x - 2*d) - e^(-4*b*x - 4*d) - 1))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.62 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=-\frac {2 \, {\left (4 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right )} e^{\left (\frac {2}{3} \, d\right )}\right ) e^{\left (-\frac {14}{3} \, d\right )} + 2 \, e^{\left (-\frac {14}{3} \, d\right )} \log \left (e^{\left (\frac {4}{3} \, b x\right )} + e^{\left (\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + e^{\left (-\frac {4}{3} \, d\right )}\right ) - 4 \, e^{\left (-\frac {14}{3} \, d\right )} \log \left ({\left | e^{\left (\frac {2}{3} \, b x\right )} - e^{\left (-\frac {2}{3} \, d\right )} \right |}\right ) + \frac {3 \, {\left (7 \, e^{\left (\frac {8}{3} \, b x + 2 \, d\right )} - 4 \, e^{\left (\frac {2}{3} \, b x\right )}\right )} e^{\left (-4 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{2}}\right )} e^{\left (\frac {5}{3} \, a + 3 \, d\right )}}{9 \, b} \] Input:

integrate(exp(5/3*b*x+5/3*a)*csch(b*x+d)^3,x, algorithm="giac")
 

Output:

-2/9*(4*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(2/3*b*x) + e^(-2/3*d))*e^(2/3*d)) 
*e^(-14/3*d) + 2*e^(-14/3*d)*log(e^(4/3*b*x) + e^(2/3*b*x - 2/3*d) + e^(-4 
/3*d)) - 4*e^(-14/3*d)*log(abs(e^(2/3*b*x) - e^(-2/3*d))) + 3*(7*e^(8/3*b* 
x + 2*d) - 4*e^(2/3*b*x))*e^(-4*d)/(e^(2*b*x + 2*d) - 1)^2)*e^(5/3*a + 3*d 
)/b
 

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.20 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=\frac {8\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (\frac {16\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{9}-\frac {16\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{9}\right )}{9\,b}-\frac {8\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}}{3\,b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}-1\right )}-\frac {2\,{\mathrm {e}}^{\frac {5\,a}{3}+d+\frac {8\,b\,x}{3}}}{b\,\left ({\mathrm {e}}^{4\,d+4\,b\,x}-2\,{\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}+\frac {8\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (\frac {16\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{9}-\frac {16\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{9}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,b}-\frac {8\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-\frac {16\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{9}-\frac {16\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{9}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,b} \] Input:

int(exp((5*a)/3 + (5*b*x)/3)/sinh(d + b*x)^3,x)
 

Output:

(8*exp(5*a - 5*d)^(1/3)*log((16*(exp(5*a)*exp(-5*d))^(1/3))/9 - (16*exp((5 
*a)/3)*exp((2*d)/3)*exp(-(5*d)/3)*exp((2*b*x)/3))/9))/(9*b) - (8*exp((5*a) 
/3 - d + (2*b*x)/3))/(3*b*(exp(2*d + 2*b*x) - 1)) - (2*exp((5*a)/3 + d + ( 
8*b*x)/3))/(b*(exp(4*d + 4*b*x) - 2*exp(2*d + 2*b*x) + 1)) + (8*exp(5*a - 
5*d)^(1/3)*log((16*((3^(1/2)*1i)/2 - 1/2)*(exp(5*a)*exp(-5*d))^(1/3))/9 - 
(16*exp((5*a)/3)*exp((2*d)/3)*exp(-(5*d)/3)*exp((2*b*x)/3))/9)*((3^(1/2)*1 
i)/2 - 1/2))/(9*b) - (8*exp(5*a - 5*d)^(1/3)*log(- (16*((3^(1/2)*1i)/2 + 1 
/2)*(exp(5*a)*exp(-5*d))^(1/3))/9 - (16*exp((5*a)/3)*exp((2*d)/3)*exp(-(5* 
d)/3)*exp((2*b*x)/3))/9)*((3^(1/2)*1i)/2 + 1/2))/(9*b)
 

Reduce [F]

\[ \int e^{\frac {5}{3} (a+b x)} \text {csch}^3(d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \mathrm {csch}\left (b x +d \right )^{3}d x \] Input:

int(exp(5/3*b*x+5/3*a)*csch(b*x+d)^3,x)
 

Output:

int(e**((5*a + 5*b*x)/3)*csch(b*x + d)**3,x)