\(\int \frac {x}{(2+2 x+x^2)^2} \, dx\) [2257]

Optimal result

Integrand size = 12, antiderivative size = 26 \[ \int \frac {x}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {2+x}{2 \left (2+2 x+x^2\right )}-\frac {1}{2} \arctan (1+x) \]

[Out]

1/2*(-2-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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {652, 631, 210} \[ \int \frac {x}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {1}{2} \arctan (x+1)-\frac {x+2}{2 \left (x^2+2 x+2\right )} \]

[In]

Int[x/(2 + 2*x + x^2)^2,x]

[Out]

-1/2*(2 + x)/(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 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 652

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)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), 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]

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

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {x}{\left (2+2 x+x^2\right )^2} \, dx=\frac {-2-x}{2 \left (2+2 x+x^2\right )}-\frac {1}{2} \arctan (1+x) \]

[In]

Integrate[x/(2 + 2*x + x^2)^2,x]

[Out]

(-2 - x)/(2*(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	\(\frac {-\frac {x}{2}-1}{x^{2}+2 x +2}-\frac {\arctan \left (1+x \right )}{2}\)	\(24\)
default	\(\frac {-2 x -4}{4 x^{2}+8 x +8}-\frac {\arctan \left (1+x \right )}{2}\)	\(25\)
parallelrisch	\(\frac {i \ln \left (x +1-i\right ) x^{2}-i \ln \left (x +1+i\right ) x^{2}+2 i \ln \left (x +1-i\right ) x -2 i \ln \left (x +1+i\right ) x -2+2 i \ln \left (x +1-i\right )-2 i \ln \left (x +1+i\right )+x^{2}}{4 x^{2}+8 x +8}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {x}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {{\left (x^{2} + 2 \, x + 2\right )} \arctan \left (x + 1\right ) + x + 2}{2 \, {\left (x^{2} + 2 \, x + 2\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {x}{\left (2+2 x+x^2\right )^2} \, dx=\frac {- x - 2}{2 x^{2} + 4 x + 4} - \frac {\operatorname {atan}{\left (x + 1 \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {x}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {x + 2}{2 \, {\left (x^{2} + 2 \, x + 2\right )}} - \frac {1}{2} \, \arctan \left (x + 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {x}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {\mathrm {atan}\left (x+1\right )}{2}-\frac {\frac {x}{2}+1}{x^2+2\,x+2} \]

[In]

int(x/(2*x + x^2 + 2)^2,x)

[Out]

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