\(\int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx\) [126]

Optimal result

Integrand size = 12, antiderivative size = 35 \[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=-\frac {1}{4} i \operatorname {PolyLog}\left (2,-\frac {i}{1+x}\right )+\frac {1}{4} i \operatorname {PolyLog}\left (2,\frac {i}{1+x}\right ) \]

[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 = 35, 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 = {5152, 12, 4941, 2438} \[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=\frac {1}{4} i \operatorname {PolyLog}\left (2,\frac {i}{x+1}\right )-\frac {1}{4} i \operatorname {PolyLog}\left (2,-\frac {i}{x+1}\right ) \]

[In]

Int[ArcCot[1 + x]/(2 + 2*x),x]

[Out]

(-1/4*I)*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 4941

Int[((a_.) + ArcCot[(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 5152

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[(f*(x/d))^m*(a + b*ArcCot[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 {\cot ^{-1}(x)}{2 x} \, dx,x,1+x\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {\cot ^{-1}(x)}{x} \, dx,x,1+x\right ) \\ & = \frac {1}{4} i \text {Subst}\left (\int \frac {\log \left (1-\frac {i}{x}\right )}{x} \, dx,x,1+x\right )-\frac {1}{4} i \text {Subst}\left (\int \frac {\log \left (1+\frac {i}{x}\right )}{x} \, dx,x,1+x\right ) \\ & = -\frac {1}{4} i \operatorname {PolyLog}\left (2,-\frac {i}{1+x}\right )+\frac {1}{4} i \operatorname {PolyLog}\left (2,\frac {i}{1+x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=-\frac {1}{4} i \operatorname {PolyLog}\left (2,-\frac {i}{1+x}\right )+\frac {1}{4} i \operatorname {PolyLog}\left (2,\frac {i}{1+x}\right ) \]

[In]

Integrate[ArcCot[1 + x]/(2 + 2*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {i \operatorname {dilog}\left (-i x -i+1\right )}{4}+\frac {\pi \ln \left (-i x -i\right )}{4}-\frac {i \operatorname {dilog}\left (i x +i+1\right )}{4}\)	\(37\)
derivativedivides	\(\frac {\ln \left (1+x \right ) \operatorname {arccot}\left (1+x \right )}{2}-\frac {i \ln \left (1+x \right ) \ln \left (1+i \left (1+x \right )\right )}{4}+\frac {i \ln \left (1+x \right ) \ln \left (1-i \left (1+x \right )\right )}{4}-\frac {i \operatorname {dilog}\left (1+i \left (1+x \right )\right )}{4}+\frac {i \operatorname {dilog}\left (1-i \left (1+x \right )\right )}{4}\)	\(68\)
default	\(\frac {\ln \left (1+x \right ) \operatorname {arccot}\left (1+x \right )}{2}-\frac {i \ln \left (1+x \right ) \ln \left (1+i \left (1+x \right )\right )}{4}+\frac {i \ln \left (1+x \right ) \ln \left (1-i \left (1+x \right )\right )}{4}-\frac {i \operatorname {dilog}\left (1+i \left (1+x \right )\right )}{4}+\frac {i \operatorname {dilog}\left (1-i \left (1+x \right )\right )}{4}\)	\(68\)
parts	\(\frac {\ln \left (1+x \right ) \operatorname {arccot}\left (1+x \right )}{2}-\frac {i \ln \left (1+x \right ) \ln \left (1+i \left (1+x \right )\right )}{4}+\frac {i \ln \left (1+x \right ) \ln \left (1-i \left (1+x \right )\right )}{4}-\frac {i \operatorname {dilog}\left (1+i \left (1+x \right )\right )}{4}+\frac {i \operatorname {dilog}\left (1-i \left (1+x \right )\right )}{4}\)	\(68\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=\int { \frac {\operatorname {arccot}\left (x + 1\right )}{2 \, {\left (x + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=\frac {\int \frac {\operatorname {acot}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \]

[In]

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

[Out]

Integral(acot(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. 64 vs. \(2 (21) = 42\).

Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.83 \[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=\frac {1}{4} \, \arctan \left (x + 1, 0\right ) \log \left (x^{2} + 2 \, x + 2\right ) + \frac {1}{2} \, \operatorname {arccot}\left (x + 1\right ) \log \left (x + 1\right ) + \frac {1}{2} \, \arctan \left (x + 1\right ) \log \left (x + 1\right ) - \frac {1}{2} \, \arctan \left (x + 1\right ) \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} i \, {\rm Li}_2\left (i \, x + i + 1\right ) - \frac {1}{4} i \, {\rm Li}_2\left (-i \, x - i + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=-\frac {1}{4} \, {\left (x + 1\right )}^{2} \arctan \left (\frac {1}{x + 1}\right ) - \frac {1}{4} \, x - \frac {1}{4} \, \arctan \left (\frac {1}{x + 1}\right ) - \frac {1}{4} \]

[In]

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

[Out]

-1/4*(x + 1)^2*arctan(1/(x + 1)) - 1/4*x - 1/4*arctan(1/(x + 1)) - 1/4

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx=\int \frac {\mathrm {acot}\left (x+1\right )}{2\,x+2} \,d x \]

[In]

int(acot(x + 1)/(2*x + 2),x)

[Out]

int(acot(x + 1)/(2*x + 2), x)