3.119 \(\int e^{-x} \tan ^{-1}(e^x) \, dx\)

Optimal. Leaf size=25 \[ x-\frac{1}{2} \log \left (e^{2 x}+1\right )-e^{-x} \tan ^{-1}\left (e^x\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0211932, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {2194, 5207, 2282, 36, 29, 31} \[ x-\frac{1}{2} \log \left (e^{2 x}+1\right )-e^{-x} \tan ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2194

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 5207

Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTan[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*ArcTan[u]), x]]

Rule 2282

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 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 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]

Rubi steps

\begin{align*} \int e^{-x} \tan ^{-1}\left (e^x\right ) \, dx &=-e^{-x} \tan ^{-1}\left (e^x\right )+\int \frac{1}{1+e^{2 x}} \, dx\\ &=-e^{-x} \tan ^{-1}\left (e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,e^{2 x}\right )\\ &=-e^{-x} \tan ^{-1}\left (e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,e^{2 x}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,e^{2 x}\right )\\ &=x-e^{-x} \tan ^{-1}\left (e^x\right )-\frac{1}{2} \log \left (1+e^{2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0172051, size = 25, normalized size = 1. \[ x-\frac{1}{2} \log \left (e^{2 x}+1\right )-e^{-x} \tan ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.04, size = 23, normalized size = 0.9 \begin{align*} -{\frac{\arctan \left ({{\rm e}^{x}} \right ) }{{{\rm e}^{x}}}}-{\frac{\ln \left ( \left ({{\rm e}^{x}} \right ) ^{2}+1 \right ) }{2}}+\ln \left ({{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(exp(x))/exp(x),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.946223, size = 26, normalized size = 1.04 \begin{align*} -\arctan \left (e^{x}\right ) e^{\left (-x\right )} - \frac{1}{2} \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.14754, size = 82, normalized size = 3.28 \begin{align*} \frac{1}{2} \,{\left (2 \, x e^{x} - e^{x} \log \left (e^{\left (2 \, x\right )} + 1\right ) - 2 \, \arctan \left (e^{x}\right )\right )} e^{\left (-x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 11.4519, size = 19, normalized size = 0.76 \begin{align*} x - \frac{\log{\left (e^{2 x} + 1 \right )}}{2} - e^{- x} \operatorname{atan}{\left (e^{x} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(exp(x))/exp(x),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.09933, size = 27, normalized size = 1.08 \begin{align*} -\arctan \left (e^{x}\right ) e^{\left (-x\right )} + x - \frac{1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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