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

Optimal. Leaf size=49 \[ \frac{1}{2 \left (x^2+2\right )}+\frac{1}{3} \log \left (x^2+2\right )+\frac{1}{3} \log (1-x)-\frac{\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{3 \sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0672271, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {1647, 1629, 635, 203, 260} \[ \frac{1}{2 \left (x^2+2\right )}+\frac{1}{3} \log \left (x^2+2\right )+\frac{1}{3} \log (1-x)-\frac{\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

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

Mathematica [A]  time = 0.026986, size = 61, normalized size = 1.24 \[ \frac{1}{2 \left ((x-1)^2+2 (x-1)+3\right )}+\frac{1}{3} \log \left ((x-1)^2+2 (x-1)+3\right )+\frac{1}{3} \log (x-1)-\frac{\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 37, normalized size = 0.8 \begin{align*}{\frac{1}{2\,{x}^{2}+4}}+{\frac{\ln \left ({x}^{2}+2 \right ) }{3}}-{\frac{\sqrt{2}}{6}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }+{\frac{\ln \left ( -1+x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(x^2+2)+1/3*ln(x^2+2)-1/6*arctan(1/2*x*2^(1/2))*2^(1/2)+1/3*ln(-1+x)

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Maxima [A]  time = 1.48141, size = 49, normalized size = 1. \begin{align*} -\frac{1}{6} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{1}{2 \,{\left (x^{2} + 2\right )}} + \frac{1}{3} \, \log \left (x^{2} + 2\right ) + \frac{1}{3} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/2/(x^2 + 2) + 1/3*log(x^2 + 2) + 1/3*log(x - 1)

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Fricas [A]  time = 0.7595, size = 154, normalized size = 3.14 \begin{align*} -\frac{\sqrt{2}{\left (x^{2} + 2\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 2 \,{\left (x^{2} + 2\right )} \log \left (x^{2} + 2\right ) - 2 \,{\left (x^{2} + 2\right )} \log \left (x - 1\right ) - 3}{6 \,{\left (x^{2} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(sqrt(2)*(x^2 + 2)*arctan(1/2*sqrt(2)*x) - 2*(x^2 + 2)*log(x^2 + 2) - 2*(x^2 + 2)*log(x - 1) - 3)/(x^2 +
2)

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Sympy [A]  time = 0.155543, size = 14, normalized size = 0.29 \begin{align*} \frac{\log{\left (x - 1 \right )}}{3} + \frac{1}{2 x^{2} + 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1)/3 + 1/(2*x**2 + 4)

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Giac [A]  time = 1.07925, size = 50, normalized size = 1.02 \begin{align*} -\frac{1}{6} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{1}{2 \,{\left (x^{2} + 2\right )}} + \frac{1}{3} \, \log \left (x^{2} + 2\right ) + \frac{1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/2/(x^2 + 2) + 1/3*log(x^2 + 2) + 1/3*log(abs(x - 1))