\(\int \frac {-5+2 x}{-5 x+x^2} \, dx\) [6548]

Optimal result

Integrand size = 15, antiderivative size = 8 \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\log ((5-x) x) \]

[Out]

ln(x*(5-x))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.25, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {642} \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\log \left (5 x-x^2\right ) \]

[In]

Int[(-5 + 2*x)/(-5*x + x^2),x]

[Out]

Log[5*x - x^2]

Rule 642

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

Rubi steps \begin{align*} \text {integral}& = \log \left (5 x-x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.12 \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\log (5-x)+\log (x) \]

[In]

Integrate[(-5 + 2*x)/(-5*x + x^2),x]

[Out]

Log[5 - x] + Log[x]

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(\ln \left (\left (-5+x \right ) x \right )\)	\(7\)
norman	\(\ln \left (x \right )+\ln \left (-5+x \right )\)	\(8\)
parallelrisch	\(\ln \left (x \right )+\ln \left (-5+x \right )\)	\(8\)
derivativedivides	\(\ln \left (x^{2}-5 x \right )\)	\(9\)
risch	\(\ln \left (x^{2}-5 x \right )\)	\(9\)
meijerg	\(\ln \left (x \right )-\ln \left (5\right )+i \pi +\ln \left (1-\frac {x}{5}\right )\)	\(18\)

[In]

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

[Out]

ln((-5+x)*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\log \left (x^{2} - 5 \, x\right ) \]

[In]

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

[Out]

log(x^2 - 5*x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\log {\left (x^{2} - 5 x \right )} \]

[In]

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

[Out]

log(x**2 - 5*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\log \left (x^{2} - 5 \, x\right ) \]

[In]

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

[Out]

log(x^2 - 5*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.12 \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\log \left ({\left | x^{2} - 5 \, x \right |}\right ) \]

[In]

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

[Out]

log(abs(x^2 - 5*x))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {-5+2 x}{-5 x+x^2} \, dx=\ln \left (x\,\left (x-5\right )\right ) \]

[In]

int(-(2*x - 5)/(5*x - x^2),x)

[Out]

log(x*(x - 5))