\(\int \frac {\log (\frac {x^2}{1+x^2})}{1+x^2} \, dx\) [278]

Optimal result

Integrand size = 20, antiderivative size = 61 \[ \int \frac {\log \left (\frac {x^2}{1+x^2}\right )}{1+x^2} \, dx=i \arctan (x)^2-2 \arctan (x) \log \left (2-\frac {2}{1-i x}\right )+\arctan (x) \log \left (\frac {x^2}{1+x^2}\right )+i \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {209, 2606, 12, 5044, 4988, 2497} \[ \int \frac {\log \left (\frac {x^2}{1+x^2}\right )}{1+x^2} \, dx=\arctan (x) \log \left (\frac {x^2}{x^2+1}\right )+i \arctan (x)^2-2 \arctan (x) \log \left (2-\frac {2}{1-i x}\right )+i \operatorname {PolyLog}\left (2,\frac {2}{1-i x}-1\right ) \]

[In]

Int[Log[x^2/(1 + x^2)]/(1 + x^2),x]

[Out]

I*ArcTan[x]^2 - 2*ArcTan[x]*Log[2 - 2/(1 - I*x)] + ArcTan[x]*Log[x^2/(1 + x^2)] + I*PolyLog[2, -1 + 2/(1 - I*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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 2606

Int[Log[(c_.)*(RFx_)^(n_.)]/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2), x]}, Simp[u*L
og[c*RFx^n], x] - Dist[n, Int[SimplifyIntegrand[u*(D[RFx, x]/RFx), x], x], x]] /; FreeQ[{c, d, e, n}, x] && Ra
tionalFunctionQ[RFx, x] &&  !PolynomialQ[RFx, x]

Rule 4988

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])
^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))
]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 5044

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*d*(p + 1))), x] + Dist[I/d, Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(239\) vs. \(2(61)=122\).

Time = 0.04 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.92 \[ \int \frac {\log \left (\frac {x^2}{1+x^2}\right )}{1+x^2} \, dx=-\frac {1}{4} i \log ^2(i-x)+i \log (i-x) \log (-i x)-\frac {1}{2} i \log (i-x) \log \left (-\frac {1}{2} i (i+x)\right )+\frac {1}{2} i \log \left (-\frac {1}{2} i (i-x)\right ) \log (i+x)-i \log (i x) \log (i+x)+\frac {1}{4} i \log ^2(i+x)-\frac {1}{2} i \log (i-x) \log \left (\frac {x^2}{1+x^2}\right )+\frac {1}{2} i \log (i+x) \log \left (\frac {x^2}{1+x^2}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {1}{2} i (i-x)\right )+i \operatorname {PolyLog}(2,-i (i-x))+\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {1}{2} i (i+x)\right )-i \operatorname {PolyLog}(2,-i (i+x)) \]

[In]

Integrate[Log[x^2/(1 + x^2)]/(1 + x^2),x]

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (57 ) = 114\).

Time = 2.06 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.39

method	result	size
default	\(-\frac {i \left (\ln \left (x -i\right ) \ln \left (\frac {x^{2}}{x^{2}+1}\right )-2 \operatorname {dilog}\left (-i x \right )-2 \ln \left (x -i\right ) \ln \left (-i x \right )+\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )+\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )+\frac {\ln \left (x -i\right )^{2}}{2}\right )}{2}+\frac {i \left (\ln \left (x +i\right ) \ln \left (\frac {x^{2}}{x^{2}+1}\right )-2 \operatorname {dilog}\left (i x \right )-2 \ln \left (x +i\right ) \ln \left (i x \right )+\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )+\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )+\frac {\ln \left (x +i\right )^{2}}{2}\right )}{2}\)	\(146\)
risch	\(-\frac {i \ln \left (x -i\right ) \ln \left (\frac {x^{2}}{x^{2}+1}\right )}{2}+i \operatorname {dilog}\left (-i x \right )+i \ln \left (x -i\right ) \ln \left (-i x \right )-\frac {i \operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )}{2}-\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{2}-\frac {i \ln \left (x -i\right )^{2}}{4}+\frac {i \ln \left (x +i\right ) \ln \left (\frac {x^{2}}{x^{2}+1}\right )}{2}-i \operatorname {dilog}\left (i x \right )-i \ln \left (x +i\right ) \ln \left (i x \right )+\frac {i \operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )}{2}+\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{2}+\frac {i \ln \left (x +i\right )^{2}}{4}\)	\(158\)
parts	\(\arctan \left (x \right ) \ln \left (\frac {x^{2}}{x^{2}+1}\right )-2 \arctan \left (x \right ) \ln \left (x \right )+\arctan \left (x \right ) \ln \left (x^{2}+1\right )-i \ln \left (x \right ) \ln \left (i x +1\right )+i \ln \left (x \right ) \ln \left (-i x +1\right )-i \operatorname {dilog}\left (i x +1\right )+i \operatorname {dilog}\left (-i x +1\right )+\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{2}-\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{2}\)	\(176\)

[In]

int(ln(x^2/(x^2+1))/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

-1/2*I*(ln(x-I)*ln(x^2/(x^2+1))-2*dilog(-I*x)-2*ln(x-I)*ln(-I*x)+dilog(-1/2*I*(x+I))+ln(x-I)*ln(-1/2*I*(x+I))+
1/2*ln(x-I)^2)+1/2*I*(ln(x+I)*ln(x^2/(x^2+1))-2*dilog(I*x)-2*ln(x+I)*ln(I*x)+dilog(1/2*I*(x-I))+ln(x+I)*ln(1/2
*I*(x-I))+1/2*ln(x+I)^2)

Fricas [F]

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

[In]

integrate(log(x^2/(x^2+1))/(x^2+1),x, algorithm="fricas")

[Out]

integral(log(x^2/(x^2 + 1))/(x^2 + 1), x)

Sympy [F]

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

[In]

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

[Out]

Integral(log(x**2/(x**2 + 1))/(x**2 + 1), x)

Maxima [F]

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

[In]

integrate(log(x^2/(x^2+1))/(x^2+1),x, algorithm="maxima")

[Out]

integrate(log(x^2/(x^2 + 1))/(x^2 + 1), x)

Giac [F]

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

[In]

integrate(log(x^2/(x^2+1))/(x^2+1),x, algorithm="giac")

[Out]

integrate(log(x^2/(x^2 + 1))/(x^2 + 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (\frac {x^2}{1+x^2}\right )}{1+x^2} \, dx=\int \frac {\ln \left (\frac {x^2}{x^2+1}\right )}{x^2+1} \,d x \]

[In]

int(log(x^2/(x^2 + 1))/(x^2 + 1),x)

[Out]

int(log(x^2/(x^2 + 1))/(x^2 + 1), x)