\(\int \frac {-e^3+2 x}{e^3 x-x^2} \, dx\) [3659]

Optimal result

Integrand size = 23, antiderivative size = 19 \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=16+\log \left (\frac {1}{5 \left (e^3-x\right ) x}\right ) \]

[Out]

ln(1/5/x/(-x+exp(3)))+16

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {642} \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=-\log \left (e^3 x-x^2\right ) \]

[In]

Int[(-E^3 + 2*x)/(E^3*x - x^2),x]

[Out]

-Log[E^3*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 (e^3 x-x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=-\log \left (e^3-x\right )-\log (x) \]

[In]

Integrate[(-E^3 + 2*x)/(E^3*x - x^2),x]

[Out]

-Log[E^3 - x] - Log[x]

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63

method	result	size
default	\(-\ln \left (x \left (-x +{\mathrm e}^{3}\right )\right )\)	\(12\)
risch	\(-\ln \left (-x \,{\mathrm e}^{3}+x^{2}\right )\)	\(13\)
norman	\(-\ln \left (x \right )-\ln \left (-x +{\mathrm e}^{3}\right )\)	\(15\)
parallelrisch	\(-\ln \left (x \right )-\ln \left (x -{\mathrm e}^{3}\right )\)	\(15\)
meijerg	\(-\ln \left (x \right )+3-i \pi -\ln \left (1-{\mathrm e}^{-3} x \right )\)	\(21\)

[In]

int((-exp(3)+2*x)/(x*exp(3)-x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(x*(-x+exp(3)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=-\log \left (x^{2} - x e^{3}\right ) \]

[In]

integrate((-exp(3)+2*x)/(x*exp(3)-x^2),x, algorithm="fricas")

[Out]

-log(x^2 - x*e^3)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=- \log {\left (x^{2} - x e^{3} \right )} \]

[In]

integrate((-exp(3)+2*x)/(x*exp(3)-x**2),x)

[Out]

-log(x**2 - x*exp(3))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=-\log \left (x^{2} - x e^{3}\right ) \]

[In]

integrate((-exp(3)+2*x)/(x*exp(3)-x^2),x, algorithm="maxima")

[Out]

-log(x^2 - x*e^3)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=-\log \left ({\left | x^{2} - x e^{3} \right |}\right ) \]

[In]

integrate((-exp(3)+2*x)/(x*exp(3)-x^2),x, algorithm="giac")

[Out]

-log(abs(x^2 - x*e^3))

Mupad [B] (verification not implemented)

Time = 8.98 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {-e^3+2 x}{e^3 x-x^2} \, dx=-\ln \left (x\,\left (x-{\mathrm {e}}^3\right )\right ) \]

[In]

int((2*x - exp(3))/(x*exp(3) - x^2),x)

[Out]

-log(x*(x - exp(3)))