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

Optimal. Leaf size=91 \[ -\frac{i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{2 b^2}+\frac{i \text{PolyLog}\left (3,i e^{a+b x}\right )}{2 b^2}+\frac{i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b}-\frac{i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0579984, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5143, 2531, 2282, 6589} \[ -\frac{i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{2 b^2}+\frac{i \text{PolyLog}\left (3,i e^{a+b x}\right )}{2 b^2}+\frac{i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b}-\frac{i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[E^(a + b*x)],x]

[Out]

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

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

Mathematica [A]  time = 0.0130628, size = 71, normalized size = 0.78 \[ \frac{i \left (b x \text{PolyLog}\left (2,-i e^{a+b x}\right )-b x \text{PolyLog}\left (2,i e^{a+b x}\right )-\text{PolyLog}\left (3,-i e^{a+b x}\right )+\text{PolyLog}\left (3,i e^{a+b x}\right )\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTan[E^(a + b*x)],x]

[Out]

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

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Maple [B]  time = 0.171, size = 349, normalized size = 3.8 \begin{align*}{\frac{{\frac{i}{2}}\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) xa}{b}}-{\frac{{\frac{i}{2}}{\it polylog} \left ( 2,i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-{\frac{{\frac{i}{2}}\ln \left ( -i \left ( -{{\rm e}^{bx+a}}+i \right ) \right ) xa}{b}}+{\frac{{\frac{i}{2}}{\it polylog} \left ( 2,-i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( -i \left ({{\rm e}^{bx+a}}+i \right ) \right ) a}{{b}^{2}}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( -i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\frac{i}{2}}{\it polylog} \left ( 3,i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{{\frac{i}{2}}\ln \left ( -i \left ({{\rm e}^{bx+a}}+i \right ) \right ) xa}{b}}-{\frac{{\frac{i}{2}}x{\it polylog} \left ( 2,i{{\rm e}^{bx+a}} \right ) }{b}}-{\frac{{\frac{i}{2}}\ln \left ( -i \left ( -{{\rm e}^{bx+a}}+i \right ) \right ){a}^{2}}{{b}^{2}}}-{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{{\frac{i}{2}}\ln \left ( -i{{\rm e}^{bx+a}} \right ) \ln \left ( -i \left ( -{{\rm e}^{bx+a}}+i \right ) \right ) a}{{b}^{2}}}+{\frac{{\frac{i}{2}}x{\it polylog} \left ( 2,-i{{\rm e}^{bx+a}} \right ) }{b}}-{\frac{{\frac{i}{2}}{\it polylog} \left ( 3,-i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{{\frac{i}{2}}\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) xa}{b}}+{\frac{{\frac{i}{2}}\ln \left ( -i \left ({{\rm e}^{bx+a}}+i \right ) \right ){a}^{2}}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I/b*ln(1+I*exp(b*x+a))*x*a-1/2*I/b^2*polylog(2,I*exp(b*x+a))*a-1/2*I/b*ln(-I*(-exp(b*x+a)+I))*x*a+1/2*I/b^
2*polylog(2,-I*exp(b*x+a))*a+1/2*I/b^2*dilog(-I*(exp(b*x+a)+I))*a+1/2*I/b^2*dilog(-I*exp(b*x+a))*a+1/2*I*polyl
og(3,I*exp(b*x+a))/b^2+1/2*I/b*ln(-I*(exp(b*x+a)+I))*x*a-1/2*I*x*polylog(2,I*exp(b*x+a))/b-1/2*I/b^2*ln(-I*(-e
xp(b*x+a)+I))*a^2-1/2*I/b^2*a^2*ln(1-I*exp(b*x+a))+1/2*I/b^2*ln(-I*exp(b*x+a))*ln(-I*(-exp(b*x+a)+I))*a+1/2*I*
x*polylog(2,-I*exp(b*x+a))/b-1/2*I*polylog(3,-I*exp(b*x+a))/b^2+1/2*I/b^2*a^2*ln(1+I*exp(b*x+a))-1/2*I/b*ln(1-
I*exp(b*x+a))*x*a+1/2*I/b^2*ln(-I*(exp(b*x+a)+I))*a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.33572, size = 431, normalized size = 4.74 \begin{align*} \frac{2 \, b^{2} x^{2} \arctan \left (e^{\left (b x + a\right )}\right ) - 2 i \, b x{\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) + 2 i \, b x{\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right ) - i \, a^{2} \log \left (e^{\left (b x + a\right )} + i\right ) + i \, a^{2} \log \left (e^{\left (b x + a\right )} - i\right ) +{\left (i \, b^{2} x^{2} - i \, a^{2}\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) +{\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) + 2 i \,{\rm polylog}\left (3, i \, e^{\left (b x + a\right )}\right ) - 2 i \,{\rm polylog}\left (3, -i \, e^{\left (b x + a\right )}\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*atan(exp(a)*exp(b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \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(x*arctan(exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arctan(e^(b*x + a)), x)