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

Optimal. Leaf size=63 \[ \frac{1}{2} i x \text{PolyLog}\left (2,-i e^x\right )-\frac{1}{2} i x \text{PolyLog}\left (2,i e^x\right )-\frac{1}{2} i \text{PolyLog}\left (3,-i e^x\right )+\frac{1}{2} i \text{PolyLog}\left (3,i e^x\right ) \]

[Out]

(I/2)*x*PolyLog[2, (-I)*E^x] - (I/2)*x*PolyLog[2, I*E^x] - (I/2)*PolyLog[3, (-I)*E^x] + (I/2)*PolyLog[3, I*E^x
]

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Rubi [A]  time = 0.0427798, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5143, 2531, 2282, 6589} \[ \frac{1}{2} i x \text{PolyLog}\left (2,-i e^x\right )-\frac{1}{2} i x \text{PolyLog}\left (2,i e^x\right )-\frac{1}{2} i \text{PolyLog}\left (3,-i e^x\right )+\frac{1}{2} i \text{PolyLog}\left (3,i e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[E^x],x]

[Out]

(I/2)*x*PolyLog[2, (-I)*E^x] - (I/2)*x*PolyLog[2, I*E^x] - (I/2)*PolyLog[3, (-I)*E^x] + (I/2)*PolyLog[3, I*E^x
]

Rule 5143

Int[ArcTan[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[I/2, Int[x^m*Log[1 - I*a - I
*b*f^(c + d*x)], x], x] - Dist[I/2, Int[x^m*Log[1 + I*a + I*b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x
] && IntegerQ[m] && m > 0

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x \tan ^{-1}\left (e^x\right ) \, dx &=\frac{1}{2} i \int x \log \left (1-i e^x\right ) \, dx-\frac{1}{2} i \int x \log \left (1+i e^x\right ) \, dx\\ &=\frac{1}{2} i x \text{Li}_2\left (-i e^x\right )-\frac{1}{2} i x \text{Li}_2\left (i e^x\right )-\frac{1}{2} i \int \text{Li}_2\left (-i e^x\right ) \, dx+\frac{1}{2} i \int \text{Li}_2\left (i e^x\right ) \, dx\\ &=\frac{1}{2} i x \text{Li}_2\left (-i e^x\right )-\frac{1}{2} i x \text{Li}_2\left (i e^x\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^x\right )+\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^x\right )\\ &=\frac{1}{2} i x \text{Li}_2\left (-i e^x\right )-\frac{1}{2} i x \text{Li}_2\left (i e^x\right )-\frac{1}{2} i \text{Li}_3\left (-i e^x\right )+\frac{1}{2} i \text{Li}_3\left (i e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0088854, size = 50, normalized size = 0.79 \[ \frac{1}{2} i \left (x \text{PolyLog}\left (2,-i e^x\right )-x \text{PolyLog}\left (2,i e^x\right )-\text{PolyLog}\left (3,-i e^x\right )+\text{PolyLog}\left (3,i e^x\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTan[E^x],x]

[Out]

(I/2)*(x*PolyLog[2, (-I)*E^x] - x*PolyLog[2, I*E^x] - PolyLog[3, (-I)*E^x] + PolyLog[3, I*E^x])

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Maple [A]  time = 0.089, size = 44, normalized size = 0.7 \begin{align*}{\frac{i}{2}}x{\it polylog} \left ( 2,-i{{\rm e}^{x}} \right ) -{\frac{i}{2}}x{\it polylog} \left ( 2,i{{\rm e}^{x}} \right ) -{\frac{i}{2}}{\it polylog} \left ( 3,-i{{\rm e}^{x}} \right ) +{\frac{i}{2}}{\it polylog} \left ( 3,i{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(exp(x)),x)

[Out]

1/2*I*x*polylog(2,-I*exp(x))-1/2*I*x*polylog(2,I*exp(x))-1/2*I*polylog(3,-I*exp(x))+1/2*I*polylog(3,I*exp(x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (e^{x}\right ) - \int \frac{x^{2} e^{x}}{2 \,{\left (e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(e^x) - integrate(1/2*x^2*e^x/(e^(2*x) + 1), x)

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Fricas [C]  time = 2.16289, size = 238, normalized size = 3.78 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (e^{x}\right ) + \frac{1}{4} i \, x^{2} \log \left (i \, e^{x} + 1\right ) - \frac{1}{4} i \, x^{2} \log \left (-i \, e^{x} + 1\right ) - \frac{1}{2} i \, x{\rm Li}_2\left (i \, e^{x}\right ) + \frac{1}{2} i \, x{\rm Li}_2\left (-i \, e^{x}\right ) + \frac{1}{2} i \,{\rm polylog}\left (3, i \, e^{x}\right ) - \frac{1}{2} i \,{\rm polylog}\left (3, -i \, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(e^x) + 1/4*I*x^2*log(I*e^x + 1) - 1/4*I*x^2*log(-I*e^x + 1) - 1/2*I*x*dilog(I*e^x) + 1/2*I*x*di
log(-I*e^x) + 1/2*I*polylog(3, I*e^x) - 1/2*I*polylog(3, -I*e^x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(exp(x)),x)

[Out]

Integral(x*atan(exp(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arctan(e^x), x)