\(\int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx\) [2797]

Optimal result

Integrand size = 48, antiderivative size = 19 \[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=\log \left (\frac {x}{e^{24}+e^{-e^2+x}+x}\right ) \]

[Out]

ln(x/(exp(x-exp(2))+x+exp(27)/exp(3)))

Rubi [F]

\[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=\int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx \]

[In]

Int[(E^27 + E^(3 - E^2 + x)*(1 - x))/(E^27*x + E^(3 - E^2 + x)*x + E^3*x^2),x]

[Out]

-x + Log[x] - E^E^2*(1 - E^24)*Defer[Int][(E^(24 + E^2) + E^x + E^E^2*x)^(-1), x] + E^E^2*Defer[Int][x/(E^(24
+ E^2) + E^x + E^E^2*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1-x}{x}+\frac {e^{e^2} \left (-1+e^{24}+x\right )}{e^{24+e^2}+e^x+e^{e^2} x}\right ) \, dx \\ & = e^{e^2} \int \frac {-1+e^{24}+x}{e^{24+e^2}+e^x+e^{e^2} x} \, dx+\int \frac {1-x}{x} \, dx \\ & = e^{e^2} \int \left (-\frac {1-e^{24}}{e^{24+e^2}+e^x+e^{e^2} x}+\frac {x}{e^{24+e^2}+e^x+e^{e^2} x}\right ) \, dx+\int \left (-1+\frac {1}{x}\right ) \, dx \\ & = -x+\log (x)+e^{e^2} \int \frac {x}{e^{24+e^2}+e^x+e^{e^2} x} \, dx-\left (e^{e^2} \left (1-e^{24}\right )\right ) \int \frac {1}{e^{24+e^2}+e^x+e^{e^2} x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=\log (x)-\log \left (e^{48+e^2}+e^{24+x}+e^{24+e^2} x\right ) \]

[In]

Integrate[(E^27 + E^(3 - E^2 + x)*(1 - x))/(E^27*x + E^(3 - E^2 + x)*x + E^3*x^2),x]

[Out]

Log[x] - Log[E^(48 + E^2) + E^(24 + x) + E^(24 + E^2)*x]

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16

method	result	size
risch	\(\ln \left (x \right )-{\mathrm e}^{2}-\ln \left ({\mathrm e}^{24}+x +{\mathrm e}^{x -{\mathrm e}^{2}}\right )\)	\(22\)
norman	\(-\ln \left ({\mathrm e}^{3} {\mathrm e}^{x -{\mathrm e}^{2}}+x \,{\mathrm e}^{3}+{\mathrm e}^{27}\right )+\ln \left (x \right )\)	\(24\)
parallelrisch	\(\ln \left (x \right )-\ln \left (\left ({\mathrm e}^{3} {\mathrm e}^{x -{\mathrm e}^{2}}+x \,{\mathrm e}^{3}+{\mathrm e}^{27}\right ) {\mathrm e}^{-3}\right )\)	\(29\)

[In]

int(((1-x)*exp(3)*exp(x-exp(2))+exp(27))/(x*exp(3)*exp(x-exp(2))+x*exp(27)+x^2*exp(3)),x,method=_RETURNVERBOSE
)

[Out]

ln(x)-exp(2)-ln(exp(24)+x+exp(x-exp(2)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=-\log \left (x e^{3} + e^{27} + e^{\left (x - e^{2} + 3\right )}\right ) + \log \left (x\right ) \]

[In]

integrate(((1-x)*exp(3)*exp(x-exp(2))+exp(27))/(x*exp(3)*exp(x-exp(2))+x*exp(27)+x^2*exp(3)),x, algorithm="fri
cas")

[Out]

-log(x*e^3 + e^27 + e^(x - e^2 + 3)) + log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=\log {\left (x \right )} - \log {\left (x + e^{x - e^{2}} + e^{24} \right )} \]

[In]

integrate(((1-x)*exp(3)*exp(x-exp(2))+exp(27))/(x*exp(3)*exp(x-exp(2))+x*exp(27)+x**2*exp(3)),x)

[Out]

log(x) - log(x + exp(x - exp(2)) + exp(24))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=-\log \left (x e^{\left (e^{2}\right )} + e^{x} + e^{\left (e^{2} + 24\right )}\right ) + \log \left (x\right ) \]

[In]

integrate(((1-x)*exp(3)*exp(x-exp(2))+exp(27))/(x*exp(3)*exp(x-exp(2))+x*exp(27)+x^2*exp(3)),x, algorithm="max
ima")

[Out]

-log(x*e^(e^2) + e^x + e^(e^2 + 24)) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=-\log \left (x e^{3} + e^{27} + e^{\left (x - e^{2} + 3\right )}\right ) + \log \left (x\right ) \]

[In]

integrate(((1-x)*exp(3)*exp(x-exp(2))+exp(27))/(x*exp(3)*exp(x-exp(2))+x*exp(27)+x^2*exp(3)),x, algorithm="gia
c")

[Out]

-log(x*e^3 + e^27 + e^(x - e^2 + 3)) + log(x)

Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{27}+e^{3-e^2+x} (1-x)}{e^{27} x+e^{3-e^2+x} x+e^3 x^2} \, dx=\ln \left (x\right )-\ln \left (x+{\mathrm {e}}^{24}+{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^x\right ) \]

[In]

int((exp(27) - exp(x - exp(2))*exp(3)*(x - 1))/(x*exp(27) + x^2*exp(3) + x*exp(x - exp(2))*exp(3)),x)

[Out]

log(x) - log(x + exp(24) + exp(-exp(2))*exp(x))