3.272 \(\int \frac {x}{2-2 x+x^2} \, dx\)

Optimal. Leaf size=22 \[ \frac {1}{2} \log \left (x^2-2 x+2\right )-\tan ^{-1}(1-x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {634, 617, 204, 628} \[ \frac {1}{2} \log \left (x^2-2 x+2\right )-\tan ^{-1}(1-x) \]

Antiderivative was successfully verified.

[In]

Int[x/(2 - 2*x + x^2),x]

[Out]

-ArcTan[1 - x] + Log[2 - 2*x + x^2]/2

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 22, normalized size = 1.00 \[ \frac {1}{2} \log \left (x^2-2 x+2\right )-\tan ^{-1}(1-x) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(2 - 2*x + x^2),x]

[Out]

-ArcTan[1 - x] + Log[2 - 2*x + x^2]/2

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fricas [A]  time = 0.40, size = 16, normalized size = 0.73 \[ \arctan \left (x - 1\right ) + \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.25, size = 16, normalized size = 0.73 \[ \arctan \left (x - 1\right ) + \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 17, normalized size = 0.77 \[ \arctan \left (x -1\right )+\frac {\ln \left (x^{2}-2 x +2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.30, size = 16, normalized size = 0.73 \[ \arctan \left (x - 1\right ) + \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.16, size = 16, normalized size = 0.73 \[ \mathrm {atan}\left (x-1\right )+\frac {\ln \left (x^2-2\,x+2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2 - 2*x + 2),x)

[Out]

atan(x - 1) + log(x^2 - 2*x + 2)/2

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sympy [A]  time = 0.11, size = 15, normalized size = 0.68 \[ \frac {\log {\left (x^{2} - 2 x + 2 \right )}}{2} + \operatorname {atan}{\left (x - 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x**2 - 2*x + 2)/2 + atan(x - 1)

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