3.65.93 \(\int \frac {e^{-\frac {e^2 x}{3 x-e^x x+\log (\log (\frac {x^2}{\log ^2(2) \log ^2(x)}))}} (-2 e^2+2 e^2 \log (x)-e^{2+x} x^2 \log (x) \log (\frac {x^2}{\log ^2(2) \log ^2(x)})-e^2 \log (x) \log (\frac {x^2}{\log ^2(2) \log ^2(x)}) \log (\log (\frac {x^2}{\log ^2(2) \log ^2(x)})))}{(9 x^2-6 e^x x^2+e^{2 x} x^2) \log (x) \log (\frac {x^2}{\log ^2(2) \log ^2(x)})+(6 x-2 e^x x) \log (x) \log (\frac {x^2}{\log ^2(2) \log ^2(x)}) \log (\log (\frac {x^2}{\log ^2(2) \log ^2(x)}))+\log (x) \log (\frac {x^2}{\log ^2(2) \log ^2(x)}) \log ^2(\log (\frac {x^2}{\log ^2(2) \log ^2(x)}))} \, dx\)

Optimal. Leaf size=33 \[ e^{\frac {e^2 x}{\left (-3+e^x\right ) x-\log \left (\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right )\right )}} \]

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Rubi [A]  time = 4.89, antiderivative size = 34, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, integrand size = 221, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6688, 6706} \begin {gather*} \exp \left (-\frac {e^2 x}{\log \left (\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right )\right )+\left (3-e^x\right ) x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^2 + 2*E^2*Log[x] - E^(2 + x)*x^2*Log[x]*Log[x^2/(Log[2]^2*Log[x]^2)] - E^2*Log[x]*Log[x^2/(Log[2]^2*
Log[x]^2)]*Log[Log[x^2/(Log[2]^2*Log[x]^2)]])/(E^((E^2*x)/(3*x - E^x*x + Log[Log[x^2/(Log[2]^2*Log[x]^2)]]))*(
(9*x^2 - 6*E^x*x^2 + E^(2*x)*x^2)*Log[x]*Log[x^2/(Log[2]^2*Log[x]^2)] + (6*x - 2*E^x*x)*Log[x]*Log[x^2/(Log[2]
^2*Log[x]^2)]*Log[Log[x^2/(Log[2]^2*Log[x]^2)]] + Log[x]*Log[x^2/(Log[2]^2*Log[x]^2)]*Log[Log[x^2/(Log[2]^2*Lo
g[x]^2)]]^2)),x]

[Out]

E^(-((E^2*x)/((3 - E^x)*x + Log[Log[x^2/(Log[2]^2*Log[x]^2)]])))

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (2+\frac {e^2 x}{\left (-3+e^x\right ) x-\log \left (\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right )\right )}\right ) \left (-2-\log (x) \left (-2+\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right ) \left (e^x x^2+\log \left (\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right )\right )\right )\right )\right )}{\log (x) \log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right ) \left (\left (-3+e^x\right ) x-\log \left (\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right )\right )\right )^2} \, dx\\ &=\exp \left (-\frac {e^2 x}{\left (3-e^x\right ) x+\log \left (\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right )\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 34, normalized size = 1.03 \begin {gather*} e^{-\frac {e^2 x}{3 x-e^x x+\log \left (\log \left (\frac {x^2}{\log ^2(2) \log ^2(x)}\right )\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^2 + 2*E^2*Log[x] - E^(2 + x)*x^2*Log[x]*Log[x^2/(Log[2]^2*Log[x]^2)] - E^2*Log[x]*Log[x^2/(Log
[2]^2*Log[x]^2)]*Log[Log[x^2/(Log[2]^2*Log[x]^2)]])/(E^((E^2*x)/(3*x - E^x*x + Log[Log[x^2/(Log[2]^2*Log[x]^2)
]]))*((9*x^2 - 6*E^x*x^2 + E^(2*x)*x^2)*Log[x]*Log[x^2/(Log[2]^2*Log[x]^2)] + (6*x - 2*E^x*x)*Log[x]*Log[x^2/(
Log[2]^2*Log[x]^2)]*Log[Log[x^2/(Log[2]^2*Log[x]^2)]] + Log[x]*Log[x^2/(Log[2]^2*Log[x]^2)]*Log[Log[x^2/(Log[2
]^2*Log[x]^2)]]^2)),x]

[Out]

E^(-((E^2*x)/(3*x - E^x*x + Log[Log[x^2/(Log[2]^2*Log[x]^2)]])))

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fricas [A]  time = 0.91, size = 38, normalized size = 1.15 \begin {gather*} e^{\left (-\frac {x e^{4}}{3 \, x e^{2} - x e^{\left (x + 2\right )} + e^{2} \log \left (\log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)*log(x)*log(x^2/log(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))-x^2*exp(2)*exp(x)*log(x)*
log(x^2/log(2)^2/log(x)^2)+2*exp(2)*log(x)-2*exp(2))*exp(-x*exp(2)/(log(log(x^2/log(2)^2/log(x)^2))-exp(x)*x+3
*x))/(log(x)*log(x^2/log(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))^2+(-2*exp(x)*x+6*x)*log(x)*log(x^2/log
(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))+(exp(x)^2*x^2-6*exp(x)*x^2+9*x^2)*log(x)*log(x^2/log(2)^2/log(
x)^2)),x, algorithm="fricas")

[Out]

e^(-x*e^4/(3*x*e^2 - x*e^(x + 2) + e^2*log(log(x^2/(log(2)^2*log(x)^2)))))

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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((-exp(2)*log(x)*log(x^2/log(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))-x^2*exp(2)*exp(x)*log(x)*
log(x^2/log(2)^2/log(x)^2)+2*exp(2)*log(x)-2*exp(2))*exp(-x*exp(2)/(log(log(x^2/log(2)^2/log(x)^2))-exp(x)*x+3
*x))/(log(x)*log(x^2/log(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))^2+(-2*exp(x)*x+6*x)*log(x)*log(x^2/log
(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))+(exp(x)^2*x^2-6*exp(x)*x^2+9*x^2)*log(x)*log(x^2/log(2)^2/log(
x)^2)),x, algorithm="giac")

[Out]

undef

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maple [C]  time = 0.74, size = 151, normalized size = 4.58

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{2} x}{{\mathrm e}^{x} x -\ln \left (-2 \ln \left (\ln \relax (2)\right )+2 \ln \relax (x )-2 \ln \left (\ln \relax (x )\right )+\frac {i \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \left (-\mathrm {csgn}\left (i \ln \relax (x )^{2}\right )+\mathrm {csgn}\left (i \ln \relax (x )\right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )^{2}}\right ) \left (-\mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )^{2}}\right )+\mathrm {csgn}\left (i x^{2}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )^{2}}\right )+\mathrm {csgn}\left (\frac {i}{\ln \relax (x )^{2}}\right )\right )}{2}\right )-3 x}}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(2)*ln(x)*ln(x^2/ln(2)^2/ln(x)^2)*ln(ln(x^2/ln(2)^2/ln(x)^2))-x^2*exp(2)*exp(x)*ln(x)*ln(x^2/ln(2)^2/
ln(x)^2)+2*exp(2)*ln(x)-2*exp(2))*exp(-x*exp(2)/(ln(ln(x^2/ln(2)^2/ln(x)^2))-exp(x)*x+3*x))/(ln(x)*ln(x^2/ln(2
)^2/ln(x)^2)*ln(ln(x^2/ln(2)^2/ln(x)^2))^2+(-2*exp(x)*x+6*x)*ln(x)*ln(x^2/ln(2)^2/ln(x)^2)*ln(ln(x^2/ln(2)^2/l
n(x)^2))+(exp(x)^2*x^2-6*exp(x)*x^2+9*x^2)*ln(x)*ln(x^2/ln(2)^2/ln(x)^2)),x,method=_RETURNVERBOSE)

[Out]

exp(exp(2)*x/(exp(x)*x-ln(-2*ln(ln(2))+2*ln(x)-2*ln(ln(x))+1/2*I*Pi*csgn(I*ln(x)^2)*(-csgn(I*ln(x)^2)+csgn(I*l
n(x)))^2-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2-1/2*I*Pi*csgn(I*x^2/ln(x)^2)*(-csgn(I*x^2/ln(x)^2)+cs
gn(I*x^2))*(-csgn(I*x^2/ln(x)^2)+csgn(I/ln(x)^2)))-3*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} e^{\left (x + 2\right )} \log \relax (x) \log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right ) + e^{2} \log \relax (x) \log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right ) \log \left (\log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right )\right ) - 2 \, e^{2} \log \relax (x) + 2 \, e^{2}\right )} e^{\left (\frac {x e^{2}}{x e^{x} - 3 \, x - \log \left (\log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right )\right )}\right )}}{2 \, {\left (x e^{x} - 3 \, x\right )} \log \relax (x) \log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right ) \log \left (\log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right )\right ) - \log \relax (x) \log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right ) \log \left (\log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right )\right )^{2} - {\left (x^{2} e^{\left (2 \, x\right )} - 6 \, x^{2} e^{x} + 9 \, x^{2}\right )} \log \relax (x) \log \left (\frac {x^{2}}{\log \relax (2)^{2} \log \relax (x)^{2}}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)*log(x)*log(x^2/log(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))-x^2*exp(2)*exp(x)*log(x)*
log(x^2/log(2)^2/log(x)^2)+2*exp(2)*log(x)-2*exp(2))*exp(-x*exp(2)/(log(log(x^2/log(2)^2/log(x)^2))-exp(x)*x+3
*x))/(log(x)*log(x^2/log(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))^2+(-2*exp(x)*x+6*x)*log(x)*log(x^2/log
(2)^2/log(x)^2)*log(log(x^2/log(2)^2/log(x)^2))+(exp(x)^2*x^2-6*exp(x)*x^2+9*x^2)*log(x)*log(x^2/log(2)^2/log(
x)^2)),x, algorithm="maxima")

[Out]

integrate((x^2*e^(x + 2)*log(x)*log(x^2/(log(2)^2*log(x)^2)) + e^2*log(x)*log(x^2/(log(2)^2*log(x)^2))*log(log
(x^2/(log(2)^2*log(x)^2))) - 2*e^2*log(x) + 2*e^2)*e^(x*e^2/(x*e^x - 3*x - log(log(x^2/(log(2)^2*log(x)^2)))))
/(2*(x*e^x - 3*x)*log(x)*log(x^2/(log(2)^2*log(x)^2))*log(log(x^2/(log(2)^2*log(x)^2))) - log(x)*log(x^2/(log(
2)^2*log(x)^2))*log(log(x^2/(log(2)^2*log(x)^2)))^2 - (x^2*e^(2*x) - 6*x^2*e^x + 9*x^2)*log(x)*log(x^2/(log(2)
^2*log(x)^2))), x)

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mupad [B]  time = 5.03, size = 33, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^2}{3\,x+\ln \left (\ln \left (x^2\right )+\ln \left (\frac {1}{{\ln \relax (x)}^2}\right )-2\,\ln \left (\ln \relax (2)\right )\right )-x\,{\mathrm {e}}^x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x*exp(2))/(3*x + log(log(x^2/(log(2)^2*log(x)^2))) - x*exp(x)))*(2*exp(2) - 2*exp(2)*log(x) + log(
log(x^2/(log(2)^2*log(x)^2)))*exp(2)*log(x^2/(log(2)^2*log(x)^2))*log(x) + x^2*exp(2)*exp(x)*log(x^2/(log(2)^2
*log(x)^2))*log(x)))/(log(x^2/(log(2)^2*log(x)^2))*log(x)*(x^2*exp(2*x) - 6*x^2*exp(x) + 9*x^2) + log(log(x^2/
(log(2)^2*log(x)^2)))^2*log(x^2/(log(2)^2*log(x)^2))*log(x) + log(log(x^2/(log(2)^2*log(x)^2)))*log(x^2/(log(2
)^2*log(x)^2))*log(x)*(6*x - 2*x*exp(x))),x)

[Out]

exp(-(x*exp(2))/(3*x + log(log(x^2) + log(1/log(x)^2) - 2*log(log(2))) - x*exp(x)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)*ln(x)*ln(x**2/ln(2)**2/ln(x)**2)*ln(ln(x**2/ln(2)**2/ln(x)**2))-x**2*exp(2)*exp(x)*ln(x)*ln
(x**2/ln(2)**2/ln(x)**2)+2*exp(2)*ln(x)-2*exp(2))*exp(-x*exp(2)/(ln(ln(x**2/ln(2)**2/ln(x)**2))-exp(x)*x+3*x))
/(ln(x)*ln(x**2/ln(2)**2/ln(x)**2)*ln(ln(x**2/ln(2)**2/ln(x)**2))**2+(-2*exp(x)*x+6*x)*ln(x)*ln(x**2/ln(2)**2/
ln(x)**2)*ln(ln(x**2/ln(2)**2/ln(x)**2))+(exp(x)**2*x**2-6*exp(x)*x**2+9*x**2)*ln(x)*ln(x**2/ln(2)**2/ln(x)**2
)),x)

[Out]

Timed out

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