\(\int e^x \text {csch}^2(2 x) \, dx\) [313]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 10, antiderivative size = 32 \[ \int e^x \text {csch}^2(2 x) \, dx=\frac {e^x}{1-e^{4 x}}-\frac {\arctan \left (e^x\right )}{2}-\frac {\text {arctanh}\left (e^x\right )}{2} \]

[Out]

exp(x)/(1-exp(4*x))-1/2*arctan(exp(x))-1/2*arctanh(exp(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2320, 12, 294, 218, 212, 209} \[ \int e^x \text {csch}^2(2 x) \, dx=-\frac {1}{2} \arctan \left (e^x\right )-\frac {\text {arctanh}\left (e^x\right )}{2}+\frac {e^x}{1-e^{4 x}} \]

[In]

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

[Out]

E^x/(1 - E^(4*x)) - ArcTan[E^x]/2 - ArcTanh[E^x]/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 294

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] - Dist[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 2320

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {4 x^4}{\left (1-x^4\right )^2} \, dx,x,e^x\right ) \\ & = 4 \text {Subst}\left (\int \frac {x^4}{\left (1-x^4\right )^2} \, dx,x,e^x\right ) \\ & = \frac {e^x}{1-e^{4 x}}-\text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,e^x\right ) \\ & = \frac {e^x}{1-e^{4 x}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right ) \\ & = \frac {e^x}{1-e^{4 x}}-\frac {\arctan \left (e^x\right )}{2}-\frac {\text {arctanh}\left (e^x\right )}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int e^x \text {csch}^2(2 x) \, dx=\frac {e^x}{1-e^{4 x}}-\frac {\arctan \left (e^x\right )}{2}-\frac {\text {arctanh}\left (e^x\right )}{2} \]

[In]

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

[Out]

E^x/(1 - E^(4*x)) - ArcTan[E^x]/2 - ArcTanh[E^x]/2

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44

method result size
risch \(-\frac {{\mathrm e}^{x}}{{\mathrm e}^{4 x}-1}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{4}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{4}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{4}\) \(46\)

[In]

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

[Out]

-exp(x)/(exp(4*x)-1)-1/4*ln(exp(x)+1)+1/4*I*ln(exp(x)-I)-1/4*I*ln(exp(x)+I)+1/4*ln(exp(x)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (23) = 46\).

Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 5.69 \[ \int e^x \text {csch}^2(2 x) \, dx=-\frac {2 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 4 \, \cosh \left (x\right ) + 4 \, \sinh \left (x\right )}{4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )}} \]

[In]

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

[Out]

-1/4*(2*(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)*arctan
(cosh(x) + sinh(x)) + (cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)
^4 - 1)*log(cosh(x) + sinh(x) + 1) - (cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh
(x)^3 + sinh(x)^4 - 1)*log(cosh(x) + sinh(x) - 1) + 4*cosh(x) + 4*sinh(x))/(cosh(x)^4 + 4*cosh(x)^3*sinh(x) +
6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)

Sympy [F]

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

[In]

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

[Out]

Integral(exp(x)*csch(2*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int e^x \text {csch}^2(2 x) \, dx=-\frac {e^{x}}{e^{\left (4 \, x\right )} - 1} - \frac {1}{2} \, \arctan \left (e^{x}\right ) - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left (e^{x} - 1\right ) \]

[In]

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

[Out]

-e^x/(e^(4*x) - 1) - 1/2*arctan(e^x) - 1/4*log(e^x + 1) + 1/4*log(e^x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int e^x \text {csch}^2(2 x) \, dx=-\frac {e^{x}}{e^{\left (4 \, x\right )} - 1} - \frac {1}{2} \, \arctan \left (e^{x}\right ) - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[In]

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

[Out]

-e^x/(e^(4*x) - 1) - 1/2*arctan(e^x) - 1/4*log(e^x + 1) + 1/4*log(abs(e^x - 1))

Mupad [B] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int e^x \text {csch}^2(2 x) \, dx=\frac {\ln \left (1-{\mathrm {e}}^x\right )}{4}-\frac {\ln \left (-{\mathrm {e}}^x-1\right )}{4}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{2}-\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{4\,x}-1} \]

[In]

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

[Out]

log(1 - exp(x))/4 - log(- exp(x) - 1)/4 - atan(exp(x))/2 - exp(x)/(exp(4*x) - 1)