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

Optimal. Leaf size=31 \[ \frac {1}{2} i \text {Li}_2\left (-i e^x\right )-\frac {1}{2} i \text {Li}_2\left (i e^x\right ) \]

[Out]

1/2*I*polylog(2,-I*exp(x))-1/2*I*polylog(2,I*exp(x))

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, 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 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]

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]

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*}

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Mathematica [A]  time = 0.05, size = 59, normalized size = 1.90 \[ x \tan ^{-1}\left (e^x\right )-\frac {1}{2} i \left (-\text {Li}_2\left (-i e^x\right )+\text {Li}_2\left (i e^x\right )+x \left (\log \left (1-i e^x\right )-\log \left (1+i e^x\right )\right )\right ) \]

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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fricas [B]  time = 0.46, size = 40, normalized size = 1.29 \[ x \arctan \left (e^{x}\right ) + \frac {1}{2} i \, x \log \left (i \, e^{x} + 1\right ) - \frac {1}{2} i \, x \log \left (-i \, e^{x} + 1\right ) - \frac {1}{2} i \, {\rm Li}_2\left (i \, e^{x}\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-i \, e^{x}\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \arctan \left (e^{x}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(e^x), x)

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maple [B]  time = 0.06, size = 59, normalized size = 1.90 \[ \ln \left ({\mathrm e}^{x}\right ) \arctan \left ({\mathrm e}^{x}\right )+\frac {i \ln \left ({\mathrm e}^{x}\right ) \ln \left (1+i {\mathrm e}^{x}\right )}{2}-\frac {i \ln \left ({\mathrm e}^{x}\right ) \ln \left (1-i {\mathrm e}^{x}\right )}{2}+\frac {i \dilog \left (1+i {\mathrm e}^{x}\right )}{2}-\frac {i \dilog \left (1-i {\mathrm e}^{x}\right )}{2} \]

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 = 0.43, size = 34, normalized size = 1.10 \[ x \arctan \left (e^{x}\right ) - \frac {1}{4} \, \pi \log \left (e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} i \, {\rm Li}_2\left (i \, e^{x} + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-i \, e^{x} + 1\right ) \]

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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mupad [B]  time = 0.69, size = 21, normalized size = 0.68 \[ \frac {\mathrm {polylog}\left (2,-{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(exp(x)),x)

[Out]

(polylog(2, -exp(x)*1i)*1i)/2 - (polylog(2, exp(x)*1i)*1i)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {atan}{\left (e^{x} \right )}\, dx \]

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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