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

Optimal. Leaf size=47 \[ \frac{3}{4} \log \left (x^2+1\right )+\frac{5}{2 (1-x)}-\frac{1}{2 (1-x)^2}-\frac{3}{2} \log (1-x)-\tan ^{-1}(x) \]

[Out]

-1/(2*(1 - x)^2) + 5/(2*(1 - x)) - ArcTan[x] - (3*Log[1 - x])/2 + (3*Log[1 + x^2])/4

________________________________________________________________________________________

Rubi [A]  time = 0.0428182, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {1629, 635, 203, 260} \[ \frac{3}{4} \log \left (x^2+1\right )+\frac{5}{2 (1-x)}-\frac{1}{2 (1-x)^2}-\frac{3}{2} \log (1-x)-\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*(1 - x)^2) + 5/(2*(1 - x)) - ArcTan[x] - (3*Log[1 - x])/2 + (3*Log[1 + x^2])/4

Rule 1629

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]

Rule 635

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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]

Rubi steps

\begin{align*} \int \frac{-2+x+3 x^2}{(-1+x)^3 \left (1+x^2\right )} \, dx &=\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)}-\tan ^{-1}(x)-\frac{3}{2} \log (1-x)+\frac{3}{4} \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.02364, size = 37, normalized size = 0.79 \[ \frac{1}{4} \left (3 \log \left (x^2+1\right )-\frac{10}{x-1}-\frac{2}{(x-1)^2}-6 \log (x-1)-4 \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[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]  time = 0.006, size = 34, normalized size = 0.7 \begin{align*}{\frac{3\,\ln \left ({x}^{2}+1 \right ) }{4}}-\arctan \left ( x \right ) -{\frac{1}{2\, \left ( -1+x \right ) ^{2}}}-{\frac{5}{2\,x-2}}-{\frac{3\,\ln \left ( -1+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.4535, size = 49, normalized size = 1.04 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 2.10564, size = 171, normalized size = 3.64 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0.149242, size = 36, normalized size = 0.77 \begin{align*} - \frac{5 x - 4}{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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.05329, size = 43, normalized size = 0.91 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))