3.148 \(\int \frac{1}{(2+2 x+x^2)^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0055748, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {614, 617, 204} \[ \frac{x+1}{2 \left (x^2+2 x+2\right )}+\frac{1}{2} \tan ^{-1}(x+1) \]

Antiderivative was successfully verified.

[In]

Int[(2 + 2*x + x^2)^(-2),x]

[Out]

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

Rule 614

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

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

Mathematica [A]  time = 0.007493, size = 23, normalized size = 0.88 \[ \frac{1}{2} \left (\frac{x+1}{x^2+2 x+2}+\tan ^{-1}(x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 2*x + x^2)^(-2),x]

[Out]

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

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Maple [A]  time = 0.001, size = 25, normalized size = 1. \begin{align*}{\frac{2\,x+2}{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(1/(x^2+2*x+2)^2,x)

[Out]

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

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Maxima [A]  time = 1.4083, size = 30, normalized size = 1.15 \begin{align*} \frac{x + 1}{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(1/(x^2+2*x+2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.2187, size = 82, normalized size = 3.15 \begin{align*} \frac{{\left (x^{2} + 2 \, x + 2\right )} \arctan \left (x + 1\right ) + x + 1}{2 \,{\left (x^{2} + 2 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.113337, size = 19, normalized size = 0.73 \begin{align*} \frac{x + 1}{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(1/(x**2+2*x+2)**2,x)

[Out]

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

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Giac [A]  time = 1.05064, size = 30, normalized size = 1.15 \begin{align*} \frac{x + 1}{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(1/(x^2+2*x+2)^2,x, algorithm="giac")

[Out]

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