\(\int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} (1-2 x+x^2)} \, dx\) [6679]

Optimal result

Integrand size = 35, antiderivative size = 25 \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\log \left (-1-e^{-e^3} (1-5 x)+2 x-x^2\right ) \]

[Out]

ln(-1-(-5*x+1)/exp(exp(3))-x^2+2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {1601} \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\log \left (e^{e^3} \left (x^2-2 x+1\right )-5 x+1\right ) \]

[In]

Int[(-5 + E^E^3*(-2 + 2*x))/(1 - 5*x + E^E^3*(1 - 2*x + x^2)),x]

[Out]

Log[1 - 5*x + E^E^3*(1 - 2*x + x^2)]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps \begin{align*} \text {integral}& = \log \left (1-5 x+e^{e^3} \left (1-2 x+x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\log \left (-4-5 (-1+x)+e^{e^3} (-1+x)^2\right ) \]

[In]

Integrate[(-5 + E^E^3*(-2 + 2*x))/(1 - 5*x + E^E^3*(1 - 2*x + x^2)),x]

[Out]

Log[-4 - 5*(-1 + x) + E^E^3*(-1 + x)^2]

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\ln \left (\left (x^{2}-2 x +1\right ) {\mathrm e}^{{\mathrm e}^{3}}-5 x +1\right )\)	\(19\)
default	\(\ln \left (x^{2} {\mathrm e}^{{\mathrm e}^{3}}-2 x \,{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{{\mathrm e}^{3}}-5 x +1\right )\)	\(23\)
norman	\(\ln \left (x^{2} {\mathrm e}^{{\mathrm e}^{3}}-2 x \,{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{{\mathrm e}^{3}}-5 x +1\right )\)	\(23\)
risch	\(\ln \left (x^{2} {\mathrm e}^{{\mathrm e}^{3}}+\left (-2 \,{\mathrm e}^{{\mathrm e}^{3}}-5\right ) x +{\mathrm e}^{{\mathrm e}^{3}}+1\right )\)	\(23\)
parallelrisch	\(\ln \left (\left (x^{2} {\mathrm e}^{{\mathrm e}^{3}}-2 x \,{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{{\mathrm e}^{3}}-5 x +1\right ) {\mathrm e}^{-{\mathrm e}^{3}}\right )\)	\(29\)

[In]

int(((-2+2*x)*exp(exp(3))-5)/((x^2-2*x+1)*exp(exp(3))-5*x+1),x,method=_RETURNVERBOSE)

[Out]

ln((x^2-2*x+1)*exp(exp(3))-5*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\log \left ({\left (x^{2} - 2 \, x + 1\right )} e^{\left (e^{3}\right )} - 5 \, x + 1\right ) \]

[In]

integrate(((-2+2*x)*exp(exp(3))-5)/((x^2-2*x+1)*exp(exp(3))-5*x+1),x, algorithm="fricas")

[Out]

log((x^2 - 2*x + 1)*e^(e^3) - 5*x + 1)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\log {\left (x^{2} e^{e^{3}} + x \left (- 2 e^{e^{3}} - 5\right ) + 1 + e^{e^{3}} \right )} \]

[In]

integrate(((-2+2*x)*exp(exp(3))-5)/((x**2-2*x+1)*exp(exp(3))-5*x+1),x)

[Out]

log(x**2*exp(exp(3)) + x*(-2*exp(exp(3)) - 5) + 1 + exp(exp(3)))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\log \left (x^{2} e^{\left (e^{3}\right )} - x {\left (2 \, e^{\left (e^{3}\right )} + 5\right )} + e^{\left (e^{3}\right )} + 1\right ) \]

[In]

integrate(((-2+2*x)*exp(exp(3))-5)/((x^2-2*x+1)*exp(exp(3))-5*x+1),x, algorithm="maxima")

[Out]

log(x^2*e^(e^3) - x*(2*e^(e^3) + 5) + e^(e^3) + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\log \left ({\left | {\left (x^{2} - 2 \, x\right )} e^{\left (e^{3}\right )} - 5 \, x + e^{\left (e^{3}\right )} + 1 \right |}\right ) \]

[In]

integrate(((-2+2*x)*exp(exp(3))-5)/((x^2-2*x+1)*exp(exp(3))-5*x+1),x, algorithm="giac")

[Out]

log(abs((x^2 - 2*x)*e^(e^3) - 5*x + e^(e^3) + 1))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-5+e^{e^3} (-2+2 x)}{1-5 x+e^{e^3} \left (1-2 x+x^2\right )} \, dx=\ln \left ({\mathrm {e}}^{{\mathrm {e}}^3}\,x^2+\left (-2\,{\mathrm {e}}^{{\mathrm {e}}^3}-5\right )\,x+{\mathrm {e}}^{{\mathrm {e}}^3}+1\right ) \]

[In]

int((exp(exp(3))*(2*x - 2) - 5)/(exp(exp(3))*(x^2 - 2*x + 1) - 5*x + 1),x)

[Out]

log(exp(exp(3)) + x^2*exp(exp(3)) - x*(2*exp(exp(3)) + 5) + 1)