\(\int \frac {x^2}{(2+2 x+x^2)^2} \, dx\) [2262]

Optimal result

Integrand size = 14, antiderivative size = 15 \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\frac {1}{2+2 x+x^2}+\arctan (1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.53, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {736, 631, 210} \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\arctan (x+1)-\frac {x (x+2)}{2 \left (x^2+2 x+2\right )} \]

[In]

Int[x^2/(2 + 2*x + x^2)^2,x]

[Out]

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

Rule 210

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

Rule 631

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 736

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]

Rubi steps \begin{align*} \text {integral}& = -\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 )}-\text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\frac {1}{2+2 x+x^2}+\arctan (1+x) \]

[In]

Integrate[x^2/(2 + 2*x + x^2)^2,x]

[Out]

(2 + 2*x + x^2)^(-1) + ArcTan[1 + x]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {1}{x^{2}+2 x +2}+\arctan \left (1+x \right )\)	\(16\)
risch	\(\frac {1}{x^{2}+2 x +2}+\arctan \left (1+x \right )\)	\(16\)
parallelrisch	\(-\frac {i \ln \left (x +1-i\right ) x^{2}-i \ln \left (x +1+i\right ) x^{2}+2 i \ln \left (x +1-i\right ) x -2 i \ln \left (x +1+i\right ) x -2+2 i \ln \left (x +1-i\right )-2 i \ln \left (x +1+i\right )}{2 \left (x^{2}+2 x +2\right )}\)	\(77\)

[In]

int(x^2/(x^2+2*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.73 \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\frac {{\left (x^{2} + 2 \, x + 2\right )} \arctan \left (x + 1\right ) + 1}{x^{2} + 2 \, x + 2} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\operatorname {atan}{\left (x + 1 \right )} + \frac {1}{x^{2} + 2 x + 2} \]

[In]

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

[Out]

atan(x + 1) + 1/(x**2 + 2*x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\frac {1}{x^{2} + 2 \, x + 2} + \arctan \left (x + 1\right ) \]

[In]

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

[Out]

1/(x^2 + 2*x + 2) + arctan(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\frac {1}{x^{2} + 2 \, x + 2} + \arctan \left (x + 1\right ) \]

[In]

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

[Out]

1/(x^2 + 2*x + 2) + arctan(x + 1)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\left (2+2 x+x^2\right )^2} \, dx=\mathrm {atan}\left (x+1\right )+\frac {1}{x^2+2\,x+2} \]

[In]

int(x^2/(2*x + x^2 + 2)^2,x)

[Out]

atan(x + 1) + 1/(2*x + x^2 + 2)