\(\int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx\) [6790]

Optimal result

Integrand size = 28, antiderivative size = 18 \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\log \left (2-(-5+x) \left (2-e^3+2 x\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1601} \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\log \left (-2 x^2+8 x-e^3 (5-x)+12\right ) \]

[In]

Int[(8 + E^3 - 4*x)/(12 + E^3*(-5 + x) + 8*x - 2*x^2),x]

[Out]

Log[12 - E^3*(5 - x) + 8*x - 2*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 (12-e^3 (5-x)+8 x-2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\log \left (12+e^3 (-5+x)+8 x-2 x^2\right ) \]

[In]

Integrate[(8 + E^3 - 4*x)/(12 + E^3*(-5 + x) + 8*x - 2*x^2),x]

[Out]

Log[12 + E^3*(-5 + x) + 8*x - 2*x^2]

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\ln \left (\left (-5+x \right ) {\mathrm e}^{3}-2 x^{2}+8 x +12\right )\)	\(18\)
parallelrisch	\(\ln \left (-\frac {x \,{\mathrm e}^{3}}{2}+x^{2}+\frac {5 \,{\mathrm e}^{3}}{2}-4 x -6\right )\)	\(19\)
default	\(\ln \left (x \,{\mathrm e}^{3}-2 x^{2}-5 \,{\mathrm e}^{3}+8 x +12\right )\)	\(20\)
norman	\(\ln \left (x \,{\mathrm e}^{3}-2 x^{2}-5 \,{\mathrm e}^{3}+8 x +12\right )\)	\(20\)
risch	\(\ln \left (2 x^{2}+x \left (-8-{\mathrm e}^{3}\right )+5 \,{\mathrm e}^{3}-12\right )\)	\(21\)

[In]

int((exp(3)-4*x+8)/((-5+x)*exp(3)-2*x^2+8*x+12),x,method=_RETURNVERBOSE)

[Out]

ln((-5+x)*exp(3)-2*x^2+8*x+12)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\log \left (2 \, x^{2} - {\left (x - 5\right )} e^{3} - 8 \, x - 12\right ) \]

[In]

integrate((exp(3)-4*x+8)/((-5+x)*exp(3)-2*x^2+8*x+12),x, algorithm="fricas")

[Out]

log(2*x^2 - (x - 5)*e^3 - 8*x - 12)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\log {\left (2 x^{2} + x \left (- e^{3} - 8\right ) - 12 + 5 e^{3} \right )} \]

[In]

integrate((exp(3)-4*x+8)/((-5+x)*exp(3)-2*x**2+8*x+12),x)

[Out]

log(2*x**2 + x*(-exp(3) - 8) - 12 + 5*exp(3))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\log \left (2 \, x^{2} - {\left (x - 5\right )} e^{3} - 8 \, x - 12\right ) \]

[In]

integrate((exp(3)-4*x+8)/((-5+x)*exp(3)-2*x^2+8*x+12),x, algorithm="maxima")

[Out]

log(2*x^2 - (x - 5)*e^3 - 8*x - 12)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\log \left ({\left | 2 \, x^{2} - x e^{3} - 8 \, x + 5 \, e^{3} - 12 \right |}\right ) \]

[In]

integrate((exp(3)-4*x+8)/((-5+x)*exp(3)-2*x^2+8*x+12),x, algorithm="giac")

[Out]

log(abs(2*x^2 - x*e^3 - 8*x + 5*e^3 - 12))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx=\ln \left (\frac {5\,{\mathrm {e}}^3}{2}-\frac {x\,\left ({\mathrm {e}}^3+8\right )}{2}+x^2-6\right ) \]

[In]

int((exp(3) - 4*x + 8)/(8*x + exp(3)*(x - 5) - 2*x^2 + 12),x)

[Out]

log((5*exp(3))/2 - (x*(exp(3) + 8))/2 + x^2 - 6)