3.24.85 \(\int \frac {e^{e^{\frac {2-2 x+2 x^2+(-5+2 e+2 x^2) \log (x)}{2 \log (x)}}+\frac {2-2 x+2 x^2+(-5+2 e+2 x^2) \log (x)}{2 \log (x)}} (-1+x-x^2+(-x+2 x^2) \log (x)+2 x^2 \log ^2(x))}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ e^{e^{-\frac {5}{2}+e+x^2+\frac {(1-x)^2+x}{\log (x)}}} \]

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Rubi [F]  time = 12.96, 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 {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) \left (-1+x-x^2+\left (-x+2 x^2\right ) \log (x)+2 x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^((2 - 2*x + 2*x^2 + (-5 + 2*E + 2*x^2)*Log[x])/(2*Log[x])) + (2 - 2*x + 2*x^2 + (-5 + 2*E + 2*x^2)*L
og[x])/(2*Log[x]))*(-1 + x - x^2 + (-x + 2*x^2)*Log[x] + 2*x^2*Log[x]^2))/(x*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 \exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x+\frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) \left (-1+x-x^2\right )}{x \log ^2(x)}+\frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) (-1+2 x)}{\log (x)}\right ) \, dx\\ &=2 \int \exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x \, dx+\int \frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) \left (-1+x-x^2\right )}{x \log ^2(x)} \, dx+\int \frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) (-1+2 x)}{\log (x)} \, dx\\ &=2 \int \exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x \, dx+\int \left (\frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )}{\log ^2(x)}-\frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )}{x \log ^2(x)}-\frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x}{\log ^2(x)}\right ) \, dx+\int \left (-\frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )}{\log (x)}+\frac {2 \exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x}{\log (x)}\right ) \, dx\\ &=2 \int \exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x \, dx+2 \int \frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x}{\log (x)} \, dx+\int \frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )}{\log ^2(x)} \, dx-\int \frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )}{x \log ^2(x)} \, dx-\int \frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right ) x}{\log ^2(x)} \, dx-\int \frac {\exp \left (\exp \left (\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )+\frac {2-2 x+2 x^2+\left (-5+2 e+2 x^2\right ) \log (x)}{2 \log (x)}\right )}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 25, normalized size = 0.96 \begin {gather*} e^{e^{-\frac {5}{2}+e+x^2+\frac {1-x+x^2}{\log (x)}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^((2 - 2*x + 2*x^2 + (-5 + 2*E + 2*x^2)*Log[x])/(2*Log[x])) + (2 - 2*x + 2*x^2 + (-5 + 2*E + 2*
x^2)*Log[x])/(2*Log[x]))*(-1 + x - x^2 + (-x + 2*x^2)*Log[x] + 2*x^2*Log[x]^2))/(x*Log[x]^2),x]

[Out]

E^E^(-5/2 + E + x^2 + (1 - x + x^2)/Log[x])

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fricas [B]  time = 0.59, size = 97, normalized size = 3.73 \begin {gather*} e^{\left (\frac {2 \, x^{2} + {\left (2 \, x^{2} + 2 \, e - 5\right )} \log \relax (x) + 2 \, e^{\left (\frac {2 \, x^{2} + {\left (2 \, x^{2} + 2 \, e - 5\right )} \log \relax (x) - 2 \, x + 2}{2 \, \log \relax (x)}\right )} \log \relax (x) - 2 \, x + 2}{2 \, \log \relax (x)} - \frac {2 \, x^{2} + {\left (2 \, x^{2} + 2 \, e - 5\right )} \log \relax (x) - 2 \, x + 2}{2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^2+(2*x^2-x)*log(x)-x^2+x-1)*exp(1/2*((2*exp(1)+2*x^2-5)*log(x)+2*x^2-2*x+2)/log(x))*ex
p(exp(1/2*((2*exp(1)+2*x^2-5)*log(x)+2*x^2-2*x+2)/log(x)))/x/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^2+(2*x^2-x)*log(x)-x^2+x-1)*exp(1/2*((2*exp(1)+2*x^2-5)*log(x)+2*x^2-2*x+2)/log(x))*ex
p(exp(1/2*((2*exp(1)+2*x^2-5)*log(x)+2*x^2-2*x+2)/log(x)))/x/log(x)^2,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.03, size = 36, normalized size = 1.38

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{\frac {2 x^{2} \ln \relax (x )+2 \,{\mathrm e} \ln \relax (x )+2 x^{2}-5 \ln \relax (x )-2 x +2}{2 \ln \relax (x )}}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.12, size = 28, normalized size = 1.08 \begin {gather*} e^{\left (e^{\left (x^{2} + \frac {x^{2}}{\log \relax (x)} - \frac {x}{\log \relax (x)} + \frac {1}{\log \relax (x)} + e - \frac {5}{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^2+(2*x^2-x)*log(x)-x^2+x-1)*exp(1/2*((2*exp(1)+2*x^2-5)*log(x)+2*x^2-2*x+2)/log(x))*ex
p(exp(1/2*((2*exp(1)+2*x^2-5)*log(x)+2*x^2-2*x+2)/log(x)))/x/log(x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.95, size = 33, normalized size = 1.27 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-\frac {5}{2}}\,{\mathrm {e}}^{-\frac {x}{\ln \relax (x)}}\,{\mathrm {e}}^{\frac {1}{\ln \relax (x)}}\,{\mathrm {e}}^{\mathrm {e}}\,{\mathrm {e}}^{\frac {x^2}{\ln \relax (x)}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(((log(x)*(2*exp(1) + 2*x^2 - 5))/2 - x + x^2 + 1)/log(x)))*exp(((log(x)*(2*exp(1) + 2*x^2 - 5))/
2 - x + x^2 + 1)/log(x))*(log(x)*(x - 2*x^2) - x - 2*x^2*log(x)^2 + x^2 + 1))/(x*log(x)^2),x)

[Out]

exp(exp(x^2)*exp(-5/2)*exp(-x/log(x))*exp(1/log(x))*exp(exp(1))*exp(x^2/log(x)))

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sympy [A]  time = 1.86, size = 29, normalized size = 1.12 \begin {gather*} e^{e^{\frac {x^{2} - x + \frac {\left (2 x^{2} - 5 + 2 e\right ) \log {\relax (x )}}{2} + 1}{\log {\relax (x )}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp((x**2 - x + (2*x**2 - 5 + 2*E)*log(x)/2 + 1)/log(x)))

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