### 3.2262 $$\int \frac{x^2}{(2+2 x+x^2)^2} \, dx$$

Optimal. Leaf size=15 $\frac{1}{x^2+2 x+2}+\tan ^{-1}(x+1)$

[Out]

(2 + 2*x + x^2)^(-1) + ArcTan[1 + x]

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Rubi [A]  time = 0.0076848, antiderivative size = 23, normalized size of antiderivative = 1.53, number of steps used = 3, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {722, 617, 204} $\tan ^{-1}(x+1)-\frac{x (x+2)}{2 \left (x^2+2 x+2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2/(2 + 2*x + x^2)^2,x]

[Out]

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

Rule 722

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

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

Mathematica [A]  time = 0.0075959, size = 15, normalized size = 1. $\frac{1}{x^2+2 x+2}+\tan ^{-1}(x+1)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2/(2 + 2*x + x^2)^2,x]

[Out]

(2 + 2*x + x^2)^(-1) + ArcTan[1 + x]

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Maple [A]  time = 0.042, size = 16, normalized size = 1.1 \begin{align*} \left ({x}^{2}+2\,x+2 \right ) ^{-1}+\arctan \left ( 1+x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.53619, size = 20, normalized size = 1.33 \begin{align*} \frac{1}{x^{2} + 2 \, x + 2} + \arctan \left (x + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(x^2 + 2*x + 2) + arctan(x + 1)

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Fricas [A]  time = 2.12864, size = 72, normalized size = 4.8 \begin{align*} \frac{{\left (x^{2} + 2 \, x + 2\right )} \arctan \left (x + 1\right ) + 1}{x^{2} + 2 \, x + 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.108789, size = 14, normalized size = 0.93 \begin{align*} \operatorname{atan}{\left (x + 1 \right )} + \frac{1}{x^{2} + 2 x + 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(x + 1) + 1/(x**2 + 2*x + 2)

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Giac [A]  time = 1.08865, size = 20, normalized size = 1.33 \begin{align*} \frac{1}{x^{2} + 2 \, x + 2} + \arctan \left (x + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(x^2 + 2*x + 2) + arctan(x + 1)