3.21 \(\int \frac {\tan ^{-1}(1+x)}{2+2 x} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{4} i \text {Li}_2(-i (x+1))-\frac {1}{4} i \text {Li}_2(i (x+1)) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5043, 12, 4848, 2391} \[ \frac {1}{4} i \text {PolyLog}(2,-i (x+1))-\frac {1}{4} i \text {PolyLog}(2,i (x+1)) \]

Antiderivative was successfully verified.

[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 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]

Rule 4848

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 5043

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

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 31, normalized size = 1.00 \[ \frac {1}{4} i \text {Li}_2(-i (x+1))-\frac {1}{4} i \text {Li}_2(i (x+1)) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[1 + x]/(2 + 2*x),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (x + 1\right )}{2 \, {\left (x + 1\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

maple [B]  time = 0.06, size = 68, normalized size = 2.19 \[ \frac {\ln \left (x +1\right ) \arctan \left (x +1\right )}{2}+\frac {i \ln \left (x +1\right ) \ln \left (1+i \left (x +1\right )\right )}{4}-\frac {i \ln \left (x +1\right ) \ln \left (1-i \left (x +1\right )\right )}{4}+\frac {i \dilog \left (1+i \left (x +1\right )\right )}{4}-\frac {i \dilog \left (1-i \left (x +1\right )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(x+1)*arctan(x+1)+1/4*I*ln(x+1)*ln(1+I*(x+1))-1/4*I*ln(x+1)*ln(1-I*(x+1))+1/4*I*dilog(1+I*(x+1))-1/4*I*d
ilog(1-I*(x+1))

________________________________________________________________________________________

maxima [B]  time = 0.46, size = 44, normalized size = 1.42 \[ -\frac {1}{4} \, \arctan \left (x + 1, 0\right ) \log \left (x^{2} + 2 \, x + 2\right ) + \frac {1}{2} \, \arctan \left (x + 1\right ) \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{4} i \, {\rm Li}_2\left (i \, x + i + 1\right ) + \frac {1}{4} i \, {\rm Li}_2\left (-i \, x - i + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 0.08, size = 25, normalized size = 0.81 \[ -\frac {{\mathrm {Li}}_{\mathrm {2}}\left (1-x\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\frac {{\mathrm {Li}}_{\mathrm {2}}\left (x\,1{}\mathrm {i}+1+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x + 1)/(2*x + 2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(x + 1)/(x + 1), x)/2

________________________________________________________________________________________