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

Optimal. Leaf size=26 \[ \frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \]

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Rubi [F]  time = 2.88, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [B]  time = 1.38, size = 67, normalized size = 2.58 \begin {gather*} e^{\left (-\frac {1}{2} \, {\left ({\left (2 \, x + 2 \, e^{2} - \log \left (\frac {2}{25} \, \log \relax (2)^{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \relax (2)}{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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.40, size = 38, normalized size = 1.46

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

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{50} \, \int 5 \, {\left (x + \frac {10 \, e^{\left (x + e^{2}\right )}}{x^{2}} - 2\right )} x e^{\left (\frac {1}{10} \, x^{2} {\left (\frac {10 \, e^{\left (x + e^{2}\right )} \log \left (\frac {2 \, \log \relax (2)}{x}\right )}{x^{2}} + 1\right )} e^{\left (-x - e^{2}\right )} - x - e^{2}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {x\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \relax (2)}{5}-\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \relax (2)}{5}+\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}}\,\ln \relax (2)}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.42, size = 36, normalized size = 1.38 \begin {gather*} e^{\frac {x^{2} \left (5 + \frac {50 e^{x + e^{2}} \log {\left (\frac {2 \log {\relax (2 )}}{x} \right )}}{x^{2}}\right ) e^{- x - e^{2}}}{50}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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