\(\int \frac {1}{15} e^{\frac {1}{15} (105+15 x+5 x^2+(-3 x-5 x^2) \log (e^{-e^x} x))} (12+5 x+e^x (3 x+5 x^2)+(-3-10 x) \log (e^{-e^x} x)) \, dx\) [4226]

Optimal result

Integrand size = 74, antiderivative size = 29 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {1}{3} x \left (x-\left (\frac {3}{5}+x\right ) \log \left (e^{-e^x} x\right )\right )} \]

[Out]

exp(7+1/3*x*(x-ln(x/exp(exp(x)))*(x+3/5))+x)

Rubi [F]

\[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\int \frac {1}{15} \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx \]

[In]

Int[(E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*(12 + 5*x + E^x*(3*x + 5*x^2) + (-3 - 10*x)*Log
[x/E^E^x]))/15,x]

[Out]

(4*Defer[Int][E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15), x])/5 + Defer[Int][E^((105 + 15*x + 5
*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*x, x]/3 + Defer[Int][E^(x + (105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x
/E^E^x])/15)*x, x]/5 + Defer[Int][E^(x + (105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*x^2, x]/3 - De
fer[Int][E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*Log[x/E^E^x], x]/5 - (2*Defer[Int][E^((105
+ 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*x*Log[x/E^E^x], x])/3

Rubi steps \begin{align*} \text {integral}& = \frac {1}{15} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx \\ & = \frac {1}{15} \int \left (12 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right )+5 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x+\exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x (3+5 x)-\exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) (3+10 x) \log \left (e^{-e^x} x\right )\right ) \, dx \\ & = \frac {1}{15} \int \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x (3+5 x) \, dx-\frac {1}{15} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) (3+10 x) \log \left (e^{-e^x} x\right ) \, dx+\frac {1}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx+\frac {4}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \, dx \\ & = \frac {1}{15} \int \left (3 \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x+5 \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x^2\right ) \, dx-\frac {1}{15} \int \left (3 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \log \left (e^{-e^x} x\right )+10 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \log \left (e^{-e^x} x\right )\right ) \, dx+\frac {1}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx+\frac {4}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \, dx \\ & = \frac {1}{5} \int \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx-\frac {1}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \log \left (e^{-e^x} x\right ) \, dx+\frac {1}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx+\frac {1}{3} \int \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x^2 \, dx-\frac {2}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \log \left (e^{-e^x} x\right ) \, dx+\frac {4}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {x^2}{3}} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (3+5 x)} \]

[In]

Integrate[(E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*(12 + 5*x + E^x*(3*x + 5*x^2) + (-3 - 10*
x)*Log[x/E^E^x]))/15,x]

[Out]

E^(7 + x + x^2/3)/(x/E^E^x)^((x*(3 + 5*x))/15)

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (-5 x^{2}-3 x \right ) \ln \left (x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )}{15}+\frac {x^{2}}{3}+x +7}\)	\(29\)
risch	\(x^{-\frac {x^{2}}{3}} x^{-\frac {x}{5}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x^{2}}{3}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x}{5}} {\mathrm e}^{7+\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} x^{2}}{6}+\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} x}{10}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) x^{2}}{6}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) x}{10}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {x^{2}}{3}+x}\)	\(235\)

[In]

int(1/15*((-10*x-3)*ln(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*exp(1/15*(-5*x^2-3*x)*ln(x/exp(exp(x)))+1/3*x
^2+x+7),x,method=_RETURNVERBOSE)

[Out]

exp(1/15*(-5*x^2-3*x)*ln(x/exp(exp(x)))+1/3*x^2+x+7)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} - \frac {1}{15} \, {\left (5 \, x^{2} + 3 \, x\right )} \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \]

[In]

integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*exp(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)
))+1/3*x^2+x+7),x, algorithm="fricas")

[Out]

e^(1/3*x^2 - 1/15*(5*x^2 + 3*x)*log(x*e^(-e^x)) + x + 7)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate(1/15*((-10*x-3)*ln(x/exp(exp(x)))+(5*x**2+3*x)*exp(x)+5*x+12)*exp(1/15*(-5*x**2-3*x)*ln(x/exp(exp(x)
))+1/3*x**2+x+7),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} e^{x} - \frac {1}{3} \, x^{2} \log \left (x\right ) + \frac {1}{3} \, x^{2} + \frac {1}{5} \, x e^{x} - \frac {1}{5} \, x \log \left (x\right ) + x + 7\right )} \]

[In]

integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*exp(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)
))+1/3*x^2+x+7),x, algorithm="maxima")

[Out]

e^(1/3*x^2*e^x - 1/3*x^2*log(x) + 1/3*x^2 + 1/5*x*e^x - 1/5*x*log(x) + x + 7)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (-\frac {1}{3} \, x^{2} \log \left (x e^{\left (-e^{x}\right )}\right ) + \frac {1}{3} \, x^{2} - \frac {1}{5} \, x \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \]

[In]

integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*exp(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)
))+1/3*x^2+x+7),x, algorithm="giac")

[Out]

e^(-1/3*x^2*log(x*e^(-e^x)) + 1/3*x^2 - 1/5*x*log(x*e^(-e^x)) + x + 7)

Mupad [B] (verification not implemented)

Time = 11.97 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^7\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{3}}\,{\mathrm {e}}^{\frac {x^2}{3}}\,{\mathrm {e}}^x}{x^{\frac {x^2}{3}+\frac {x}{5}}} \]

[In]

int((exp(x - (log(x*exp(-exp(x)))*(3*x + 5*x^2))/15 + x^2/3 + 7)*(5*x - log(x*exp(-exp(x)))*(10*x + 3) + exp(x
)*(3*x + 5*x^2) + 12))/15,x)

[Out]

(exp((x*exp(x))/5)*exp(7)*exp((x^2*exp(x))/3)*exp(x^2/3)*exp(x))/x^(x/5 + x^2/3)