\(\int \frac {\arctan (1+x)}{2+2 x} \, dx\) [21]

Optimal result

Integrand size = 12, antiderivative size = 31 \[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=\frac {1}{4} i \operatorname {PolyLog}(2,-i (1+x))-\frac {1}{4} i \operatorname {PolyLog}(2,i (1+x)) \]

[Out]

1/4*I*polylog(2,-I*(1+x))-1/4*I*polylog(2,I*(1+x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5151, 12, 4940, 2438} \[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=\frac {1}{4} i \operatorname {PolyLog}(2,-i (x+1))-\frac {1}{4} i \operatorname {PolyLog}(2,i (x+1)) \]

[In]

Int[ArcTan[1 + x]/(2 + 2*x),x]

[Out]

(I/4)*PolyLog[2, (-I)*(1 + x)] - (I/4)*PolyLog[2, I*(1 + x)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2438

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 4940

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 5151

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[(f*(x/d))^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\arctan (x)}{2 x} \, dx,x,1+x\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {\arctan (x)}{x} \, dx,x,1+x\right ) \\ & = \frac {1}{4} i \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,1+x\right )-\frac {1}{4} i \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,1+x\right ) \\ & = \frac {1}{4} i \operatorname {PolyLog}(2,-i (1+x))-\frac {1}{4} i \operatorname {PolyLog}(2,i (1+x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=\frac {1}{4} i \operatorname {PolyLog}(2,-i (1+x))-\frac {1}{4} i \operatorname {PolyLog}(2,i (1+x)) \]

[In]

Integrate[ArcTan[1 + x]/(2 + 2*x),x]

[Out]

(I/4)*PolyLog[2, (-I)*(1 + x)] - (I/4)*PolyLog[2, I*(1 + x)]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {i \operatorname {dilog}\left (-i x -i+1\right )}{4}+\frac {i \operatorname {dilog}\left (i x +i+1\right )}{4}\)	\(26\)
derivativedivides	\(\frac {\ln \left (1+x \right ) \arctan \left (1+x \right )}{2}+\frac {i \ln \left (1+x \right ) \ln \left (1+i \left (1+x \right )\right )}{4}-\frac {i \ln \left (1+x \right ) \ln \left (1-i \left (1+x \right )\right )}{4}+\frac {i \operatorname {dilog}\left (1+i \left (1+x \right )\right )}{4}-\frac {i \operatorname {dilog}\left (1-i \left (1+x \right )\right )}{4}\)	\(68\)
default	\(\frac {\ln \left (1+x \right ) \arctan \left (1+x \right )}{2}+\frac {i \ln \left (1+x \right ) \ln \left (1+i \left (1+x \right )\right )}{4}-\frac {i \ln \left (1+x \right ) \ln \left (1-i \left (1+x \right )\right )}{4}+\frac {i \operatorname {dilog}\left (1+i \left (1+x \right )\right )}{4}-\frac {i \operatorname {dilog}\left (1-i \left (1+x \right )\right )}{4}\)	\(68\)
parts	\(\frac {\ln \left (1+x \right ) \arctan \left (1+x \right )}{2}+\frac {i \ln \left (1+x \right ) \ln \left (1+i \left (1+x \right )\right )}{4}-\frac {i \ln \left (1+x \right ) \ln \left (1-i \left (1+x \right )\right )}{4}+\frac {i \operatorname {dilog}\left (1+i \left (1+x \right )\right )}{4}-\frac {i \operatorname {dilog}\left (1-i \left (1+x \right )\right )}{4}\)	\(68\)

[In]

int(arctan(1+x)/(2+2*x),x,method=_RETURNVERBOSE)

[Out]

-1/4*I*dilog(-I*x+1-I)+1/4*I*dilog(I*x+1+I)

Fricas [F]

\[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=\int { \frac {\arctan \left (x + 1\right )}{2 \, {\left (x + 1\right )}} \,d x } \]

[In]

integrate(arctan(1+x)/(2+2*x),x, algorithm="fricas")

[Out]

integral(1/2*arctan(x + 1)/(x + 1), x)

Sympy [F]

\[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=\frac {\int \frac {\operatorname {atan}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \]

[In]

integrate(atan(1+x)/(2+2*x),x)

[Out]

Integral(atan(x + 1)/(x + 1), x)/2

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (17) = 34\).

Time = 0.32 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=-\frac {1}{4} \, \arctan \left (x + 1, 0\right ) \log \left (x^{2} + 2 \, x + 2\right ) + \frac {1}{2} \, \arctan \left (x + 1\right ) \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{4} i \, {\rm Li}_2\left (i \, x + i + 1\right ) + \frac {1}{4} i \, {\rm Li}_2\left (-i \, x - i + 1\right ) \]

[In]

integrate(arctan(1+x)/(2+2*x),x, algorithm="maxima")

[Out]

-1/4*arctan2(x + 1, 0)*log(x^2 + 2*x + 2) + 1/2*arctan(x + 1)*log(abs(x + 1)) - 1/4*I*dilog(I*x + I + 1) + 1/4
*I*dilog(-I*x - I + 1)

Giac [F]

\[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=\int { \frac {\arctan \left (x + 1\right )}{2 \, {\left (x + 1\right )}} \,d x } \]

[In]

integrate(arctan(1+x)/(2+2*x),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {\arctan (1+x)}{2+2 x} \, dx=-\frac {{\mathrm {Li}}_{\mathrm {2}}\left (1-x\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\frac {{\mathrm {Li}}_{\mathrm {2}}\left (x\,1{}\mathrm {i}+1+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

[In]

int(atan(x + 1)/(2*x + 2),x)

[Out]

(dilog(x*1i + (1 + 1i))*1i)/4 - (dilog((1 - 1i) - x*1i)*1i)/4