3.218 \(\int \cot ^{-1}(e^x) \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{2} i \text {Li}_2\left (i e^{-x}\right )-\frac {1}{2} i \text {Li}_2\left (-i e^{-x}\right ) \]

[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 = 35, 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, 4849, 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[ArcCot[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 4849

Int[((a_.) + ArcCot[(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 \cot ^{-1}\left (e^x\right ) \, dx &=\operatorname {Subst}\left (\int \frac {\cot ^{-1}(x)}{x} \, dx,x,e^x\right )\\ &=\frac {1}{2} i \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i}{x}\right )}{x} \, dx,x,e^x\right )-\frac {1}{2} i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i}{x}\right )}{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 = 0.08, size = 59, normalized size = 1.69 \[ x \cot ^{-1}\left (e^x\right )+\frac {1}{2} i \left (-\text {Li}_2\left (-i e^x\right )+\text {Li}_2\left (i e^x\right )+x \left (\log \left (1-i e^x\right )-\log \left (1+i e^x\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[E^x],x]

[Out]

x*ArcCot[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 = 0.49, size = 40, normalized size = 1.14 \[ x \operatorname {arccot}\left (e^{x}\right ) - \frac {1}{2} i \, x \log \left (i \, e^{x} + 1\right ) + \frac {1}{2} i \, x \log \left (-i \, e^{x} + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (i \, e^{x}\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-i \, e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccot(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 \[ \int \operatorname {arccot}\left (e^{x}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccot(e^x), x)

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maple [B]  time = 0.06, size = 59, normalized size = 1.69 \[ \ln \left ({\mathrm e}^{x}\right ) \mathrm {arccot}\left ({\mathrm e}^{x}\right )-\frac {i \ln \left ({\mathrm e}^{x}\right ) \ln \left (1+i {\mathrm e}^{x}\right )}{2}+\frac {i \ln \left ({\mathrm e}^{x}\right ) \ln \left (1-i {\mathrm e}^{x}\right )}{2}-\frac {i \dilog \left (1+i {\mathrm e}^{x}\right )}{2}+\frac {i \dilog \left (1-i {\mathrm e}^{x}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(exp(x)),x)

[Out]

ln(exp(x))*arccot(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 [A]  time = 0.45, size = 34, normalized size = 0.97 \[ x \operatorname {arccot}\left (e^{x}\right ) + \frac {1}{4} \, \pi \log \left (e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (i \, e^{x} + 1\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-i \, e^{x} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccot(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 [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \mathrm {acot}\left ({\mathrm {e}}^x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(exp(x)),x)

[Out]

int(acot(exp(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acot(exp(x)), x)

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