\(\int \frac {\log (c (1+x^2)^n)}{1+x^2} \, dx\) [277]

Optimal result

Integrand size = 18, antiderivative size = 60 \[ \int \frac {\log \left (c \left (1+x^2\right )^n\right )}{1+x^2} \, dx=i n \arctan (x)^2+2 n \arctan (x) \log \left (\frac {2}{1+i x}\right )+\arctan (x) \log \left (c \left (1+x^2\right )^n\right )+i n \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {209, 2520, 5040, 4964, 2449, 2352} \[ \int \frac {\log \left (c \left (1+x^2\right )^n\right )}{1+x^2} \, dx=\arctan (x) \log \left (c \left (x^2+1\right )^n\right )+i n \arctan (x)^2+2 n \arctan (x) \log \left (\frac {2}{1+i x}\right )+i n \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right ) \]

[In]

Int[Log[c*(1 + x^2)^n]/(1 + x^2),x]

[Out]

I*n*ArcTan[x]^2 + 2*n*ArcTan[x]*Log[2/(1 + I*x)] + ArcTan[x]*Log[c*(1 + x^2)^n] + I*n*PolyLog[2, 1 - 2/(1 + I*
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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2520

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[u*(x^(n - 1)/(d + e*x^n)
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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 5040

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*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.03 \[ \int \frac {\log \left (c \left (1+x^2\right )^n\right )}{1+x^2} \, dx=i n \arctan (x)^2+2 n \arctan (x) \log \left (\frac {2 i}{i-x}\right )+\arctan (x) \log \left (c \left (1+x^2\right )^n\right )+i n \operatorname {PolyLog}\left (2,\frac {i+x}{-i+x}\right ) \]

[In]

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

[Out]

I*n*ArcTan[x]^2 + 2*n*ArcTan[x]*Log[(2*I)/(I - x)] + ArcTan[x]*Log[c*(1 + x^2)^n] + I*n*PolyLog[2, (I + x)/(-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. 128 vs. \(2 (56 ) = 112\).

Time = 1.38 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.15

method	result	size
parts	\(\arctan \left (x \right ) \ln \left (c \left (x^{2}+1\right )^{n}\right )-2 n \left (\frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{2}+\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{4}-\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{4}\right )\)	\(129\)
risch	\(\ln \left (\left (x^{2}+1\right )^{n}\right ) \arctan \left (x \right )-n \arctan \left (x \right ) \ln \left (x^{2}+1\right )-\frac {i n \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{2}+\frac {i n \ln \left (x -i\right )^{2}}{4}+\frac {i n \operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )}{2}+\frac {i n \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{2}+\frac {i n \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{2}-\frac {i n \ln \left (x +i\right )^{2}}{4}-\frac {i n \operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )}{2}-\frac {i n \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{2}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (x^{2}+1\right )^{n}\right ) {\operatorname {csgn}\left (i c \left (x^{2}+1\right )^{n}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (x^{2}+1\right )^{n}\right ) \operatorname {csgn}\left (i c \left (x^{2}+1\right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (x^{2}+1\right )^{n}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (x^{2}+1\right )^{n}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \arctan \left (x \right )\)	\(242\)

[In]

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

[Out]

arctan(x)*ln(c*(x^2+1)^n)-2*n*(1/2*arctan(x)*ln(x^2+1)+1/4*I*(ln(x-I)*ln(x^2+1)-1/2*ln(x-I)^2-dilog(-1/2*I*(x+
I))-ln(x-I)*ln(-1/2*I*(x+I)))-1/4*I*(ln(x+I)*ln(x^2+1)-1/2*ln(x+I)^2-dilog(1/2*I*(x-I))-ln(x+I)*ln(1/2*I*(x-I)
)))

Fricas [F]

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

[In]

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

[Out]

integral(log((x^2 + 1)^n*c)/(x^2 + 1), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(log((x^2 + 1)^n*c)/(x^2 + 1), x)

Giac [F]

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

[In]

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

[Out]

integrate(log((x^2 + 1)^n*c)/(x^2 + 1), x)

Mupad [F(-1)]

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

[In]

int(log(c*(x^2 + 1)^n)/(x^2 + 1),x)

[Out]

int(log(c*(x^2 + 1)^n)/(x^2 + 1), x)