\(\int \frac {2+2 x+x^2}{2+x} \, dx\) [2175]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.01 (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 = {712} \[ \int \frac {2+2 x+x^2}{2+x} \, dx=\frac {x^2}{2}+2 \log (x+2) \]

[In]

Int[(2 + 2*x + x^2)/(2 + x),x]

[Out]

x^2/2 + 2*Log[2 + x]

Rule 712

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] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {2+2 x+x^2}{2+x} \, dx=\frac {1}{2} \left (-4+x^2+4 \log (2+x)\right ) \]

[In]

Integrate[(2 + 2*x + x^2)/(2 + x),x]

[Out]

(-4 + x^2 + 4*Log[2 + x])/2

Maple [A] (verified)

Time = 16.17 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {x^{2}}{2}+2 \ln \left (2+x \right )\)	\(13\)
norman	\(\frac {x^{2}}{2}+2 \ln \left (2+x \right )\)	\(13\)
risch	\(\frac {x^{2}}{2}+2 \ln \left (2+x \right )\)	\(13\)
parallelrisch	\(\frac {x^{2}}{2}+2 \ln \left (2+x \right )\)	\(13\)
meijerg	\(2 \ln \left (\frac {x}{2}+1\right )-\frac {x \left (-\frac {3 x}{2}+6\right )}{3}+2 x\)	\(21\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {2+2 x+x^2}{2+x} \, dx=\frac {1}{2} \, x^{2} + 2 \, \log \left (x + 2\right ) \]

[In]

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

[Out]

1/2*x^2 + 2*log(x + 2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {2+2 x+x^2}{2+x} \, dx=\frac {x^{2}}{2} + 2 \log {\left (x + 2 \right )} \]

[In]

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

[Out]

x**2/2 + 2*log(x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {2+2 x+x^2}{2+x} \, dx=\frac {1}{2} \, x^{2} + 2 \, \log \left (x + 2\right ) \]

[In]

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

[Out]

1/2*x^2 + 2*log(x + 2)

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

1/2*x^2 + 2*log(abs(x + 2))

Mupad [B] (verification not implemented)

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

[In]

int((2*x + x^2 + 2)/(x + 2),x)

[Out]

2*log(x + 2) + x^2/2