\(\int \frac {2+2 x+x^2}{(1+x)^3} \, dx\) [2177]

Optimal result

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

[Out]

-1/2/(1+x)^2+ln(1+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {697} \[ \int \frac {2+2 x+x^2}{(1+x)^3} \, dx=\log (x+1)-\frac {1}{2 (x+1)^2} \]

[In]

Int[(2 + 2*x + x^2)/(1 + x)^3,x]

[Out]

-1/2*1/(1 + x)^2 + Log[1 + x]

Rule 697

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{(1+x)^3}+\frac {1}{1+x}\right ) \, dx \\ & = -\frac {1}{2 (1+x)^2}+\log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {2+2 x+x^2}{(1+x)^3} \, dx=-\frac {1}{2 (1+x)^2}+\log (1+x) \]

[In]

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

[Out]

-1/2*1/(1 + x)^2 + Log[1 + x]

Maple [A] (verified)

Time = 15.88 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {1}{2 \left (1+x \right )^{2}}+\ln \left (1+x \right )\)	\(13\)
norman	\(-\frac {1}{2 \left (1+x \right )^{2}}+\ln \left (1+x \right )\)	\(13\)
risch	\(-\frac {1}{2 \left (1+x \right )^{2}}+\ln \left (1+x \right )\)	\(13\)
parallelrisch	\(\frac {2 \ln \left (1+x \right ) x^{2}-1+4 \ln \left (1+x \right ) x +2 \ln \left (1+x \right )}{2 \left (1+x \right )^{2}}\)	\(32\)
meijerg	\(\frac {x \left (2+x \right )}{\left (1+x \right )^{2}}-\frac {x \left (9 x +6\right )}{6 \left (1+x \right )^{2}}+\ln \left (1+x \right )+\frac {x^{2}}{\left (1+x \right )^{2}}\)	\(38\)

[In]

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

[Out]

-1/2/(1+x)^2+ln(1+x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (12) = 24\).

Time = 0.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 2.00 \[ \int \frac {2+2 x+x^2}{(1+x)^3} \, dx=\frac {2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) - 1}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {2+2 x+x^2}{(1+x)^3} \, dx=\log {\left (x + 1 \right )} - \frac {1}{2 x^{2} + 4 x + 2} \]

[In]

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

[Out]

log(x + 1) - 1/(2*x**2 + 4*x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {2+2 x+x^2}{(1+x)^3} \, dx=-\frac {1}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} + \log \left (x + 1\right ) \]

[In]

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

[Out]

-1/2/(x^2 + 2*x + 1) + log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {2+2 x+x^2}{(1+x)^3} \, dx=-\frac {1}{2 \, {\left (x + 1\right )}^{2}} + \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

-1/2/(x + 1)^2 + log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {2+2 x+x^2}{(1+x)^3} \, dx=\ln \left (x+1\right )-\frac {1}{2\,{\left (x+1\right )}^2} \]

[In]

int((2*x + x^2 + 2)/(x + 1)^3,x)

[Out]

log(x + 1) - 1/(2*(x + 1)^2)