\(\int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 (5+\frac {50 e^{e^2+x} \log (\frac {\log (4)}{x})}{x^2})} (10-\frac {50 e^{e^2+x}}{x^2}-5 x) x \, dx\) [1368]

Optimal result

Integrand size = 72, antiderivative size = 26 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \]

[Out]

exp(5/exp(ln(50/x^2)+x+exp(2))+ln(2*ln(2)/x))

Rubi [F]

\[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\int \frac {1}{50} \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx \]

[In]

Int[(E^(-E^2 - x + (E^(-E^2 - x)*x^2*(5 + (50*E^(E^2 + x)*Log[Log[4]/x])/x^2))/50)*(10 - (50*E^(E^2 + x))/x^2
- 5*x)*x)/50,x]

[Out]

(Log[4]*Defer[Int][E^(-E^2 - x + (E^(-E^2 - x)*x^2)/10), x])/5 - Log[4]*Defer[Int][E^((E^(-E^2 - x)*x^2)/10)/x
^2, x] - (Log[4]*Defer[Int][E^(-E^2 - x + (E^(-E^2 - x)*x^2)/10)*x, x])/10

Rubi steps \begin{align*} \text {integral}& = \frac {1}{50} \int \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx \\ & = \frac {1}{50} \int \left (-\frac {50 \exp \left (\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right )}{x}-5 \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) (-2+x) x\right ) \, dx \\ & = -\left (\frac {1}{10} \int \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) (-2+x) x \, dx\right )-\int \frac {\exp \left (\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right )}{x} \, dx \\ & = -\left (\frac {1}{10} \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} (-2+x) \log (4) \, dx\right )-\int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x^2} \, dx \\ & = -\left (\frac {1}{10} \log (4) \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} (-2+x) \, dx\right )-\log (4) \int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2}}{x^2} \, dx \\ & = -\left (\frac {1}{10} \log (4) \int \left (-2 e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2}+e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} x\right ) \, dx\right )-\log (4) \int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2}}{x^2} \, dx \\ & = -\left (\frac {1}{10} \log (4) \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} x \, dx\right )+\frac {1}{5} \log (4) \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} \, dx-\log (4) \int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2}}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \]

[In]

Integrate[(E^(-E^2 - x + (E^(-E^2 - x)*x^2*(5 + (50*E^(E^2 + x)*Log[Log[4]/x])/x^2))/50)*(10 - (50*E^(E^2 + x)
)/x^2 - 5*x)*x)/50,x]

[Out]

(E^((E^(-E^2 - x)*x^2)/10)*Log[4])/x

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46

method	result	size
default	\({\mathrm e}^{\frac {\left (\ln \left (\frac {2 \ln \left (2\right )}{x}\right ) {\mathrm e}^{\ln \left (\frac {50}{x^{2}}\right )+x +{\mathrm e}^{2}}+5\right ) x^{2} {\mathrm e}^{-x -{\mathrm e}^{2}}}{50}}\)	\(38\)
parallelrisch	\({\mathrm e}^{\frac {\left (\ln \left (\frac {2 \ln \left (2\right )}{x}\right ) {\mathrm e}^{\ln \left (\frac {50}{x^{2}}\right )+x +{\mathrm e}^{2}}+5\right ) x^{2} {\mathrm e}^{-x -{\mathrm e}^{2}}}{50}}\)	\(38\)
risch	\(\frac {2 \ln \left (2\right ) {\mathrm e}^{-\frac {x^{2} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{2}-x -{\mathrm e}^{2}}}{10}}}{x}\)	\(54\)

[In]

int((-exp(ln(50/x^2)+x+exp(2))-5*x+10)*exp((ln(2*ln(2)/x)*exp(ln(50/x^2)+x+exp(2))+5)/exp(ln(50/x^2)+x+exp(2))
)/x/exp(ln(50/x^2)+x+exp(2)),x,method=_RETURNVERBOSE)

[Out]

exp((ln(2*ln(2)/x)*exp(ln(50/x^2)+x+exp(2))+5)/exp(ln(50/x^2)+x+exp(2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (29) = 58\).

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.58 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=e^{\left (-\frac {1}{2} \, {\left ({\left (2 \, x + 2 \, e^{2} - \log \left (\frac {2}{25} \, \log \left (2\right )^{2}\right ) + \log \left (\frac {50}{x^{2}}\right )\right )} e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} + x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \]

[In]

integrate((-exp(log(50/x^2)+x+exp(2))-5*x+10)*exp((log(2*log(2)/x)*exp(log(50/x^2)+x+exp(2))+5)/exp(log(50/x^2
)+x+exp(2)))/x/exp(log(50/x^2)+x+exp(2)),x, algorithm="fricas")

[Out]

e^(-1/2*((2*x + 2*e^2 - log(2/25*log(2)^2) + log(50/x^2))*e^(x + e^2 + log(50/x^2)) - 10)*e^(-x - e^2 - log(50
/x^2)) + x + e^2 + log(50/x^2))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=e^{\frac {x^{2} \cdot \left (5 + \frac {50 e^{x + e^{2}} \log {\left (\frac {2 \log {\left (2 \right )}}{x} \right )}}{x^{2}}\right ) e^{- x - e^{2}}}{50}} \]

[In]

integrate((-exp(ln(50/x**2)+x+exp(2))-5*x+10)*exp((ln(2*ln(2)/x)*exp(ln(50/x**2)+x+exp(2))+5)/exp(ln(50/x**2)+
x+exp(2)))/x/exp(ln(50/x**2)+x+exp(2)),x)

[Out]

exp(x**2*(5 + 50*exp(x + exp(2))*log(2*log(2)/x)/x**2)*exp(-x - exp(2))/50)

Maxima [F]

\[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\int { -\frac {{\left (5 \, x + e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left ({\left (e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \log \left (\frac {2 \, \log \left (2\right )}{x}\right ) + 5\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} - x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )}}{x} \,d x } \]

[In]

integrate((-exp(log(50/x^2)+x+exp(2))-5*x+10)*exp((log(2*log(2)/x)*exp(log(50/x^2)+x+exp(2))+5)/exp(log(50/x^2
)+x+exp(2)))/x/exp(log(50/x^2)+x+exp(2)),x, algorithm="maxima")

[Out]

-1/50*integrate(5*(x + 10*e^(x + e^2)/x^2 - 2)*x*e^(1/10*x^2*(10*e^(x + e^2)*log(2*log(2)/x)/x^2 + 1)*e^(-x -
e^2) - x - e^2), x)

Giac [F]

\[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=\int { -\frac {{\left (5 \, x + e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left ({\left (e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \log \left (\frac {2 \, \log \left (2\right )}{x}\right ) + 5\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} - x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )}}{x} \,d x } \]

[In]

integrate((-exp(log(50/x^2)+x+exp(2))-5*x+10)*exp((log(2*log(2)/x)*exp(log(50/x^2)+x+exp(2))+5)/exp(log(50/x^2
)+x+exp(2)))/x/exp(log(50/x^2)+x+exp(2)),x, algorithm="giac")

[Out]

integrate(-(5*x + e^(x + e^2 + log(50/x^2)) - 10)*e^((e^(x + e^2 + log(50/x^2))*log(2*log(2)/x) + 5)*e^(-x - e
^2 - log(50/x^2)) - x - e^2 - log(50/x^2))/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{50} e^{-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )} \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx=-\int \frac {x\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \left (2\right )}{5}-\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \left (2\right )}{5}+\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}}\,\ln \left (2\right )}{x^2} \,d x \]

[In]

int(-(exp(exp(- x - exp(2) - log(50/x^2))*(log((2*log(2))/x)*exp(x + exp(2) + log(50/x^2)) + 5))*exp(- x - exp
(2) - log(50/x^2))*(5*x + exp(x + exp(2) + log(50/x^2)) - 10))/x,x)

[Out]

-int((x*exp((x^2*exp(-exp(2))*exp(-x))/10 - exp(2) - x)*log(2))/5 - (2*exp((x^2*exp(-exp(2))*exp(-x))/10 - exp
(2) - x)*log(2))/5 + (2*exp((x^2*exp(-exp(2))*exp(-x))/10)*log(2))/x^2, x)