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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0284399, antiderivative size = 35, 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, 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 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]

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 \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*}

Mathematica [A]  time = 0.0334939, size = 59, normalized size = 1.69 \[ x \cot ^{-1}\left (e^x\right )+\frac{1}{2} i \left (-\text{PolyLog}\left (2,-i e^x\right )+\text{PolyLog}\left (2,i e^x\right )+x \left (\log \left (1-i e^x\right )-\log \left (1+i e^x\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[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])

________________________________________________________________________________________

Maple [B]  time = 0.033, size = 59, normalized size = 1.7 \begin{align*} \ln \left ({{\rm e}^{x}} \right ){\rm arccot} \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(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))

________________________________________________________________________________________

Maxima [A]  time = 1.61049, size = 46, normalized size = 1.31 \begin{align*} 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 ) \end{align*}

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)

________________________________________________________________________________________

Fricas [B]  time = 2.15755, size = 147, normalized size = 4.2 \begin{align*} 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 ) \end{align*}

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acot(exp(x)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccot(e^x), x)