\(\int \frac {e^{-x} (9 e^x x+e^{e^{-x} (e^x (5-e^2)-x+e^x \log (x^2))} (-2 e^x+x-x^2))}{x} \, dx\) [3648]

Optimal result

Integrand size = 60, antiderivative size = 27 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=3+9 x-e^{5-e^2-e^{-x} x} x^2 \]

[Out]

3+9*x-exp(5+ln(x^2)-x/exp(x)-exp(1)^2)

Rubi [F]

\[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=\int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx \]

[In]

Int[(9*E^x*x + E^((E^x*(5 - E^2) - x + E^x*Log[x^2])/E^x)*(-2*E^x + x - x^2))/(E^x*x),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \left (9+e^{5 \left (1-\frac {e^2}{5}\right )-x-e^{-x} x} x \left (-2 e^x+x-x^2\right )\right ) \, dx \\ & = 9 x+\int e^{5 \left (1-\frac {e^2}{5}\right )-x-e^{-x} x} x \left (-2 e^x+x-x^2\right ) \, dx \\ & = 9 x+\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x \left (-2 e^x+x-x^2\right ) \, dx \\ & = 9 x+\int \left (-2 \exp \left (x-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x-\exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) (-1+x) x^2\right ) \, dx \\ & = 9 x-2 \int \exp \left (x-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x \, dx-\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) (-1+x) x^2 \, dx \\ & = 9 x-2 \int e^{5 \left (1-\frac {e^2}{5}\right )-e^{-x} x} x \, dx-\int \left (-\exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^2+\exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^3\right ) \, dx \\ & = 9 x-2 \int e^{-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x\right )} x \, dx+\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^2 \, dx-\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^3 \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9 x-e^{5-e^2-e^{-x} x} x^2 \]

[In]

Integrate[(9*E^x*x + E^((E^x*(5 - E^2) - x + E^x*Log[x^2])/E^x)*(-2*E^x + x - x^2))/(E^x*x),x]

[Out]

9*x - E^(5 - E^2 - x/E^x)*x^2

Maple [A] (verified)

Time = 2.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30

method	result	size
parallelrisch	\(9 x -{\mathrm e}^{-\left ({\mathrm e}^{2} {\mathrm e}^{x}-{\mathrm e}^{x} \ln \left (x^{2}\right )-5 \,{\mathrm e}^{x}+x \right ) {\mathrm e}^{-x}}\)	\(35\)
default	\(\left (9 \,{\mathrm e}^{x} x -{\mathrm e}^{x} {\mathrm e}^{\left ({\mathrm e}^{x} \ln \left (x^{2}\right )+\left (5-{\mathrm e}^{2}\right ) {\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}}\right ) {\mathrm e}^{-x}\)	\(44\)
norman	\(\left (9 \,{\mathrm e}^{x} x -{\mathrm e}^{x} {\mathrm e}^{\left ({\mathrm e}^{x} \ln \left (x^{2}\right )+\left (5-{\mathrm e}^{2}\right ) {\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}}\right ) {\mathrm e}^{-x}\)	\(44\)
risch	\(9 x -{\mathrm e}^{\frac {\left (-i {\mathrm e}^{x} \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i {\mathrm e}^{x} \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i {\mathrm e}^{x} \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 \,{\mathrm e}^{x} \ln \left (x \right )+10 \,{\mathrm e}^{x}-2 \,{\mathrm e}^{2+x}-2 x \right ) {\mathrm e}^{-x}}{2}}\)	\(89\)

[In]

int(((-2*exp(x)-x^2+x)*exp((exp(x)*ln(x^2)+(-exp(1)^2+5)*exp(x)-x)/exp(x))+9*exp(x)*x)/exp(x)/x,x,method=_RETU
RNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9 \, x - e^{\left (-{\left ({\left (e^{2} - 5\right )} e^{x} - e^{x} \log \left (x^{2}\right ) + x\right )} e^{\left (-x\right )}\right )} \]

[In]

integrate(((-2*exp(x)-x^2+x)*exp((exp(x)*log(x^2)+(-exp(1)^2+5)*exp(x)-x)/exp(x))+9*exp(x)*x)/exp(x)/x,x, algo
rithm="fricas")

[Out]

9*x - e^(-((e^2 - 5)*e^x - e^x*log(x^2) + x)*e^(-x))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9 x - e^{\left (- x + e^{x} \log {\left (x^{2} \right )} + \left (5 - e^{2}\right ) e^{x}\right ) e^{- x}} \]

[In]

integrate(((-2*exp(x)-x**2+x)*exp((exp(x)*ln(x**2)+(-exp(1)**2+5)*exp(x)-x)/exp(x))+9*exp(x)*x)/exp(x)/x,x)

[Out]

9*x - exp((-x + exp(x)*log(x**2) + (5 - exp(2))*exp(x))*exp(-x))

Maxima [F]

\[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=\int { -\frac {{\left ({\left (x^{2} - x + 2 \, e^{x}\right )} e^{\left (-{\left ({\left (e^{2} - 5\right )} e^{x} - e^{x} \log \left (x^{2}\right ) + x\right )} e^{\left (-x\right )}\right )} - 9 \, x e^{x}\right )} e^{\left (-x\right )}}{x} \,d x } \]

[In]

integrate(((-2*exp(x)-x^2+x)*exp((exp(x)*log(x^2)+(-exp(1)^2+5)*exp(x)-x)/exp(x))+9*exp(x)*x)/exp(x)/x,x, algo
rithm="maxima")

[Out]

9*x - integrate((x^3*e^5 - x^2*e^5 + 2*x*e^(x + 5))*e^(-x*e^(-x) - x - e^2), x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=-{\left (x^{2} e^{\left (-x e^{\left (-x\right )} - x - e^{2} + 5\right )} - 9 \, x e^{\left (-x\right )}\right )} e^{x} \]

[In]

integrate(((-2*exp(x)-x^2+x)*exp((exp(x)*log(x^2)+(-exp(1)^2+5)*exp(x)-x)/exp(x))+9*exp(x)*x)/exp(x)/x,x, algo
rithm="giac")

[Out]

-(x^2*e^(-x*e^(-x) - x - e^2 + 5) - 9*x*e^(-x))*e^x

Mupad [B] (verification not implemented)

Time = 9.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9\,x-x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{-x}} \]

[In]

int(-(exp(-x)*(exp(-exp(-x)*(x + exp(x)*(exp(2) - 5) - log(x^2)*exp(x)))*(2*exp(x) - x + x^2) - 9*x*exp(x)))/x
,x)

[Out]

9*x - x^2*exp(-exp(2))*exp(5)*exp(-x*exp(-x))