\(\int \frac {2+x^2}{2+x} \, dx\) [448]

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int \frac {2+x^2}{2+x} \, dx=-2 x+\frac {x^2}{2}+6 \log (2+x) \]

[Out]

-2*x+1/2*x^2+6*ln(2+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {711} \[ \int \frac {2+x^2}{2+x} \, dx=\frac {x^2}{2}-2 x+6 \log (x+2) \]

[In]

Int[(2 + x^2)/(2 + x),x]

[Out]

-2*x + x^2/2 + 6*Log[2 + x]

Rule 711

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-2+x+\frac {6}{2+x}\right ) \, dx \\ & = -2 x+\frac {x^2}{2}+6 \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {2+x^2}{2+x} \, dx=-6-2 x+\frac {x^2}{2}+6 \log (2+x) \]

[In]

Integrate[(2 + x^2)/(2 + x),x]

[Out]

-6 - 2*x + x^2/2 + 6*Log[2 + x]

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(-2 x +\frac {x^{2}}{2}+6 \ln \left (x +2\right )\)	\(16\)
norman	\(-2 x +\frac {x^{2}}{2}+6 \ln \left (x +2\right )\)	\(16\)
risch	\(-2 x +\frac {x^{2}}{2}+6 \ln \left (x +2\right )\)	\(16\)
parallelrisch	\(-2 x +\frac {x^{2}}{2}+6 \ln \left (x +2\right )\)	\(16\)
meijerg	\(6 \ln \left (1+\frac {x}{2}\right )-\frac {x \left (-\frac {3 x}{2}+6\right )}{3}\)	\(18\)

[In]

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

[Out]

-2*x+1/2*x^2+6*ln(x+2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2+x^2}{2+x} \, dx=\frac {1}{2} \, x^{2} - 2 \, x + 6 \, \log \left (x + 2\right ) \]

[In]

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

[Out]

1/2*x^2 - 2*x + 6*log(x + 2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2+x^2}{2+x} \, dx=\frac {x^{2}}{2} - 2 x + 6 \log {\left (x + 2 \right )} \]

[In]

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

[Out]

x**2/2 - 2*x + 6*log(x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2+x^2}{2+x} \, dx=\frac {1}{2} \, x^{2} - 2 \, x + 6 \, \log \left (x + 2\right ) \]

[In]

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

[Out]

1/2*x^2 - 2*x + 6*log(x + 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {2+x^2}{2+x} \, dx=\frac {1}{2} \, x^{2} - 2 \, x + 6 \, \log \left ({\left | x + 2 \right |}\right ) \]

[In]

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

[Out]

1/2*x^2 - 2*x + 6*log(abs(x + 2))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2+x^2}{2+x} \, dx=6\,\ln \left (x+2\right )-2\,x+\frac {x^2}{2} \]

[In]

int((x^2 + 2)/(x + 2),x)

[Out]

6*log(x + 2) - 2*x + x^2/2