\(\int \frac {-2+x+3 x^2}{(-1+x)^3 (1+x^2)} \, dx\) [112]

Optimal result

Integrand size = 21, antiderivative size = 47 \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=-\frac {1}{2 (1-x)^2}+\frac {5}{2 (1-x)}-\arctan (x)-\frac {3}{2} \log (1-x)+\frac {3}{4} \log \left (1+x^2\right ) \]

[Out]

-1/2/(1-x)^2+5/2/(1-x)-arctan(x)-3/2*ln(1-x)+3/4*ln(x^2+1)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1643, 649, 209, 266} \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=-\arctan (x)+\frac {3}{4} \log \left (x^2+1\right )+\frac {5}{2 (1-x)}-\frac {1}{2 (1-x)^2}-\frac {3}{2} \log (1-x) \]

[In]

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

[Out]

-1/2*1/(1 - x)^2 + 5/(2*(1 - x)) - ArcTan[x] - (3*Log[1 - x])/2 + (3*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 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 {1}{(-1+x)^3}+\frac {5}{2 (-1+x)^2}-\frac {3}{2 (-1+x)}+\frac {-2+3 x}{2 \left (1+x^2\right )}\right ) \, dx \\ & = -\frac {1}{2 (1-x)^2}+\frac {5}{2 (1-x)}-\frac {3}{2} \log (1-x)+\frac {1}{2} \int \frac {-2+3 x}{1+x^2} \, dx \\ & = -\frac {1}{2 (1-x)^2}+\frac {5}{2 (1-x)}-\frac {3}{2} \log (1-x)+\frac {3}{2} \int \frac {x}{1+x^2} \, dx-\int \frac {1}{1+x^2} \, dx \\ & = -\frac {1}{2 (1-x)^2}+\frac {5}{2 (1-x)}-\arctan (x)-\frac {3}{2} \log (1-x)+\frac {3}{4} \log \left (1+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=\frac {1}{4} \left (-\frac {2}{(-1+x)^2}-\frac {10}{-1+x}-4 \arctan (x)-6 \log (-1+x)+3 \log \left (1+x^2\right )\right ) \]

[In]

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

[Out]

(-2/(-1 + x)^2 - 10/(-1 + x) - 4*ArcTan[x] - 6*Log[-1 + x] + 3*Log[1 + x^2])/4

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {-\frac {5 x}{2}+2}{\left (-1+x \right )^{2}}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\arctan \left (x \right )-\frac {3 \ln \left (-1+x \right )}{2}\)	\(31\)
default	\(-\frac {1}{2 \left (-1+x \right )^{2}}-\frac {5}{2 \left (-1+x \right )}-\frac {3 \ln \left (-1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\arctan \left (x \right )\)	\(34\)
parallelrisch	\(-\frac {-2 i \ln \left (x -i\right ) x^{2}+2 i \ln \left (x +i\right )+6 \ln \left (-1+x \right ) x^{2}+2 i \ln \left (x +i\right ) x^{2}-3 x^{2} \ln \left (x -i\right )+4 i \ln \left (x -i\right ) x -3 \ln \left (x +i\right ) x^{2}-3-12 \ln \left (-1+x \right ) x -2 i \ln \left (x -i\right )+6 \ln \left (x -i\right ) x -4 i \ln \left (x +i\right ) x +6 \ln \left (x +i\right ) x +5 x^{2}+6 \ln \left (-1+x \right )-3 \ln \left (x -i\right )-3 \ln \left (x +i\right )}{4 \left (-1+x \right )^{2}}\)	\(143\)

[In]

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

[Out]

(-5/2*x+2)/(-1+x)^2+3/4*ln(x^2+1)-arctan(x)-3/2*ln(-1+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=-\frac {4 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (x\right ) - 3 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (x^{2} + 1\right ) + 6 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (x - 1\right ) + 10 \, x - 8}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

[In]

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

[Out]

-1/4*(4*(x^2 - 2*x + 1)*arctan(x) - 3*(x^2 - 2*x + 1)*log(x^2 + 1) + 6*(x^2 - 2*x + 1)*log(x - 1) + 10*x - 8)/
(x^2 - 2*x + 1)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=\frac {4 - 5 x}{2 x^{2} - 4 x + 2} - \frac {3 \log {\left (x - 1 \right )}}{2} + \frac {3 \log {\left (x^{2} + 1 \right )}}{4} - \operatorname {atan}{\left (x \right )} \]

[In]

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

[Out]

(4 - 5*x)/(2*x**2 - 4*x + 2) - 3*log(x - 1)/2 + 3*log(x**2 + 1)/4 - atan(x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=-\frac {5 \, x - 4}{2 \, {\left (x^{2} - 2 \, x + 1\right )}} - \arctan \left (x\right ) + \frac {3}{4} \, \log \left (x^{2} + 1\right ) - \frac {3}{2} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

-1/2*(5*x - 4)/(x^2 - 2*x + 1) - arctan(x) + 3/4*log(x^2 + 1) - 3/2*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.68 \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=-\frac {5 \, x - 4}{2 \, {\left (x - 1\right )}^{2}} - \arctan \left (x\right ) + \frac {3}{4} \, \log \left (x^{2} + 1\right ) - \frac {3}{2} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

-1/2*(5*x - 4)/(x - 1)^2 - arctan(x) + 3/4*log(x^2 + 1) - 3/2*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \frac {-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx=-\frac {3\,\ln \left (x-1\right )}{2}-\frac {\frac {5\,x}{2}-2}{x^2-2\,x+1}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {3}{4}+\frac {1}{2}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {3}{4}-\frac {1}{2}{}\mathrm {i}\right ) \]

[In]

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

[Out]

log(x - 1i)*(3/4 + 1i/2) - (3*log(x - 1))/2 + log(x + 1i)*(3/4 - 1i/2) - ((5*x)/2 - 2)/(x^2 - 2*x + 1)