\(\int \frac {1+x}{(2+2 x+x^2)^3} \, dx\) [5]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=-\frac {1}{4 \left (2+2 x+x^2\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {643} \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=-\frac {1}{4 \left (x^2+2 x+2\right )^2} \]

[In]

Int[(1 + x)/(2 + 2*x + x^2)^3,x]

[Out]

-1/4*1/(2 + 2*x + x^2)^2

Rule 643

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

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{4 \left (2+2 x+x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=-\frac {1}{4 \left (2+2 x+x^2\right )^2} \]

[In]

Integrate[(1 + x)/(2 + 2*x + x^2)^3,x]

[Out]

-1/4*1/(2 + 2*x + x^2)^2

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
gosper	\(-\frac {1}{4 \left (x^{2}+2 x +2\right )^{2}}\)	\(13\)
default	\(-\frac {1}{4 \left (x^{2}+2 x +2\right )^{2}}\)	\(13\)
norman	\(-\frac {1}{4 \left (x^{2}+2 x +2\right )^{2}}\)	\(13\)
risch	\(-\frac {1}{4 \left (x^{2}+2 x +2\right )^{2}}\)	\(13\)
parallelrisch	\(-\frac {1}{4 \left (x^{2}+2 x +2\right )^{2}}\)	\(13\)

[In]

int((1+x)/(x^2+2*x+2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.57 \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (x^{4} + 4 \, x^{3} + 8 \, x^{2} + 8 \, x + 4\right )}} \]

[In]

integrate((1+x)/(x^2+2*x+2)^3,x, algorithm="fricas")

[Out]

-1/4/(x^4 + 4*x^3 + 8*x^2 + 8*x + 4)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.57 \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=- \frac {1}{4 x^{4} + 16 x^{3} + 32 x^{2} + 32 x + 16} \]

[In]

integrate((1+x)/(x**2+2*x+2)**3,x)

[Out]

-1/(4*x**4 + 16*x**3 + 32*x**2 + 32*x + 16)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (x^{2} + 2 \, x + 2\right )}^{2}} \]

[In]

integrate((1+x)/(x^2+2*x+2)^3,x, algorithm="maxima")

[Out]

-1/4/(x^2 + 2*x + 2)^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (x^{2} + 2 \, x + 2\right )}^{2}} \]

[In]

integrate((1+x)/(x^2+2*x+2)^3,x, algorithm="giac")

[Out]

-1/4/(x^2 + 2*x + 2)^2

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1+x}{\left (2+2 x+x^2\right )^3} \, dx=-\frac {1}{4\,{\left (x^2+2\,x+2\right )}^2} \]

[In]

int((x + 1)/(2*x + x^2 + 2)^3,x)

[Out]

-1/(4*(2*x + x^2 + 2)^2)