\(\int \frac {e^{2-x} (-141+e^x (-48-x)-3 x+(-3-e^x) \log (3+e^x))}{(46+x+\log (3+e^x)) (138+3 x+e^x (46+x)+(3+e^x) \log (3+e^x))} \, dx\) [1071]

Optimal result

Integrand size = 73, antiderivative size = 19 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2-x}}{46+x+\log \left (3+e^x\right )} \]

[Out]

exp(-ln(ln(3+exp(x))+x+46)+2-x)

Rubi [F]

\[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx \]

[In]

Int[(E^(2 - x)*(-141 + E^x*(-48 - x) - 3*x + (-3 - E^x)*Log[3 + E^x]))/((46 + x + Log[3 + E^x])*(138 + 3*x + E
^x*(46 + x) + (3 + E^x)*Log[3 + E^x])),x]

[Out]

Defer[Int][E^(2 - x)/(-46 - x - Log[3 + E^x]), x] - 2*Defer[Int][E^(2 - x)/(46 + x + Log[3 + E^x])^2, x] + 3*D
efer[Int][E^(2 - x)/((3 + E^x)*(46 + x + Log[3 + E^x])^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2-x} \left (-3 (47+x)-e^x (48+x)-\left (3+e^x\right ) \log \left (3+e^x\right )\right )}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx \\ & = \int \left (\frac {3 e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2}+\frac {e^{2-x} \left (-48-x-\log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right )^2}\right ) \, dx \\ & = 3 \int \frac {e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx+\int \frac {e^{2-x} \left (-48-x-\log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right )^2} \, dx \\ & = 3 \int \frac {e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx+\int \left (\frac {e^{2-x}}{-46-x-\log \left (3+e^x\right )}-\frac {2 e^{2-x}}{\left (46+x+\log \left (3+e^x\right )\right )^2}\right ) \, dx \\ & = -\left (2 \int \frac {e^{2-x}}{\left (46+x+\log \left (3+e^x\right )\right )^2} \, dx\right )+3 \int \frac {e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx+\int \frac {e^{2-x}}{-46-x-\log \left (3+e^x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.53 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2-x}}{46+x+\log \left (3+e^x\right )} \]

[In]

Integrate[(E^(2 - x)*(-141 + E^x*(-48 - x) - 3*x + (-3 - E^x)*Log[3 + E^x]))/((46 + x + Log[3 + E^x])*(138 + 3
*x + E^x*(46 + x) + (3 + E^x)*Log[3 + E^x])),x]

[Out]

E^(2 - x)/(46 + x + Log[3 + E^x])

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {{\mathrm e}^{2-x}}{\ln \left (3+{\mathrm e}^{x}\right )+x +46}\)	\(18\)
parallelrisch	\({\mathrm e}^{-\ln \left (\ln \left (3+{\mathrm e}^{x}\right )+x +46\right )+2-x}\)	\(18\)

[In]

int(((-exp(x)-3)*ln(3+exp(x))+(-x-48)*exp(x)-3*x-141)*exp(-ln(ln(3+exp(x))+x+46)+2-x)/((3+exp(x))*ln(3+exp(x))
+(x+46)*exp(x)+3*x+138),x,method=_RETURNVERBOSE)

[Out]

1/(ln(3+exp(x))+x+46)*exp(2-x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{{\left (x + 46\right )} e^{x} + e^{x} \log \left (e^{x} + 3\right )} \]

[In]

integrate(((-exp(x)-3)*log(3+exp(x))+(-x-48)*exp(x)-3*x-141)*exp(-log(log(3+exp(x))+x+46)+2-x)/((3+exp(x))*log
(3+exp(x))+(x+46)*exp(x)+3*x+138),x, algorithm="fricas")

[Out]

e^2/((x + 46)*e^x + e^x*log(e^x + 3))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{x e^{x} + e^{x} \log {\left (e^{x} + 3 \right )} + 46 e^{x}} \]

[In]

integrate(((-exp(x)-3)*ln(3+exp(x))+(-x-48)*exp(x)-3*x-141)*exp(-ln(ln(3+exp(x))+x+46)+2-x)/((3+exp(x))*ln(3+e
xp(x))+(x+46)*exp(x)+3*x+138),x)

[Out]

exp(2)/(x*exp(x) + exp(x)*log(exp(x) + 3) + 46*exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{{\left (x + 46\right )} e^{x} + e^{x} \log \left (e^{x} + 3\right )} \]

[In]

integrate(((-exp(x)-3)*log(3+exp(x))+(-x-48)*exp(x)-3*x-141)*exp(-log(log(3+exp(x))+x+46)+2-x)/((3+exp(x))*log
(3+exp(x))+(x+46)*exp(x)+3*x+138),x, algorithm="maxima")

[Out]

e^2/((x + 46)*e^x + e^x*log(e^x + 3))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{x e^{x} + e^{x} \log \left (e^{x} + 3\right ) + 46 \, e^{x}} \]

[In]

integrate(((-exp(x)-3)*log(3+exp(x))+(-x-48)*exp(x)-3*x-141)*exp(-log(log(3+exp(x))+x+46)+2-x)/((3+exp(x))*log
(3+exp(x))+(x+46)*exp(x)+3*x+138),x, algorithm="giac")

[Out]

e^2/(x*e^x + e^x*log(e^x + 3) + 46*e^x)

Mupad [B] (verification not implemented)

Time = 9.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^2}{x+\ln \left ({\mathrm {e}}^x+3\right )+46} \]

[In]

int(-(exp(2 - log(x + log(exp(x) + 3) + 46) - x)*(3*x + exp(x)*(x + 48) + log(exp(x) + 3)*(exp(x) + 3) + 141))
/(3*x + exp(x)*(x + 46) + log(exp(x) + 3)*(exp(x) + 3) + 138),x)

[Out]

(exp(-x)*exp(2))/(x + log(exp(x) + 3) + 46)