3.33 \(\int \tan ^{-1}(2 \tan (x)) \, dx\)

Optimal. Leaf size=80 \[ -\frac{1}{4} \text{PolyLog}\left (2,\frac{1}{3} e^{2 i x}\right )+\frac{1}{4} \text{PolyLog}\left (2,3 e^{2 i x}\right )+\frac{1}{2} i x \log \left (1-3 e^{2 i x}\right )-\frac{1}{2} i x \log \left (1-\frac{1}{3} e^{2 i x}\right )+x \tan ^{-1}(2 \tan (x)) \]

[Out]

x*ArcTan[2*Tan[x]] + (I/2)*x*Log[1 - 3*E^((2*I)*x)] - (I/2)*x*Log[1 - E^((2*I)*x)/3] - PolyLog[2, E^((2*I)*x)/
3]/4 + PolyLog[2, 3*E^((2*I)*x)]/4

________________________________________________________________________________________

Rubi [A]  time = 0.0832127, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 5, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5167, 2190, 2279, 2391} \[ -\frac{1}{4} \text{PolyLog}\left (2,\frac{1}{3} e^{2 i x}\right )+\frac{1}{4} \text{PolyLog}\left (2,3 e^{2 i x}\right )+\frac{1}{2} i x \log \left (1-3 e^{2 i x}\right )-\frac{1}{2} i x \log \left (1-\frac{1}{3} e^{2 i x}\right )+x \tan ^{-1}(2 \tan (x)) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[2*Tan[x]],x]

[Out]

x*ArcTan[2*Tan[x]] + (I/2)*x*Log[1 - 3*E^((2*I)*x)] - (I/2)*x*Log[1 - E^((2*I)*x)/3] - PolyLog[2, E^((2*I)*x)/
3]/4 + PolyLog[2, 3*E^((2*I)*x)]/4

Rule 5167

Int[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcTan[c + d*Tan[a + b*x]], x] + (Dist[
b*(1 - I*c - d), Int[(x*E^(2*I*a + 2*I*b*x))/(1 - I*c + d + (1 - I*c - d)*E^(2*I*a + 2*I*b*x)), x], x] - Dist[
b*(1 + I*c + d), Int[(x*E^(2*I*a + 2*I*b*x))/(1 + I*c - d + (1 + I*c + d)*E^(2*I*a + 2*I*b*x)), x], x]) /; Fre
eQ[{a, b, c, d}, x] && NeQ[(c + I*d)^2, -1]

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

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}(2 \tan (x)) \, dx &=x \tan ^{-1}(2 \tan (x))-3 \int \frac{e^{2 i x} x}{-1+3 e^{2 i x}} \, dx-\int \frac{e^{2 i x} x}{3-e^{2 i x}} \, dx\\ &=x \tan ^{-1}(2 \tan (x))+\frac{1}{2} i x \log \left (1-3 e^{2 i x}\right )-\frac{1}{2} i x \log \left (1-\frac{1}{3} e^{2 i x}\right )-\frac{1}{2} i \int \log \left (1-3 e^{2 i x}\right ) \, dx+\frac{1}{2} i \int \log \left (1-\frac{1}{3} e^{2 i x}\right ) \, dx\\ &=x \tan ^{-1}(2 \tan (x))+\frac{1}{2} i x \log \left (1-3 e^{2 i x}\right )-\frac{1}{2} i x \log \left (1-\frac{1}{3} e^{2 i x}\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log (1-3 x)}{x} \, dx,x,e^{2 i x}\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{3}\right )}{x} \, dx,x,e^{2 i x}\right )\\ &=x \tan ^{-1}(2 \tan (x))+\frac{1}{2} i x \log \left (1-3 e^{2 i x}\right )-\frac{1}{2} i x \log \left (1-\frac{1}{3} e^{2 i x}\right )-\frac{1}{4} \text{Li}_2\left (\frac{1}{3} e^{2 i x}\right )+\frac{1}{4} \text{Li}_2\left (3 e^{2 i x}\right )\\ \end{align*}

Mathematica [B]  time = 0.235709, size = 262, normalized size = 3.28 \[ x \tan ^{-1}(2 \tan (x))-\frac{1}{4} i \left (i \left (\text{PolyLog}\left (2,\frac{2 \tan (x)-i}{6 \tan (x)+3 i}\right )-\text{PolyLog}\left (2,\frac{6 \tan (x)-3 i}{2 \tan (x)+i}\right )\right )+2 i \cos ^{-1}\left (\frac{5}{3}\right ) \tan ^{-1}(2 \tan (x))+4 i x \tan ^{-1}\left (\frac{\cot (x)}{2}\right )-\log \left (\frac{-4 \tan (x)+4 i}{2 \tan (x)+i}\right ) \left (\cos ^{-1}\left (\frac{5}{3}\right )-2 \tan ^{-1}(2 \tan (x))\right )-\log \left (\frac{4 (\tan (x)+i)}{6 \tan (x)+3 i}\right ) \left (2 \tan ^{-1}(2 \tan (x))+\cos ^{-1}\left (\frac{5}{3}\right )\right )+\log \left (\frac{2 i \sqrt{\frac{2}{3}} e^{-i x}}{\sqrt{3 \cos (2 x)-5}}\right ) \left (2 \tan ^{-1}(2 \tan (x))+2 \tan ^{-1}\left (\frac{\cot (x)}{2}\right )+\cos ^{-1}\left (\frac{5}{3}\right )\right )+\log \left (\frac{2 i \sqrt{\frac{2}{3}} e^{i x}}{\sqrt{3 \cos (2 x)-5}}\right ) \left (-2 \tan ^{-1}(2 \tan (x))-2 \tan ^{-1}\left (\frac{\cot (x)}{2}\right )+\cos ^{-1}\left (\frac{5}{3}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[2*Tan[x]],x]

[Out]

x*ArcTan[2*Tan[x]] - (I/4)*((4*I)*x*ArcTan[Cot[x]/2] + (2*I)*ArcCos[5/3]*ArcTan[2*Tan[x]] + (ArcCos[5/3] + 2*A
rcTan[Cot[x]/2] + 2*ArcTan[2*Tan[x]])*Log[((2*I)*Sqrt[2/3])/(E^(I*x)*Sqrt[-5 + 3*Cos[2*x]])] + (ArcCos[5/3] -
2*ArcTan[Cot[x]/2] - 2*ArcTan[2*Tan[x]])*Log[((2*I)*Sqrt[2/3]*E^(I*x))/Sqrt[-5 + 3*Cos[2*x]]] - (ArcCos[5/3] -
2*ArcTan[2*Tan[x]])*Log[(4*I - 4*Tan[x])/(I + 2*Tan[x])] - (ArcCos[5/3] + 2*ArcTan[2*Tan[x]])*Log[(4*(I + Tan
[x]))/(3*I + 6*Tan[x])] + I*(-PolyLog[2, (-3*I + 6*Tan[x])/(I + 2*Tan[x])] + PolyLog[2, (-I + 2*Tan[x])/(3*I +
6*Tan[x])]))

________________________________________________________________________________________

Maple [A]  time = 0.097, size = 113, normalized size = 1.4 \begin{align*} \arctan \left ( 2\,\tan \left ( x \right ) \right ) \arctan \left ( \tan \left ( x \right ) \right ) -{\frac{i}{2}}\arctan \left ( \tan \left ( x \right ) \right ) \ln \left ( 1-{\frac{ \left ( 1+i\tan \left ( x \right ) \right ) ^{2}}{3\, \left ( \tan \left ( x \right ) \right ) ^{2}+3}} \right ) -{\frac{1}{4}{\it polylog} \left ( 2,{\frac{ \left ( 1+i\tan \left ( x \right ) \right ) ^{2}}{3\, \left ( \tan \left ( x \right ) \right ) ^{2}+3}} \right ) }+{\frac{i}{2}}\arctan \left ( \tan \left ( x \right ) \right ) \ln \left ( 1-3\,{\frac{ \left ( 1+i\tan \left ( x \right ) \right ) ^{2}}{ \left ( \tan \left ( x \right ) \right ) ^{2}+1}} \right ) +{\frac{1}{4}{\it polylog} \left ( 2,3\,{\frac{ \left ( 1+i\tan \left ( x \right ) \right ) ^{2}}{ \left ( \tan \left ( x \right ) \right ) ^{2}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(2*tan(x)),x)

[Out]

arctan(2*tan(x))*arctan(tan(x))-1/2*I*arctan(tan(x))*ln(1-1/3*(1+I*tan(x))^2/(tan(x)^2+1))-1/4*polylog(2,1/3*(
1+I*tan(x))^2/(tan(x)^2+1))+1/2*I*arctan(tan(x))*ln(1-3*(1+I*tan(x))^2/(tan(x)^2+1))+1/4*polylog(2,3*(1+I*tan(
x))^2/(tan(x)^2+1))

________________________________________________________________________________________

Maxima [A]  time = 1.48328, size = 113, normalized size = 1.41 \begin{align*} x \arctan \left (2 \, \tan \left (x\right )\right ) - \frac{1}{8} \, \log \left (4 \, \tan \left (x\right )^{2} + 4\right ) \log \left (4 \, \tan \left (x\right )^{2} + 1\right ) + \frac{1}{8} \, \log \left (4 \, \tan \left (x\right )^{2} + 1\right ) \log \left (\frac{4}{9} \, \tan \left (x\right )^{2} + \frac{4}{9}\right ) - \frac{1}{4} \,{\rm Li}_2\left (2 i \, \tan \left (x\right ) - 1\right ) + \frac{1}{4} \,{\rm Li}_2\left (\frac{2}{3} i \, \tan \left (x\right ) + \frac{1}{3}\right ) + \frac{1}{4} \,{\rm Li}_2\left (-\frac{2}{3} i \, \tan \left (x\right ) + \frac{1}{3}\right ) - \frac{1}{4} \,{\rm Li}_2\left (-2 i \, \tan \left (x\right ) - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(2*tan(x)),x, algorithm="maxima")

[Out]

x*arctan(2*tan(x)) - 1/8*log(4*tan(x)^2 + 4)*log(4*tan(x)^2 + 1) + 1/8*log(4*tan(x)^2 + 1)*log(4/9*tan(x)^2 +
4/9) - 1/4*dilog(2*I*tan(x) - 1) + 1/4*dilog(2/3*I*tan(x) + 1/3) + 1/4*dilog(-2/3*I*tan(x) + 1/3) - 1/4*dilog(
-2*I*tan(x) - 1)

________________________________________________________________________________________

Fricas [B]  time = 2.47376, size = 713, normalized size = 8.91 \begin{align*} x \arctan \left (2 \, \tan \left (x\right )\right ) - \frac{1}{4} i \, x \log \left (\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} + 3 i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1}\right ) + \frac{1}{4} i \, x \log \left (\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} + i \, \tan \left (x\right ) + 1\right )}}{3 \,{\left (\tan \left (x\right )^{2} + 1\right )}}\right ) - \frac{1}{4} i \, x \log \left (\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} - i \, \tan \left (x\right ) + 1\right )}}{3 \,{\left (\tan \left (x\right )^{2} + 1\right )}}\right ) + \frac{1}{4} i \, x \log \left (\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} - 3 i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1}\right ) + \frac{1}{8} \,{\rm Li}_2\left (-\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} + 3 i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1} + 1\right ) - \frac{1}{8} \,{\rm Li}_2\left (-\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} + i \, \tan \left (x\right ) + 1\right )}}{3 \,{\left (\tan \left (x\right )^{2} + 1\right )}} + 1\right ) - \frac{1}{8} \,{\rm Li}_2\left (-\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} - i \, \tan \left (x\right ) + 1\right )}}{3 \,{\left (\tan \left (x\right )^{2} + 1\right )}} + 1\right ) + \frac{1}{8} \,{\rm Li}_2\left (-\frac{2 \,{\left (2 \, \tan \left (x\right )^{2} - 3 i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(2*tan(x)),x, algorithm="fricas")

[Out]

x*arctan(2*tan(x)) - 1/4*I*x*log(2*(2*tan(x)^2 + 3*I*tan(x) - 1)/(tan(x)^2 + 1)) + 1/4*I*x*log(2/3*(2*tan(x)^2
+ I*tan(x) + 1)/(tan(x)^2 + 1)) - 1/4*I*x*log(2/3*(2*tan(x)^2 - I*tan(x) + 1)/(tan(x)^2 + 1)) + 1/4*I*x*log(2
*(2*tan(x)^2 - 3*I*tan(x) - 1)/(tan(x)^2 + 1)) + 1/8*dilog(-2*(2*tan(x)^2 + 3*I*tan(x) - 1)/(tan(x)^2 + 1) + 1
) - 1/8*dilog(-2/3*(2*tan(x)^2 + I*tan(x) + 1)/(tan(x)^2 + 1) + 1) - 1/8*dilog(-2/3*(2*tan(x)^2 - I*tan(x) + 1
)/(tan(x)^2 + 1) + 1) + 1/8*dilog(-2*(2*tan(x)^2 - 3*I*tan(x) - 1)/(tan(x)^2 + 1) + 1)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{atan}{\left (2 \tan{\left (x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(2*tan(x)),x)

[Out]

Integral(atan(2*tan(x)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \arctan \left (2 \, \tan \left (x\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(2*tan(x)),x, algorithm="giac")

[Out]

integrate(arctan(2*tan(x)), x)