\(\int \frac {1}{(2+2 x+x^2)^2} \, dx\) [148]

Optimal result

Integrand size = 10, antiderivative size = 26 \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {1+x}{2 \left (2+2 x+x^2\right )}+\frac {1}{2} \arctan (1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {628, 631, 210} \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {1}{2} \arctan (x+1)+\frac {x+1}{2 \left (x^2+2 x+2\right )} \]

[In]

Int[(2 + 2*x + x^2)^(-2),x]

[Out]

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

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 628

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 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]

Rubi steps \begin{align*} \text {integral}& = \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} \text {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} \arctan (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {1}{2} \left (\frac {1+x}{2+2 x+x^2}+\arctan (1+x)\right ) \]

[In]

Integrate[(2 + 2*x + x^2)^(-2),x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {\frac {1}{2}+\frac {x}{2}}{x^{2}+2 x +2}+\frac {\arctan \left (1+x \right )}{2}\)	\(24\)
default	\(\frac {2 x +2}{4 x^{2}+8 x +8}+\frac {\arctan \left (1+x \right )}{2}\)	\(25\)
parallelrisch	\(-\frac {i \ln \left (x +1-i\right ) x^{2}-i \ln \left (x +1+i\right ) x^{2}+2 i \ln \left (x +1-i\right ) x -2 i \ln \left (x +1+i\right ) x +2 i \ln \left (x +1-i\right )-2 i \ln \left (x +1+i\right )+x^{2}}{4 \left (x^{2}+2 x +2\right )}\)	\(79\)

[In]

int(1/(x^2+2*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {{\left (x^{2} + 2 \, x + 2\right )} \arctan \left (x + 1\right ) + x + 1}{2 \, {\left (x^{2} + 2 \, x + 2\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {x + 1}{2 x^{2} + 4 x + 4} + \frac {\operatorname {atan}{\left (x + 1 \right )}}{2} \]

[In]

integrate(1/(x**2+2*x+2)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x + 2\right )}} + \frac {1}{2} \, \arctan \left (x + 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x + 2\right )}} + \frac {1}{2} \, \arctan \left (x + 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (2+2 x+x^2\right )^2} \, dx=\frac {\mathrm {atan}\left (x+1\right )}{2}+\frac {\frac {x}{2}+\frac {1}{2}}{x^2+2\,x+2} \]

[In]

int(1/(2*x + x^2 + 2)^2,x)

[Out]

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