\(\int e^x \text {csch}^2(3 x) \, dx\) [317]

Optimal result

Integrand size = 10, antiderivative size = 91 \[ \int e^x \text {csch}^2(3 x) \, dx=\frac {2 e^x}{3 \left (1-e^{6 x}\right )}+\frac {\arctan \left (\frac {1-2 e^x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\arctan \left (\frac {1+2 e^x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2 \text {arctanh}\left (e^x\right )}{9}-\frac {1}{9} \text {arctanh}\left (\frac {e^x}{1+e^{2 x}}\right ) \] Output:

2*exp(x)/(3-3*exp(6*x))+1/9*arctan(1/3*(1-2*exp(x))*3^(1/2))*3^(1/2)-1/9*a 
rctan(1/3*(1+2*exp(x))*3^(1/2))*3^(1/2)-2/9*arctanh(exp(x))-1/9*arctanh(ex 
p(x)/(1+exp(2*x)))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.37 \[ \int e^x \text {csch}^2(3 x) \, dx=\frac {2}{3} e^x \left (\frac {1}{1-e^{6 x}}-\operatorname {Hypergeometric2F1}\left (\frac {1}{6},1,\frac {7}{6},e^{6 x}\right )\right ) \] Input:

Integrate[E^x*Csch[3*x]^2,x]
 

Output:

(2*E^x*((1 - E^(6*x))^(-1) - Hypergeometric2F1[1/6, 1, 7/6, E^(6*x)]))/3
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.30, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, Rules used = {2720, 27, 817, 754, 27, 219, 1142, 25, 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.

\(\displaystyle \int e^x \text {csch}^2(3 x) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \int \frac {4 e^{6 x}}{\left (1-e^{6 x}\right )^2}de^x\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \int \frac {e^{6 x}}{\left (1-e^{6 x}\right )^2}de^x\)

\(\Big \downarrow \) 817

\(\displaystyle 4 \left (\frac {e^x}{6 \left (1-e^{6 x}\right )}-\frac {1}{6} \int \frac {1}{1-e^{6 x}}de^x\right )\)

\(\Big \downarrow \) 754

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 219

\(\displaystyle 4 \left (\frac {1}{6} \left (-\frac {1}{6} \int \frac {2-e^x}{1-e^x+e^{2 x}}de^x-\frac {1}{6} \int \frac {2+e^x}{1+e^x+e^{2 x}}de^x-\frac {1}{3} \text {arctanh}\left (e^x\right )\right )+\frac {e^x}{6 \left (1-e^{6 x}\right )}\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle 4 \left (\frac {1}{6} \left (\frac {1}{6} \left (\frac {1}{2} \int -\frac {1-2 e^x}{1-e^x+e^{2 x}}de^x-\frac {3}{2} \int \frac {1}{1-e^x+e^{2 x}}de^x\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{1+e^x+e^{2 x}}de^x-\frac {1}{2} \int \frac {1+2 e^x}{1+e^x+e^{2 x}}de^x\right )-\frac {1}{3} \text {arctanh}\left (e^x\right )\right )+\frac {e^x}{6 \left (1-e^{6 x}\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 4 \left (\frac {1}{6} \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{1-e^x+e^{2 x}}de^x-\frac {1}{2} \int \frac {1-2 e^x}{1-e^x+e^{2 x}}de^x\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{1+e^x+e^{2 x}}de^x-\frac {1}{2} \int \frac {1+2 e^x}{1+e^x+e^{2 x}}de^x\right )-\frac {1}{3} \text {arctanh}\left (e^x\right )\right )+\frac {e^x}{6 \left (1-e^{6 x}\right )}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle 4 \left (\frac {1}{6} \left (\frac {1}{6} \left (3 \int \frac {1}{-3-e^{2 x}}d\left (-1+2 e^x\right )-\frac {1}{2} \int \frac {1-2 e^x}{1-e^x+e^{2 x}}de^x\right )+\frac {1}{6} \left (3 \int \frac {1}{-3-e^{2 x}}d\left (1+2 e^x\right )-\frac {1}{2} \int \frac {1+2 e^x}{1+e^x+e^{2 x}}de^x\right )-\frac {1}{3} \text {arctanh}\left (e^x\right )\right )+\frac {e^x}{6 \left (1-e^{6 x}\right )}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle 4 \left (\frac {1}{6} \left (\frac {1}{6} \left (-\frac {1}{2} \int \frac {1-2 e^x}{1-e^x+e^{2 x}}de^x-\sqrt {3} \arctan \left (\frac {2 e^x-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\frac {1}{2} \int \frac {1+2 e^x}{1+e^x+e^{2 x}}de^x-\sqrt {3} \arctan \left (\frac {2 e^x+1}{\sqrt {3}}\right )\right )-\frac {1}{3} \text {arctanh}\left (e^x\right )\right )+\frac {e^x}{6 \left (1-e^{6 x}\right )}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle 4 \left (\frac {1}{6} \left (\frac {1}{6} \left (\frac {1}{2} \log \left (-e^x+e^{2 x}+1\right )-\sqrt {3} \arctan \left (\frac {2 e^x-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {2 e^x+1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (e^x+e^{2 x}+1\right )\right )-\frac {1}{3} \text {arctanh}\left (e^x\right )\right )+\frac {e^x}{6 \left (1-e^{6 x}\right )}\right )\)

Input:

Int[E^x*Csch[3*x]^2,x]
 

Output:

4*(E^x/(6*(1 - E^(6*x))) + (-1/3*ArcTanh[E^x] + (-(Sqrt[3]*ArcTan[(-1 + 2* 
E^x)/Sqrt[3]]) + Log[1 - E^x + E^(2*x)]/2)/6 + (-(Sqrt[3]*ArcTan[(1 + 2*E^ 
x)/Sqrt[3]]) - Log[1 + E^x + E^(2*x)]/2)/6)/6)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 754

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-a 
/b, n]], s = Denominator[Rt[-a/b, n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k* 
Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r + s*Cos[(2 
*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n)) 
 Int[1/(r^2 - s^2*x^2), x] + 2*(r/(a*n))   Sum[u, {k, 1, (n - 2)/4}], x]] / 
; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]
 

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 [A] (verified)

Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27

method	result	size
default	\(\frac {\tanh \left (\frac {x}{2}\right )}{18}+\frac {-\frac {2 \tanh \left (\frac {x}{2}\right )}{3}+\frac {2}{3}}{9 \tanh \left (\frac {x}{2}\right )^{2}+3}+\frac {\ln \left (3 \tanh \left (\frac {x}{2}\right )^{2}+1\right )}{18}-\frac {\sqrt {3}\, \arctan \left (\tanh \left (\frac {x}{2}\right ) \sqrt {3}\right )}{9}-\frac {1}{18 \tanh \left (\frac {x}{2}\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{9}-\frac {-2 \tanh \left (\frac {x}{2}\right )-6}{9 \left (\tanh \left (\frac {x}{2}\right )^{2}+3\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3\right )}{18}-\frac {\sqrt {3}\, \arctan \left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {3}}{3}\right )}{9}\)	\(116\)
risch	\(-\frac {2 \,{\mathrm e}^{x}}{3 \left (-1+{\mathrm e}^{6 x}\right )}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{9}-\frac {\ln \left ({\mathrm e}^{x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{18}+\frac {i \sqrt {3}\, \ln \left ({\mathrm e}^{x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{18}-\frac {\ln \left ({\mathrm e}^{x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{18}-\frac {i \sqrt {3}\, \ln \left ({\mathrm e}^{x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{18}+\frac {\ln \left ({\mathrm e}^{x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{18}+\frac {i \sqrt {3}\, \ln \left ({\mathrm e}^{x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{18}+\frac {\ln \left ({\mathrm e}^{x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{18}-\frac {i \sqrt {3}\, \ln \left ({\mathrm e}^{x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{18}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{9}\)	\(148\)

Input:

int(exp(x)*csch(3*x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/18*tanh(1/2*x)+1/9*(-2/3*tanh(1/2*x)+2/3)/(tanh(1/2*x)^2+1/3)+1/18*ln(3* 
tanh(1/2*x)^2+1)-1/9*3^(1/2)*arctan(tanh(1/2*x)*3^(1/2))-1/18/tanh(1/2*x)+ 
1/9*ln(tanh(1/2*x))-1/9*(-2*tanh(1/2*x)-6)/(tanh(1/2*x)^2+3)-1/18*ln(tanh( 
1/2*x)^2+3)-1/9*3^(1/2)*arctan(1/3*tanh(1/2*x)*3^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 560 vs. \(2 (66) = 132\).

Time = 0.11 (sec) , antiderivative size = 560, normalized size of antiderivative = 6.15 \[ \int e^x \text {csch}^2(3 x) \, dx=\text {Too large to display} \] Input:

integrate(exp(x)*csch(3*x)^2,x, algorithm="fricas")
 

Output:

-1/18*(2*(sqrt(3)*cosh(x)^6 + 6*sqrt(3)*cosh(x)^5*sinh(x) + 15*sqrt(3)*cos 
h(x)^4*sinh(x)^2 + 20*sqrt(3)*cosh(x)^3*sinh(x)^3 + 15*sqrt(3)*cosh(x)^2*s 
inh(x)^4 + 6*sqrt(3)*cosh(x)*sinh(x)^5 + sqrt(3)*sinh(x)^6 - sqrt(3))*arct 
an(2/3*sqrt(3)*cosh(x) + 2/3*sqrt(3)*sinh(x) + 1/3*sqrt(3)) + 2*(sqrt(3)*c 
osh(x)^6 + 6*sqrt(3)*cosh(x)^5*sinh(x) + 15*sqrt(3)*cosh(x)^4*sinh(x)^2 + 
20*sqrt(3)*cosh(x)^3*sinh(x)^3 + 15*sqrt(3)*cosh(x)^2*sinh(x)^4 + 6*sqrt(3 
)*cosh(x)*sinh(x)^5 + sqrt(3)*sinh(x)^6 - sqrt(3))*arctan(2/3*sqrt(3)*cosh 
(x) + 2/3*sqrt(3)*sinh(x) - 1/3*sqrt(3)) + (cosh(x)^6 + 6*cosh(x)^5*sinh(x 
) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x) 
^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 - 1)*log((2*cosh(x) + 1)/(cosh(x) - s 
inh(x))) - (cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20* 
cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh( 
x)^6 - 1)*log((2*cosh(x) - 1)/(cosh(x) - sinh(x))) + 2*(cosh(x)^6 + 6*cosh 
(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh( 
x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 - 1)*log(cosh(x) + sinh(x 
) + 1) - 2*(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20* 
cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh( 
x)^6 - 1)*log(cosh(x) + sinh(x) - 1) + 12*cosh(x) + 12*sinh(x))/(cosh(x)^6 
 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 
 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 - 1)
 

Sympy [F]

\[ \int e^x \text {csch}^2(3 x) \, dx=\int e^{x} \operatorname {csch}^{2}{\left (3 x \right )}\, dx \] Input:

integrate(exp(x)*csch(3*x)**2,x)
 

Output:

Integral(exp(x)*csch(3*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.93 \[ \int e^x \text {csch}^2(3 x) \, dx=-\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} + 1\right )}\right ) - \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} - 1\right )}\right ) - \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} - 1\right )}} - \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac {1}{9} \, \log \left (e^{x} + 1\right ) + \frac {1}{9} \, \log \left (e^{x} - 1\right ) \] Input:

integrate(exp(x)*csch(3*x)^2,x, algorithm="maxima")
 

Output:

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^x + 1)) - 1/9*sqrt(3)*arctan(1/3*sqrt 
(3)*(2*e^x - 1)) - 2/3*e^x/(e^(6*x) - 1) - 1/18*log(e^(2*x) + e^x + 1) + 1 
/18*log(e^(2*x) - e^x + 1) - 1/9*log(e^x + 1) + 1/9*log(e^x - 1)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int e^x \text {csch}^2(3 x) \, dx=-\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} + 1\right )}\right ) - \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} - 1\right )}\right ) - \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} - 1\right )}} - \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac {1}{9} \, \log \left (e^{x} + 1\right ) + \frac {1}{9} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \] Input:

integrate(exp(x)*csch(3*x)^2,x, algorithm="giac")
 

Output:

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^x + 1)) - 1/9*sqrt(3)*arctan(1/3*sqrt 
(3)*(2*e^x - 1)) - 2/3*e^x/(e^(6*x) - 1) - 1/18*log(e^(2*x) + e^x + 1) + 1 
/18*log(e^(2*x) - e^x + 1) - 1/9*log(e^x + 1) + 1/9*log(abs(e^x - 1))
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int e^x \text {csch}^2(3 x) \, dx=\frac {\ln \left (\frac {2}{3}-\frac {2\,{\mathrm {e}}^x}{3}\right )}{9}-\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x}{3}-\frac {2}{3}\right )}{9}+\frac {\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}-\frac {1}{3}\right )}^2+\frac {1}{3}\right )}{18}-\frac {\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}+\frac {1}{3}\right )}^2+\frac {1}{3}\right )}{18}-\frac {2\,{\mathrm {e}}^x}{3\,\left ({\mathrm {e}}^{6\,x}-1\right )}-\frac {\sqrt {3}\,\mathrm {atan}\left (\sqrt {3}\,\left (\frac {2\,{\mathrm {e}}^x}{3}-\frac {1}{3}\right )\right )}{9}-\frac {\sqrt {3}\,\mathrm {atan}\left (\sqrt {3}\,\left (\frac {2\,{\mathrm {e}}^x}{3}+\frac {1}{3}\right )\right )}{9} \] Input:

int(exp(x)/sinh(3*x)^2,x)
 

Output:

log(2/3 - (2*exp(x))/3)/9 - log(- (2*exp(x))/3 - 2/3)/9 + log(((2*exp(x))/ 
3 - 1/3)^2 + 1/3)/18 - log(((2*exp(x))/3 + 1/3)^2 + 1/3)/18 - (2*exp(x))/( 
3*(exp(6*x) - 1)) - (3^(1/2)*atan(3^(1/2)*((2*exp(x))/3 - 1/3)))/9 - (3^(1 
/2)*atan(3^(1/2)*((2*exp(x))/3 + 1/3)))/9
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.21 \[ \int e^x \text {csch}^2(3 x) \, dx=\frac {-2 e^{6 x} \sqrt {3}\, \mathit {atan} \left (\frac {2 e^{x}-1}{\sqrt {3}}\right )+2 \sqrt {3}\, \mathit {atan} \left (\frac {2 e^{x}-1}{\sqrt {3}}\right )-2 e^{6 x} \sqrt {3}\, \mathit {atan} \left (\frac {2 e^{x}+1}{\sqrt {3}}\right )+2 \sqrt {3}\, \mathit {atan} \left (\frac {2 e^{x}+1}{\sqrt {3}}\right )-e^{6 x} \mathrm {log}\left (e^{2 x}+e^{x}+1\right )+e^{6 x} \mathrm {log}\left (e^{2 x}-e^{x}+1\right )+2 e^{6 x} \mathrm {log}\left (e^{x}-1\right )-2 e^{6 x} \mathrm {log}\left (e^{x}+1\right )-12 e^{x}+\mathrm {log}\left (e^{2 x}+e^{x}+1\right )-\mathrm {log}\left (e^{2 x}-e^{x}+1\right )-2 \,\mathrm {log}\left (e^{x}-1\right )+2 \,\mathrm {log}\left (e^{x}+1\right )}{18 e^{6 x}-18} \] Input:

int(exp(x)*csch(3*x)^2,x)
 

Output:

( - 2*e**(6*x)*sqrt(3)*atan((2*e**x - 1)/sqrt(3)) + 2*sqrt(3)*atan((2*e**x 
 - 1)/sqrt(3)) - 2*e**(6*x)*sqrt(3)*atan((2*e**x + 1)/sqrt(3)) + 2*sqrt(3) 
*atan((2*e**x + 1)/sqrt(3)) - e**(6*x)*log(e**(2*x) + e**x + 1) + e**(6*x) 
*log(e**(2*x) - e**x + 1) + 2*e**(6*x)*log(e**x - 1) - 2*e**(6*x)*log(e**x 
 + 1) - 12*e**x + log(e**(2*x) + e**x + 1) - log(e**(2*x) - e**x + 1) - 2* 
log(e**x - 1) + 2*log(e**x + 1))/(18*(e**(6*x) - 1))