3.221 \(\int \cot ^{-1}(e^{a+b x}) \, dx\)

Optimal. Leaf size=51 \[ \frac {i \text {Li}_2\left (i e^{-a-b x}\right )}{2 b}-\frac {i \text {Li}_2\left (-i e^{-a-b x}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2282, 4849, 2391} \[ \frac {i \text {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b}-\frac {i \text {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[E^(a + b*x)],x]

[Out]

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

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

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Mathematica [A]  time = 0.18, size = 83, normalized size = 1.63 \[ x \cot ^{-1}\left (e^{a+b x}\right )+\frac {i \left (-\text {Li}_2\left (-i e^{a+b x}\right )+\text {Li}_2\left (i e^{a+b x}\right )+b x \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[E^(a + b*x)],x]

[Out]

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

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fricas [B]  time = 0.75, size = 103, normalized size = 2.02 \[ \frac {2 \, b x \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) - i \, a \log \left (e^{\left (b x + a\right )} + i\right ) + i \, a \log \left (e^{\left (b x + a\right )} - i\right ) + {\left (-i \, b x - i \, a\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) + {\left (i \, b x + i \, a\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) + i \, {\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) - i \, {\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(exp(b*x+a)),x, algorithm="fricas")

[Out]

1/2*(2*b*x*arccot(e^(b*x + a)) - I*a*log(e^(b*x + a) + I) + I*a*log(e^(b*x + a) - I) + (-I*b*x - I*a)*log(I*e^
(b*x + a) + 1) + (I*b*x + I*a)*log(-I*e^(b*x + a) + 1) + I*dilog(I*e^(b*x + a)) - I*dilog(-I*e^(b*x + a)))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arccot}\left (e^{\left (b x + a\right )}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccot(e^(b*x + a)), x)

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maple [B]  time = 0.06, size = 106, normalized size = 2.08 \[ \frac {\ln \left ({\mathrm e}^{b x +a}\right ) \mathrm {arccot}\left ({\mathrm e}^{b x +a}\right )}{b}-\frac {i \ln \left ({\mathrm e}^{b x +a}\right ) \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2 b}+\frac {i \ln \left ({\mathrm e}^{b x +a}\right ) \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2 b}-\frac {i \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{2 b}+\frac {i \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(exp(b*x+a)),x)

[Out]

1/b*ln(exp(b*x+a))*arccot(exp(b*x+a))-1/2*I/b*ln(exp(b*x+a))*ln(1+I*exp(b*x+a))+1/2*I/b*ln(exp(b*x+a))*ln(1-I*
exp(b*x+a))-1/2*I/b*dilog(1+I*exp(b*x+a))+1/2*I/b*dilog(1-I*exp(b*x+a))

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maxima [A]  time = 0.46, size = 63, normalized size = 1.24 \[ \frac {{\left (b x + a\right )} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right )}{b} + \frac {\pi \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 i \, {\rm Li}_2\left (i \, e^{\left (b x + a\right )} + 1\right ) - 2 i \, {\rm Li}_2\left (-i \, e^{\left (b x + a\right )} + 1\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(exp(b*x+a)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {acot}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(exp(a + b*x)),x)

[Out]

int(acot(exp(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(exp(b*x+a)),x)

[Out]

Integral(acot(exp(a + b*x)), x)

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