\(\int \frac {x}{(-1+x) (1+x^2)} \, dx\) [15]

Optimal result

Integrand size = 14, antiderivative size = 27 \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {\arctan (x)}{2}+\frac {1}{2} \log (1-x)-\frac {1}{4} \log \left (1+x^2\right ) \]

[Out]

1/2*arctan(x)+1/2*ln(1-x)-1/4*ln(x^2+1)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {815, 649, 209, 266} \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {\arctan (x)}{2}-\frac {1}{4} \log \left (x^2+1\right )+\frac {1}{2} \log (1-x) \]

[In]

Int[x/((-1 + x)*(1 + x^2)),x]

[Out]

ArcTan[x]/2 + Log[1 - x]/2 - Log[1 + x^2]/4

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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2 (-1+x)}+\frac {1-x}{2 \left (1+x^2\right )}\right ) \, dx \\ & = \frac {1}{2} \log (1-x)+\frac {1}{2} \int \frac {1-x}{1+x^2} \, dx \\ & = \frac {1}{2} \log (1-x)+\frac {1}{2} \int \frac {1}{1+x^2} \, dx-\frac {1}{2} \int \frac {x}{1+x^2} \, dx \\ & = \frac {\arctan (x)}{2}+\frac {1}{2} \log (1-x)-\frac {1}{4} \log \left (1+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {\arctan (x)}{2}+\frac {1}{2} \log (1-x)-\frac {1}{4} \log \left (1+x^2\right ) \]

[In]

Integrate[x/((-1 + x)*(1 + x^2)),x]

[Out]

ArcTan[x]/2 + Log[1 - x]/2 - Log[1 + x^2]/4

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}+\frac {\arctan \left (x \right )}{2}\)	\(20\)
risch	\(\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}+\frac {\arctan \left (x \right )}{2}\)	\(20\)
parallelrisch	\(\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (x -i\right )}{4}-\frac {i \ln \left (x -i\right )}{4}-\frac {\ln \left (x +i\right )}{4}+\frac {i \ln \left (x +i\right )}{4}\)	\(38\)

[In]

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

[Out]

1/2*ln(-1+x)-1/4*ln(x^2+1)+1/2*arctan(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

1/2*arctan(x) - 1/4*log(x^2 + 1) + 1/2*log(x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {\log {\left (x - 1 \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{4} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

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

[Out]

log(x - 1)/2 - log(x**2 + 1)/4 + atan(x)/2

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

1/2*arctan(x) - 1/4*log(x^2 + 1) + 1/2*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

1/2*arctan(x) - 1/4*log(x^2 + 1) + 1/2*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {x}{(-1+x) \left (1+x^2\right )} \, dx=\frac {\ln \left (x-1\right )}{2}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]

[In]

int(x/((x^2 + 1)*(x - 1)),x)

[Out]

log(x - 1)/2 - log(x - 1i)*(1/4 + 1i/4) - log(x + 1i)*(1/4 - 1i/4)