### 3.2257 $$\int \frac{x}{(2+2 x+x^2)^2} \, dx$$

Optimal. Leaf size=26 $-\frac{x+2}{2 \left (x^2+2 x+2\right )}-\frac{1}{2} \tan ^{-1}(x+1)$

[Out]

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

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Rubi [A]  time = 0.0065948, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {638, 617, 204} $-\frac{x+2}{2 \left (x^2+2 x+2\right )}-\frac{1}{2} \tan ^{-1}(x+1)$

Antiderivative was successfully veriﬁed.

[In]

Int[x/(2 + 2*x + x^2)^2,x]

[Out]

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

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0115417, size = 28, normalized size = 1.08 $\frac{-x-2}{2 \left (x^2+2 x+2\right )}-\frac{1}{2} \tan ^{-1}(x+1)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x/(2 + 2*x + x^2)^2,x]

[Out]

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

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Maple [A]  time = 0.04, size = 25, normalized size = 1. \begin{align*}{\frac{-2\,x-4}{4\,{x}^{2}+8\,x+8}}-{\frac{\arctan \left ( 1+x \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45028, size = 30, normalized size = 1.15 \begin{align*} -\frac{x + 2}{2 \,{\left (x^{2} + 2 \, x + 2\right )}} - \frac{1}{2} \, \arctan \left (x + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.58163, size = 84, normalized size = 3.23 \begin{align*} -\frac{{\left (x^{2} + 2 \, x + 2\right )} \arctan \left (x + 1\right ) + x + 2}{2 \,{\left (x^{2} + 2 \, x + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.11191, size = 20, normalized size = 0.77 \begin{align*} - \frac{x + 2}{2 x^{2} + 4 x + 4} - \frac{\operatorname{atan}{\left (x + 1 \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x + 2)/(2*x**2 + 4*x + 4) - atan(x + 1)/2

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Giac [A]  time = 1.13224, size = 30, normalized size = 1.15 \begin{align*} -\frac{x + 2}{2 \,{\left (x^{2} + 2 \, x + 2\right )}} - \frac{1}{2} \, \arctan \left (x + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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