\(\int \frac {\log (1+x)}{1+x^2} \, dx\) [11]

Optimal result

Integrand size = 12, antiderivative size = 89 \[ \int \frac {\log (1+x)}{1+x^2} \, dx=-\frac {1}{2} i \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) (i-x)\right ) \log (1+x)+\frac {1}{2} i \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) (i+x)\right ) \log (1+x)-\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) (1+x)\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) (1+x)\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1+x)}{1+x^2} \, dx=-\frac {1}{2} i \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) (i-x)\right ) \log (1+x)+\frac {1}{2} i \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) (i+x)\right ) \log (1+x)-\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) (1+x)\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) (1+x)\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2856, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\log (x+1)}{x^2+1} \, dx\)

\(\Big \downarrow \) 2856

\(\displaystyle \int \left (\frac {i \log (x+1)}{2 (-x+i)}+\frac {i \log (x+1)}{2 (x+i)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) (x+1)\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) (x+1)\right )-\frac {1}{2} i \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) (-x+i)\right ) \log (x+1)+\frac {1}{2} i \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) (x+i)\right ) \log (x+1)\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2856

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) 
^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I 
GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}\)	\(70\)
default	\(-\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}\)	\(70\)
risch	\(-\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}\)	\(70\)
parts	\(-\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \ln \left (x +1\right ) \ln \left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}+\frac {i \left (x +1\right )}{2}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {x}{2}-\frac {i \left (x +1\right )}{2}\right )}{2}\)	\(70\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63 \[ \int \frac {\log (1+x)}{1+x^2} \, dx=\frac {1}{2} \, \arctan \left (\frac {1}{2} \, x + \frac {1}{2}, \frac {1}{2} \, x + \frac {1}{2}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \arctan \left (x\right ) \log \left (\frac {1}{2} \, x^{2} + x + \frac {1}{2}\right ) + \arctan \left (x\right ) \log \left (x + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, x + \frac {1}{2} i + \frac {1}{2}\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, x - \frac {1}{2} i + \frac {1}{2}\right ) \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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