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

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.51, size = 44, normalized size = 1.19 \[ \frac {4 \, x^{2} + 8 \, {\left (x - 1\right )} \arctan \relax (x) + 3 \, {\left (x - 1\right )} \log \left (x^{2} + 1\right ) + 2 \, {\left (x - 1\right )} \log \left (x - 1\right ) - 4 \, x - 10}{4 \, {\left (x - 1\right )}} \]

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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giac [B]  time = 0.29, size = 60, normalized size = 1.62 \[ \frac {1}{2} \, \pi - 2 \, \pi \left \lfloor \frac {\pi + 4 \, \arctan \relax (x)}{4 \, \pi } + \frac {1}{2} \right \rfloor + x - \frac {5}{2 \, {\left (x - 1\right )}} + 2 \, \arctan \relax (x) + \frac {3}{4} \, \log \left (\frac {2}{x - 1} + \frac {2}{{\left (x - 1\right )}^{2}} + 1\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) - 1 \]

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

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maple [A]  time = 0.01, size = 28, normalized size = 0.76 \[ x +2 \arctan \relax (x )+\frac {\ln \left (x -1\right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {5}{2 \left (x -1\right )} \]

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

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maxima [A]  time = 2.22, size = 27, normalized size = 0.73 \[ x - \frac {5}{2 \, {\left (x - 1\right )}} + 2 \, \arctan \relax (x) + \frac {3}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right ) \]

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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mupad [B]  time = 2.14, size = 35, normalized size = 0.95 \[ x+\frac {\ln \left (x-1\right )}{2}-\frac {5}{2\,\left (x-1\right )}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {3}{4}-\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {3}{4}+1{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.16, size = 29, normalized size = 0.78 \[ x + \frac {\log {\left (x - 1 \right )}}{2} + \frac {3 \log {\left (x^{2} + 1 \right )}}{4} + 2 \operatorname {atan}{\relax (x )} - \frac {5}{2 x - 2} \]

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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