\(\int \frac {5-4 x+3 x^2}{(-1+x) (1+x^2)} \, dx\) [201]

Optimal result

Integrand size = 23, antiderivative size = 23 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \arctan (x)+2 \log (1-x)+\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

-3*arctan(x)+2*ln(1-x)+1/2*ln(x^2+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1643, 649, 209, 266} \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \arctan (x)+\frac {1}{2} \log \left (x^2+1\right )+2 \log (1-x) \]

[In]

Int[(5 - 4*x + 3*x^2)/((-1 + x)*(1 + x^2)),x]

[Out]

-3*ArcTan[x] + 2*Log[1 - x] + Log[1 + x^2]/2

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 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{-1+x}+\frac {-3+x}{1+x^2}\right ) \, dx \\ & = 2 \log (1-x)+\int \frac {-3+x}{1+x^2} \, dx \\ & = 2 \log (1-x)-3 \int \frac {1}{1+x^2} \, dx+\int \frac {x}{1+x^2} \, dx \\ & = -3 \arctan (x)+2 \log (1-x)+\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \arctan (x)+\frac {1}{2} \log \left (2+2 (-1+x)+(-1+x)^2\right )+2 \log (-1+x) \]

[In]

Integrate[(5 - 4*x + 3*x^2)/((-1 + x)*(1 + x^2)),x]

[Out]

-3*ArcTan[x] + Log[2 + 2*(-1 + x) + (-1 + x)^2]/2 + 2*Log[-1 + x]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(2 \ln \left (-1+x \right )+\frac {\ln \left (x^{2}+1\right )}{2}-3 \arctan \left (x \right )\)	\(20\)
risch	\(2 \ln \left (-1+x \right )+\frac {\ln \left (9 x^{2}+9\right )}{2}-3 \arctan \left (x \right )\)	\(22\)
parallelrisch	\(2 \ln \left (-1+x \right )+\frac {\ln \left (x -i\right )}{2}+\frac {3 i \ln \left (x -i\right )}{2}+\frac {\ln \left (x +i\right )}{2}-\frac {3 i \ln \left (x +i\right )}{2}\)	\(38\)

[In]

int((3*x^2-4*x+5)/(-1+x)/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

2*ln(-1+x)+1/2*ln(x^2+1)-3*arctan(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) + 2 \, \log \left (x - 1\right ) \]

[In]

integrate((3*x^2-4*x+5)/(-1+x)/(x^2+1),x, algorithm="fricas")

[Out]

-3*arctan(x) + 1/2*log(x^2 + 1) + 2*log(x - 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=2 \log {\left (x - 1 \right )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} - 3 \operatorname {atan}{\left (x \right )} \]

[In]

integrate((3*x**2-4*x+5)/(-1+x)/(x**2+1),x)

[Out]

2*log(x - 1) + log(x**2 + 1)/2 - 3*atan(x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) + 2 \, \log \left (x - 1\right ) \]

[In]

integrate((3*x^2-4*x+5)/(-1+x)/(x^2+1),x, algorithm="maxima")

[Out]

-3*arctan(x) + 1/2*log(x^2 + 1) + 2*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate((3*x^2-4*x+5)/(-1+x)/(x^2+1),x, algorithm="giac")

[Out]

-3*arctan(x) + 1/2*log(x^2 + 1) + 2*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=2\,\ln \left (x-1\right )+\ln \left (x-\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {3}{2}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {1}{2}-\frac {3}{2}{}\mathrm {i}\right ) \]

[In]

int((3*x^2 - 4*x + 5)/((x^2 + 1)*(x - 1)),x)

[Out]

2*log(x - 1) + log(x - 1i)*(1/2 + 3i/2) + log(x + 1i)*(1/2 - 3i/2)