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

Optimal. Leaf size=45 \[ \frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b}-\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.027565, antiderivative size = 45, normalized size of antiderivative = 1., 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, 4848, 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[ArcTan[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 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^{a+b x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{2 b}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{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*}

Mathematica [A]  time = 0.100697, size = 83, normalized size = 1.84 \[ x \tan ^{-1}\left (e^{a+b x}\right )-\frac{i \left (-\text{PolyLog}\left (2,-i e^{a+b x}\right )+\text{PolyLog}\left (2,i e^{a+b x}\right )+b x \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[E^(a + b*x)],x]

[Out]

x*ArcTan[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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Maple [B]  time = 0.049, size = 106, normalized size = 2.4 \begin{align*}{\frac{\ln \left ({{\rm e}^{bx+a}} \right ) \arctan \left ({{\rm e}^{bx+a}} \right ) }{b}}+{\frac{{\frac{i}{2}}\ln \left ({{\rm e}^{bx+a}} \right ) \ln \left ( 1+i{{\rm e}^{bx+a}} \right ) }{b}}-{\frac{{\frac{i}{2}}\ln \left ({{\rm e}^{bx+a}} \right ) \ln \left ( 1-i{{\rm e}^{bx+a}} \right ) }{b}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{b}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*ln(exp(b*x+a))*arctan(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 [B]  time = 1.49925, size = 85, normalized size = 1.89 \begin{align*} \frac{{\left (b x + a\right )} \arctan \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)*arctan(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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Fricas [B]  time = 2.23242, size = 297, normalized size = 6.6 \begin{align*} \frac{2 \, b x \arctan \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x*arctan(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{atan}{\left (e^{a + b x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(exp(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(e^(b*x + a)), x)