\(\int \frac {5+4 x+x^2}{-2+x} \, dx\) [2176]

Optimal result

Integrand size = 14, antiderivative size = 19 \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=6 x+\frac {x^2}{2}+17 \log (2-x) \]

[Out]

6*x+1/2*x^2+17*ln(2-x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, 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} \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=\frac {x^2}{2}+6 x+17 \log (2-x) \]

[In]

Int[(5 + 4*x + x^2)/(-2 + x),x]

[Out]

6*x + x^2/2 + 17*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 (6+\frac {17}{-2+x}+x\right ) \, dx \\ & = 6 x+\frac {x^2}{2}+17 \log (2-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=-14+6 x+\frac {x^2}{2}+17 \log (-2+x) \]

[In]

Integrate[(5 + 4*x + x^2)/(-2 + x),x]

[Out]

-14 + 6*x + x^2/2 + 17*Log[-2 + x]

Maple [A] (verified)

Time = 20.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
default	\(6 x +\frac {x^{2}}{2}+17 \ln \left (-2+x \right )\)	\(16\)
norman	\(6 x +\frac {x^{2}}{2}+17 \ln \left (-2+x \right )\)	\(16\)
risch	\(6 x +\frac {x^{2}}{2}+17 \ln \left (-2+x \right )\)	\(16\)
parallelrisch	\(6 x +\frac {x^{2}}{2}+17 \ln \left (-2+x \right )\)	\(16\)
meijerg	\(17 \ln \left (1-\frac {x}{2}\right )+\frac {x \left (\frac {3 x}{2}+6\right )}{3}+4 x\)	\(21\)

[In]

int((x^2+4*x+5)/(-2+x),x,method=_RETURNVERBOSE)

[Out]

6*x+1/2*x^2+17*ln(-2+x)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=\frac {1}{2} \, x^{2} + 6 \, x + 17 \, \log \left (x - 2\right ) \]

[In]

integrate((x^2+4*x+5)/(-2+x),x, algorithm="fricas")

[Out]

1/2*x^2 + 6*x + 17*log(x - 2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=\frac {x^{2}}{2} + 6 x + 17 \log {\left (x - 2 \right )} \]

[In]

integrate((x**2+4*x+5)/(-2+x),x)

[Out]

x**2/2 + 6*x + 17*log(x - 2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=\frac {1}{2} \, x^{2} + 6 \, x + 17 \, \log \left (x - 2\right ) \]

[In]

integrate((x^2+4*x+5)/(-2+x),x, algorithm="maxima")

[Out]

1/2*x^2 + 6*x + 17*log(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=\frac {1}{2} \, x^{2} + 6 \, x + 17 \, \log \left ({\left | x - 2 \right |}\right ) \]

[In]

integrate((x^2+4*x+5)/(-2+x),x, algorithm="giac")

[Out]

1/2*x^2 + 6*x + 17*log(abs(x - 2))

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {5+4 x+x^2}{-2+x} \, dx=6\,x+17\,\ln \left (x-2\right )+\frac {x^2}{2} \]

[In]

int((4*x + x^2 + 5)/(x - 2),x)

[Out]

6*x + 17*log(x - 2) + x^2/2