\(\int \frac {8-12 x}{-x+2 x^2} \, dx\) [8061]

Optimal result

Integrand size = 17, antiderivative size = 15 \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=\log \left (\frac {(1-2 x)^2}{e^3 x^8}\right ) \]

[Out]

ln((1-2*x)^2/x^8/exp(3))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {645} \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=2 \log (1-2 x)-8 \log (x) \]

[In]

Int[(8 - 12*x)/(-x + 2*x^2),x]

[Out]

2*Log[1 - 2*x] - 8*Log[x]

Rule 645

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8}{x}+\frac {4}{-1+2 x}\right ) \, dx \\ & = 2 \log (1-2 x)-8 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=2 \log (1-2 x)-8 \log (x) \]

[In]

Integrate[(8 - 12*x)/(-x + 2*x^2),x]

[Out]

2*Log[1 - 2*x] - 8*Log[x]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method	result	size
parallelrisch	\(-8 \ln \left (x \right )+2 \ln \left (x -\frac {1}{2}\right )\)	\(12\)
default	\(2 \ln \left (-1+2 x \right )-8 \ln \left (x \right )\)	\(14\)
norman	\(2 \ln \left (-1+2 x \right )-8 \ln \left (x \right )\)	\(14\)
risch	\(2 \ln \left (-1+2 x \right )-8 \ln \left (x \right )\)	\(14\)
meijerg	\(-8 \ln \left (x \right )-8 \ln \left (2\right )-8 i \pi +2 \ln \left (1-2 x \right )\)	\(22\)

[In]

int((-12*x+8)/(2*x^2-x),x,method=_RETURNVERBOSE)

[Out]

-8*ln(x)+2*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=2 \, \log \left (2 \, x - 1\right ) - 8 \, \log \left (x\right ) \]

[In]

integrate((-12*x+8)/(2*x^2-x),x, algorithm="fricas")

[Out]

2*log(2*x - 1) - 8*log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=- 8 \log {\left (x \right )} + 2 \log {\left (x - \frac {1}{2} \right )} \]

[In]

integrate((-12*x+8)/(2*x**2-x),x)

[Out]

-8*log(x) + 2*log(x - 1/2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=2 \, \log \left (2 \, x - 1\right ) - 8 \, \log \left (x\right ) \]

[In]

integrate((-12*x+8)/(2*x^2-x),x, algorithm="maxima")

[Out]

2*log(2*x - 1) - 8*log(x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=2 \, \log \left ({\left | 2 \, x - 1 \right |}\right ) - 8 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((-12*x+8)/(2*x^2-x),x, algorithm="giac")

[Out]

2*log(abs(2*x - 1)) - 8*log(abs(x))

Mupad [B] (verification not implemented)

Time = 13.44 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {8-12 x}{-x+2 x^2} \, dx=2\,\ln \left (x-\frac {1}{2}\right )-8\,\ln \left (x\right ) \]

[In]

int((12*x - 8)/(x - 2*x^2),x)

[Out]

2*log(x - 1/2) - 8*log(x)