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

Optimal result

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

[Out]

-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))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2456, 2441, 2440, 2438} \[ \int \frac {\log (1+x)}{1+x^2} \, dx=-\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) (x+1)\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) (x+1)\right )-\frac {1}{2} i \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) (-x+i)\right ) \log (x+1)+\frac {1}{2} i \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) (x+i)\right ) \log (x+1) \]

[In]

Int[Log[1 + x]/(1 + x^2),x]

[Out]

(-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)]

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 2440

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

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2456

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[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]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

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

Mathematica [A] (verified)

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

[In]

Integrate[Log[1 + x]/(1 + x^2),x]

[Out]

(-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)]

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.79

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(log(x + 1)/(x^2 + 1), x)

Sympy [F]

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

[In]

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

[Out]

Integral(log(x + 1)/(x**2 + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63 \[ \int \frac {\log (1+x)}{1+x^2} \, dx=\frac {1}{2} \, \arctan \left (\frac {1}{2} \, x + \frac {1}{2}, \frac {1}{2} \, x + \frac {1}{2}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \arctan \left (x\right ) \log \left (\frac {1}{2} \, x^{2} + x + \frac {1}{2}\right ) + \arctan \left (x\right ) \log \left (x + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, x + \frac {1}{2} i + \frac {1}{2}\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, x - \frac {1}{2} i + \frac {1}{2}\right ) \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

integrate(log(x + 1)/(x^2 + 1), x)

Mupad [F(-1)]

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

[In]

int(log(x + 1)/(x^2 + 1),x)

[Out]

int(log(x + 1)/(x^2 + 1), x)