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

Optimal. Leaf size=31 \[ \frac{1}{2} i \text{PolyLog}\left (2,-i e^x\right )-\frac{1}{2} i \text{PolyLog}\left (2,i e^x\right ) \]

[Out]

(I/2)*PolyLog[2, (-I)*E^x] - (I/2)*PolyLog[2, I*E^x]

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Rubi [A]  time = 0.0255403, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 4, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {2282, 4848, 2391} \[ \frac{1}{2} i \text{PolyLog}\left (2,-i e^x\right )-\frac{1}{2} i \text{PolyLog}\left (2,i e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[E^x],x]

[Out]

(I/2)*PolyLog[2, (-I)*E^x] - (I/2)*PolyLog[2, I*E^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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \tan ^{-1}\left (e^x\right ) \, dx &=\operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{x} \, dx,x,e^x\right )\\ &=\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^x\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^x\right )\\ &=\frac{1}{2} i \text{Li}_2\left (-i e^x\right )-\frac{1}{2} i \text{Li}_2\left (i e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.043881, size = 59, normalized size = 1.9 \[ x \tan ^{-1}\left (e^x\right )-\frac{1}{2} i \left (-\text{PolyLog}\left (2,-i e^x\right )+\text{PolyLog}\left (2,i e^x\right )+x \left (\log \left (1-i e^x\right )-\log \left (1+i e^x\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[E^x],x]

[Out]

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

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Maple [B]  time = 0.049, size = 59, normalized size = 1.9 \begin{align*} \ln \left ({{\rm e}^{x}} \right ) \arctan \left ({{\rm e}^{x}} \right ) +{\frac{i}{2}}\ln \left ({{\rm e}^{x}} \right ) \ln \left ( 1+i{{\rm e}^{x}} \right ) -{\frac{i}{2}}\ln \left ({{\rm e}^{x}} \right ) \ln \left ( 1-i{{\rm e}^{x}} \right ) +{\frac{i}{2}}{\it dilog} \left ( 1+i{{\rm e}^{x}} \right ) -{\frac{i}{2}}{\it dilog} \left ( 1-i{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.50996, size = 46, normalized size = 1.48 \begin{align*} x \arctan \left (e^{x}\right ) - \frac{1}{4} \, \pi \log \left (e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{2} i \,{\rm Li}_2\left (i \, e^{x} + 1\right ) + \frac{1}{2} i \,{\rm Li}_2\left (-i \, e^{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.16789, size = 147, normalized size = 4.74 \begin{align*} x \arctan \left (e^{x}\right ) + \frac{1}{2} i \, x \log \left (i \, e^{x} + 1\right ) - \frac{1}{2} i \, x \log \left (-i \, e^{x} + 1\right ) - \frac{1}{2} i \,{\rm Li}_2\left (i \, e^{x}\right ) + \frac{1}{2} i \,{\rm Li}_2\left (-i \, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{atan}{\left (e^{x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(exp(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \arctan \left (e^{x}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(e^x), x)