\(\int e^{-x} \cot ^{-1}(e^x) \, dx\) [227]

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

Optimal result

Integrand size = 10, antiderivative size = 27 \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=-x-e^{-x} \cot ^{-1}\left (e^x\right )+\frac {1}{2} \log \left (1+e^{2 x}\right ) \]

[Out]

-x-arccot(exp(x))/exp(x)+1/2*ln(1+exp(2*x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2225, 5316, 2320, 36, 29, 31} \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=-x+\frac {1}{2} \log \left (e^{2 x}+1\right )-e^{-x} \cot ^{-1}\left (e^x\right ) \]

[In]

Int[ArcCot[E^x]/E^x,x]

[Out]

-x - ArcCot[E^x]/E^x + Log[1 + E^(2*x)]/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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]]]

Rule 5316

Int[((a_.) + ArcCot[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcCot[u], w, x] + Dist
[b, Int[SimplifyIntegrand[w*(D[u, x]/(1 + u^2)), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b}, x]
 && InverseFunctionFreeQ[u, x] &&  !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; FreeQ[{c, d, m}, x]] && FalseQ[Functi
onOfLinear[v*(a + b*ArcCot[u]), x]]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=-x-e^{-x} \cot ^{-1}\left (e^x\right )+\frac {1}{2} \log \left (1+e^{2 x}\right ) \]

[In]

Integrate[ArcCot[E^x]/E^x,x]

[Out]

-x - ArcCot[E^x]/E^x + Log[1 + E^(2*x)]/2

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93

method result size
derivativedivides \(-\operatorname {arccot}\left ({\mathrm e}^{x}\right ) {\mathrm e}^{-x}-\ln \left ({\mathrm e}^{x}\right )+\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{2}\) \(25\)
default \(-\operatorname {arccot}\left ({\mathrm e}^{x}\right ) {\mathrm e}^{-x}-\ln \left ({\mathrm e}^{x}\right )+\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{2}\) \(25\)
parallelrisch \(\frac {\left (\ln \left (1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{x}-2 x \,{\mathrm e}^{x}-2 \,\operatorname {arccot}\left ({\mathrm e}^{x}\right )\right ) {\mathrm e}^{-x}}{2}\) \(28\)
risch \(-\frac {i {\mathrm e}^{-x} \ln \left (1+i {\mathrm e}^{x}\right )}{2}+\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{2}-x +\frac {i {\mathrm e}^{-x} \ln \left (1-i {\mathrm e}^{x}\right )}{2}-\frac {{\mathrm e}^{-x} \pi }{2}\) \(51\)

[In]

int(arccot(exp(x))/exp(x),x,method=_RETURNVERBOSE)

[Out]

-arccot(exp(x))/exp(x)-ln(exp(x))+1/2*ln(exp(x)^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=-\frac {1}{2} \, {\left (2 \, x e^{x} - e^{x} \log \left (e^{\left (2 \, x\right )} + 1\right ) + 2 \, \operatorname {arccot}\left (e^{x}\right )\right )} e^{\left (-x\right )} \]

[In]

integrate(arccot(exp(x))/exp(x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=- x + \frac {\log {\left (e^{2 x} + 1 \right )}}{2} - e^{- x} \operatorname {acot}{\left (e^{x} \right )} \]

[In]

integrate(acot(exp(x))/exp(x),x)

[Out]

-x + log(exp(2*x) + 1)/2 - exp(-x)*acot(exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=-\operatorname {arccot}\left (e^{x}\right ) e^{\left (-x\right )} + \frac {1}{2} \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

[In]

integrate(arccot(exp(x))/exp(x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=-\arctan \left (e^{\left (-x\right )}\right ) e^{\left (-x\right )} + \frac {1}{2} \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

[In]

integrate(arccot(exp(x))/exp(x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int e^{-x} \cot ^{-1}\left (e^x\right ) \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,x}+1\right )}{2}-x-\mathrm {acot}\left ({\mathrm {e}}^x\right )\,{\mathrm {e}}^{-x} \]

[In]

int(acot(exp(x))*exp(-x),x)

[Out]

log(exp(2*x) + 1)/2 - x - acot(exp(x))*exp(-x)