\(\int \frac {1}{-2 x+x^2} \, dx\) [214]

Optimal result

Integrand size = 9, antiderivative size = 17 \[ \int \frac {1}{-2 x+x^2} \, dx=\frac {1}{2} \log (2-x)-\frac {\log (x)}{2} \]

[Out]

1/2*ln(2-x)-1/2*ln(x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {629} \[ \int \frac {1}{-2 x+x^2} \, dx=\frac {1}{2} \log (2-x)-\frac {\log (x)}{2} \]

[In]

Int[(-2*x + x^2)^(-1),x]

[Out]

Log[2 - x]/2 - Log[x]/2

Rule 629

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[Log[x]/b, x] - Simp[Log[RemoveContent[b + c*x, x]]/b,
x] /; FreeQ[{b, c}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \log (2-x)-\frac {\log (x)}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-2 x+x^2} \, dx=\frac {1}{2} \log (2-x)-\frac {\log (x)}{2} \]

[In]

Integrate[(-2*x + x^2)^(-1),x]

[Out]

Log[2 - x]/2 - Log[x]/2

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71

method	result	size
default	\(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-2+x \right )}{2}\)	\(12\)
norman	\(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-2+x \right )}{2}\)	\(12\)
risch	\(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-2+x \right )}{2}\)	\(12\)
parallelrisch	\(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-2+x \right )}{2}\)	\(12\)
meijerg	\(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (2\right )}{2}-\frac {i \pi }{2}+\frac {\ln \left (1-\frac {x}{2}\right )}{2}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{-2 x+x^2} \, dx=\frac {1}{2} \, \log \left (x - 2\right ) - \frac {1}{2} \, \log \left (x\right ) \]

[In]

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

[Out]

1/2*log(x - 2) - 1/2*log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {1}{-2 x+x^2} \, dx=- \frac {\log {\left (x \right )}}{2} + \frac {\log {\left (x - 2 \right )}}{2} \]

[In]

integrate(1/(x**2-2*x),x)

[Out]

-log(x)/2 + log(x - 2)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{-2 x+x^2} \, dx=\frac {1}{2} \, \log \left (x - 2\right ) - \frac {1}{2} \, \log \left (x\right ) \]

[In]

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

[Out]

1/2*log(x - 2) - 1/2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {1}{-2 x+x^2} \, dx=\frac {1}{2} \, \log \left ({\left | x - 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/2*log(abs(x - 2)) - 1/2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.35 \[ \int \frac {1}{-2 x+x^2} \, dx=-\mathrm {atanh}\left (x-1\right ) \]

[In]

int(-1/(2*x - x^2),x)

[Out]

-atanh(x - 1)