\(\int e^{\frac {5}{3} (a+b x)} \text {csch}(d+b x) \, dx\) [108]

Optimal result

Integrand size = 18, antiderivative size = 144 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}(d+b x) \, dx=\frac {3 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{b}-\frac {\sqrt {3} e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1+2 e^{\frac {2}{3} (d+b x)}}{\sqrt {3}}\right )}{b}+\frac {e^{\frac {5 (a-d)}{3}} \log \left (1-e^{\frac {2}{3} (d+b x)}\right )}{b}-\frac {e^{\frac {5 (a-d)}{3}} \log \left (1+e^{\frac {2}{3} (d+b x)}+e^{\frac {4}{3} (d+b x)}\right )}{2 b} \] Output:

3*exp(5/3*a-d+2/3*b*x)/b-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+exp(5/3*a-5/3*d)*ln(1-exp(2/3*b*x+2/3*d))/b-1/2*exp 
(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 = 0.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.28 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}(d+b x) \, dx=\frac {e^{5 a/3} \left (-\left (\left (-9 e^{\frac {2 b x}{3}}+\text {RootSum}\left [-\cosh \left (\frac {d}{2}\right )+\sinh \left (\frac {d}{2}\right )+\cosh \left (\frac {d}{2}\right ) \text {$\#$1}^3+\sinh \left (\frac {d}{2}\right ) \text {$\#$1}^3\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ] (\cosh (d)-\sinh (d))\right ) (\cosh (d)-\sinh (d))\right )+\text {RootSum}\left [\cosh \left (\frac {d}{2}\right )-\sinh \left (\frac {d}{2}\right )+\cosh \left (\frac {d}{2}\right ) \text {$\#$1}^3+\sinh \left (\frac {d}{2}\right ) \text {$\#$1}^3\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ] (\cosh (2 d)-2 \cosh (d) \sinh (d))\right )}{3 b} \] Input:

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

Output:

(E^((5*a)/3)*(-((-9*E^((2*b*x)/3) + 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 & ]*(Cos 
h[d] - Sinh[d]))*(Cosh[d] - Sinh[d])) + 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 & ]*( 
Cosh[2*d] - 2*Cosh[d]*Sinh[d])))/(3*b)
 

Rubi [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.60, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {2720, 27, 807, 843, 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.

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

\(\Big \downarrow \) 2720

\(\displaystyle \frac {3 \int -\frac {2 e^{\frac {5 a}{3}+\frac {7 b x}{3}}}{1-e^{2 b x}}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {6 e^{5 a/3} \int \frac {e^{\frac {7 b x}{3}}}{1-e^{2 b x}}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 750

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(-3*E^((5*a)/3)*(-E^((2*b*x)/3) - Log[1 - E^((2*b*x)/3)]/3 + (Sqrt[3]*ArcT 
an[(1 + 2*E^((2*b*x)/3))/Sqrt[3]] + Log[1 + 2*E^((2*b*x)/3)]/2)/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 843

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*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 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 = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}}}{b}-\frac {\ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}+\frac {i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}-\frac {\ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}-\frac {i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}+\frac {\ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-1\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{b}\)	\(175\)

Input:

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

Output:

3*exp(5/3*a-d+2/3*b*x)/b-1/2*ln(exp(2/3*b*x+2/3*d)+1/2-1/2*I*3^(1/2))/b*ex 
p(5/3*a-5/3*d)+1/2*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)-1/2*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)-1/2*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)+ln(exp(2/3*b*x+2/3*d)-1)/b*exp(5/3*a-5/3*d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (113) = 226\).

Time = 0.10 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.52 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}(d+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*(6*cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) + 6*(cosh(-5/3*a + 5/3 
*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 - 6*cosh(1/3*b*x + 1/3 
*d)^2*sinh(-5/3*a + 5/3*d) + 2*(sqrt(3)*cosh(-5/3*a + 5/3*d) - sqrt(3)*sin 
h(-5/3*a + 5/3*d))*arctan(-1/3*(3*sqrt(3)*cosh(1/3*b*x + 1/3*d) + sqrt(3)* 
sinh(1/3*b*x + 1/3*d))/(cosh(1/3*b*x + 1/3*d) - sinh(1/3*b*x + 1/3*d))) - 
(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*log((2*cosh(1/3*b*x + 1/3*d) 
^2 + 2*sinh(1/3*b*x + 1/3*d)^2 + 1)/(cosh(1/3*b*x + 1/3*d)^2 - 2*cosh(1/3* 
b*x + 1/3*d)*sinh(1/3*b*x + 1/3*d) + sinh(1/3*b*x + 1/3*d)^2)) + 2*(cosh(- 
5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*log(2*sinh(1/3*b*x + 1/3*d)/(cosh(1 
/3*b*x + 1/3*d) - sinh(1/3*b*x + 1/3*d))) + 12*(cosh(1/3*b*x + 1/3*d)*cosh 
(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b* 
x + 1/3*d))/b
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.42 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}(d+b x) \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{b} + \frac {\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{b} - \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{2 \, b} + \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + 1\right )}{b} + \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 1\right )}{b} - \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{2 \, b} + \frac {3 \, e^{\left (\frac {2}{3} \, b x + \frac {5}{3} \, a - d\right )}}{b} \] Input:

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

Output:

-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 + 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 - 1/2*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 + e^(5/3*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) + 1)/b + e^(5/3 
*a - 5/3*d)*log(e^(-1/3*b*x - 1/3*d) - 1)/b - 1/2*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 + 3*e^(2/3*b*x + 5/3*a - 
 d)/b
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.69 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}(d+b x) \, dx=-\frac {{\left (2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right )} e^{\left (\frac {2}{3} \, d\right )}\right ) e^{\left (-\frac {8}{3} \, d\right )} + e^{\left (-\frac {8}{3} \, d\right )} \log \left (e^{\left (\frac {4}{3} \, b x\right )} + e^{\left (\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + e^{\left (-\frac {4}{3} \, d\right )}\right ) - 2 \, e^{\left (-\frac {8}{3} \, d\right )} \log \left ({\left | e^{\left (\frac {2}{3} \, b x\right )} - e^{\left (-\frac {2}{3} \, d\right )} \right |}\right ) - 6 \, e^{\left (\frac {2}{3} \, b x - 2 \, d\right )}\right )} e^{\left (\frac {5}{3} \, a + d\right )}}{2 \, b} \] Input:

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

Output:

-1/2*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(2/3*b*x) + e^(-2/3*d))*e^(2/3*d)) 
*e^(-8/3*d) + e^(-8/3*d)*log(e^(4/3*b*x) + e^(2/3*b*x - 2/3*d) + e^(-4/3*d 
)) - 2*e^(-8/3*d)*log(abs(e^(2/3*b*x) - e^(-2/3*d))) - 6*e^(2/3*b*x - 2*d) 
)*e^(5/3*a + d)/b
 

Mupad [B] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.35 \[ \int e^{\frac {5}{3} (a+b x)} \text {csch}(d+b x) \, dx=\frac {3\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}}{b}+\frac {{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (2\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}-2\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}\right )}{b}+\frac {{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}-2\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b}-\frac {{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}-2\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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