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

Optimal. Leaf size=37 \[ \frac{3}{4} \log \left (x^2+1\right )+x+\frac{5}{2 (1-x)}+\frac{1}{2} \log (1-x)+2 \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0397911, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1629, 635, 203, 260} \[ \frac{3}{4} \log \left (x^2+1\right )+x+\frac{5}{2 (1-x)}+\frac{1}{2} \log (1-x)+2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0226373, size = 33, normalized size = 0.89 \[ \frac{3}{4} \log \left (x^2+1\right )+x+\frac{5}{2-2 x}+\frac{1}{2} \log (x-1)+2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 28, normalized size = 0.8 \begin{align*} x+{\frac{3\,\ln \left ({x}^{2}+1 \right ) }{4}}+2\,\arctan \left ( x \right ) -{\frac{5}{2\,x-2}}+{\frac{\ln \left ( x-1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.51328, size = 36, normalized size = 0.97 \begin{align*} x - \frac{5}{2 \,{\left (x - 1\right )}} + 2 \, \arctan \left (x\right ) + \frac{3}{4} \, \log \left (x^{2} + 1\right ) + \frac{1}{2} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.60854, size = 138, normalized size = 3.73 \begin{align*} \frac{4 \, x^{2} + 8 \,{\left (x - 1\right )} \arctan \left (x\right ) + 3 \,{\left (x - 1\right )} \log \left (x^{2} + 1\right ) + 2 \,{\left (x - 1\right )} \log \left (x - 1\right ) - 4 \, x - 10}{4 \,{\left (x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.144112, size = 29, normalized size = 0.78 \begin{align*} x + \frac{\log{\left (x - 1 \right )}}{2} + \frac{3 \log{\left (x^{2} + 1 \right )}}{4} + 2 \operatorname{atan}{\left (x \right )} - \frac{5}{2 x - 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1681, size = 81, normalized size = 2.19 \begin{align*} \frac{1}{2} \, \pi - 2 \, \pi \left \lfloor \frac{\pi + 4 \, \arctan \left (x\right )}{4 \, \pi } + \frac{1}{2} \right \rfloor + x - \frac{5}{2 \,{\left (x - 1\right )}} + 2 \, \arctan \left (x\right ) + \frac{3}{4} \, \log \left (\frac{2}{x - 1} + \frac{2}{{\left (x - 1\right )}^{2}} + 1\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) - 1 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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