3.52.44 \(\int \frac {e^{-e^{e+x}} (4 x \log (x)+e^{e^e} (-x+e^{e+x} x^2) \log ^2(x)+(-4 x^2+4 e^{e+x} x^3) \log ^2(x)+(-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} (-1-e^{e+x} x \log (x))) \log (\frac {1}{4} (e^{e^e}+4 x)))}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ e^{-e^{e+x}} \left (-x+\frac {\log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \]

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Rubi [F]  time = 4.17, 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 {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4*x*Log[x] + E^E^E*(-x + E^(E + x)*x^2)*Log[x]^2 + (-4*x^2 + 4*E^(E + x)*x^3)*Log[x]^2 + (-4*x - 4*E^(E +
x)*x^2*Log[x] + E^E^E*(-1 - E^(E + x)*x*Log[x]))*Log[(E^E^E + 4*x)/4])/(E^E^(E + x)*(E^E^E*x*Log[x]^2 + 4*x^2
*Log[x]^2)),x]

[Out]

-ExpIntegralEi[-E^(E + x)] + Defer[Int][E^(E - E^(E + x) + x)*x, x] + 4*Defer[Int][1/(E^E^(E + x)*(E^E^E + 4*x
)*Log[x]), x] - Defer[Int][Log[E^E^E/4 + x]/(E^E^(E + x)*x*Log[x]^2), x] - Defer[Int][(E^(E - E^(E + x) + x)*L
og[E^E^E/4 + x])/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{-e^{e+x}} \left (-1+e^{e+x} x-\frac {\log \left (\frac {e^{e^e}}{4}+x\right )}{x \log ^2(x)}+\frac {\frac {4}{e^{e^e}+4 x}-e^{e+x} \log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \, dx\\ &=\int \left (-e^{-e^{e+x}}+e^{e-e^{e+x}+x} x-\frac {e^{-e^{e+x}} \log \left (\frac {e^{e^e}}{4}+x\right )}{x \log ^2(x)}-\frac {e^{-e^{e+x}} \left (-4+e^{e+e^e+x} \log \left (\frac {e^{e^e}}{4}+x\right )+4 e^{e+x} x \log \left (\frac {e^{e^e}}{4}+x\right )\right )}{\left (e^{e^e}+4 x\right ) \log (x)}\right ) \, dx\\ &=-\int e^{-e^{e+x}} \, dx+\int e^{e-e^{e+x}+x} x \, dx-\int \frac {e^{-e^{e+x}} \log \left (\frac {e^{e^e}}{4}+x\right )}{x \log ^2(x)} \, dx-\int \frac {e^{-e^{e+x}} \left (-4+e^{e+e^e+x} \log \left (\frac {e^{e^e}}{4}+x\right )+4 e^{e+x} x \log \left (\frac {e^{e^e}}{4}+x\right )\right )}{\left (e^{e^e}+4 x\right ) \log (x)} \, dx\\ &=\int e^{e-e^{e+x}+x} x \, dx-\int \frac {e^{-e^{e+x}} \log \left (\frac {e^{e^e}}{4}+x\right )}{x \log ^2(x)} \, dx-\int \frac {e^{-e^{e+x}} \left (-4+e^{e+x} \left (e^{e^e}+4 x\right ) \log \left (\frac {e^{e^e}}{4}+x\right )\right )}{\left (e^{e^e}+4 x\right ) \log (x)} \, dx-\operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,e^{e+x}\right )\\ &=-\text {Ei}\left (-e^{e+x}\right )+\int e^{e-e^{e+x}+x} x \, dx-\int \frac {e^{-e^{e+x}} \log \left (\frac {e^{e^e}}{4}+x\right )}{x \log ^2(x)} \, dx-\int \left (-\frac {4 e^{-e^{e+x}}}{\left (e^{e^e}+4 x\right ) \log (x)}+\frac {e^{e-e^{e+x}+x} \log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \, dx\\ &=-\text {Ei}\left (-e^{e+x}\right )+4 \int \frac {e^{-e^{e+x}}}{\left (e^{e^e}+4 x\right ) \log (x)} \, dx+\int e^{e-e^{e+x}+x} x \, dx-\int \frac {e^{-e^{e+x}} \log \left (\frac {e^{e^e}}{4}+x\right )}{x \log ^2(x)} \, dx-\int \frac {e^{e-e^{e+x}+x} \log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.68, size = 31, normalized size = 1.00 \begin {gather*} e^{-e^{e+x}} \left (-x+\frac {\log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x*Log[x] + E^E^E*(-x + E^(E + x)*x^2)*Log[x]^2 + (-4*x^2 + 4*E^(E + x)*x^3)*Log[x]^2 + (-4*x - 4*
E^(E + x)*x^2*Log[x] + E^E^E*(-1 - E^(E + x)*x*Log[x]))*Log[(E^E^E + 4*x)/4])/(E^E^(E + x)*(E^E^E*x*Log[x]^2 +
4*x^2*Log[x]^2)),x]

[Out]

(-x + Log[E^E^E/4 + x]/Log[x])/E^E^(E + x)

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fricas [A]  time = 0.66, size = 38, normalized size = 1.23 \begin {gather*} -\frac {x e^{\left (-e^{\left (x + e\right )}\right )} \log \relax (x) - e^{\left (-e^{\left (x + e\right )}\right )} \log \left (x + \frac {1}{4} \, e^{\left (e^{e}\right )}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*log(x)-4*x)*log(1/4*exp(exp(exp(1
)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2*exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log
(x)^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm="fricas")

[Out]

-(x*e^(-e^(x + e))*log(x) - e^(-e^(x + e))*log(x + 1/4*e^(e^e)))/log(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x^{2} e^{\left (x + e\right )} - x\right )} e^{\left (e^{e}\right )} \log \relax (x)^{2} + 4 \, {\left (x^{3} e^{\left (x + e\right )} - x^{2}\right )} \log \relax (x)^{2} - {\left (4 \, x^{2} e^{\left (x + e\right )} \log \relax (x) + {\left (x e^{\left (x + e\right )} \log \relax (x) + 1\right )} e^{\left (e^{e}\right )} + 4 \, x\right )} \log \left (x + \frac {1}{4} \, e^{\left (e^{e}\right )}\right ) + 4 \, x \log \relax (x)\right )} e^{\left (-e^{\left (x + e\right )}\right )}}{4 \, x^{2} \log \relax (x)^{2} + x e^{\left (e^{e}\right )} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*log(x)-4*x)*log(1/4*exp(exp(exp(1
)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2*exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log
(x)^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm="giac")

[Out]

integrate(((x^2*e^(x + e) - x)*e^(e^e)*log(x)^2 + 4*(x^3*e^(x + e) - x^2)*log(x)^2 - (4*x^2*e^(x + e)*log(x) +
(x*e^(x + e)*log(x) + 1)*e^(e^e) + 4*x)*log(x + 1/4*e^(e^e)) + 4*x*log(x))*e^(-e^(x + e))/(4*x^2*log(x)^2 + x
*e^(e^e)*log(x)^2), x)

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maple [A]  time = 0.15, size = 31, normalized size = 1.00

method	result	size
risch	\(-\frac {\left (x \ln \relax (x )-\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}}}}{4}+x \right )\right ) {\mathrm e}^{-{\mathrm e}^{x +{\mathrm e}}}}{\ln \relax (x )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x*exp(x+exp(1))*ln(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*ln(x)-4*x)*ln(1/4*exp(exp(exp(1)))+x)+(x
^2*exp(x+exp(1))-x)*ln(x)^2*exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*ln(x)^2+4*x*ln(x))/(x*ln(x)^2*exp(exp
(exp(1)))+4*x^2*ln(x)^2)/exp(exp(x+exp(1))),x,method=_RETURNVERBOSE)

[Out]

-(x*ln(x)-ln(1/4*exp(exp(exp(1)))+x))/ln(x)*exp(-exp(x+exp(1)))

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maxima [A]  time = 0.53, size = 44, normalized size = 1.42 \begin {gather*} -\frac {{\left (x \log \relax (x) + 2 \, \log \relax (2)\right )} e^{\left (-e^{\left (x + e\right )}\right )} - e^{\left (-e^{\left (x + e\right )}\right )} \log \left (4 \, x + e^{\left (e^{e}\right )}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*log(x)-4*x)*log(1/4*exp(exp(exp(1
)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2*exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log
(x)^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.55, size = 28, normalized size = 0.90 \begin {gather*} \frac {{\mathrm {e}}^{-{\mathrm {e}}^{\mathrm {e}}\,{\mathrm {e}}^x}\,\left (\ln \left (x+\frac {{\mathrm {e}}^{{\mathrm {e}}^{\mathrm {e}}}}{4}\right )-x\,\ln \relax (x)\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-exp(x + exp(1)))*(log(x)^2*(4*x^2 - 4*x^3*exp(x + exp(1))) + log(x + exp(exp(exp(1)))/4)*(4*x + exp
(exp(exp(1)))*(x*exp(x + exp(1))*log(x) + 1) + 4*x^2*exp(x + exp(1))*log(x)) - 4*x*log(x) + exp(exp(exp(1)))*l
og(x)^2*(x - x^2*exp(x + exp(1)))))/(4*x^2*log(x)^2 + x*exp(exp(exp(1)))*log(x)^2),x)

[Out]

(exp(-exp(exp(1))*exp(x))*(log(x + exp(exp(exp(1)))/4) - x*log(x)))/log(x)

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sympy [A]  time = 0.54, size = 27, normalized size = 0.87 \begin {gather*} \frac {\left (- x \log {\relax (x )} + \log {\left (x + \frac {e^{e^{e}}}{4} \right )}\right ) e^{- e^{x + e}}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(x+exp(1))*ln(x)-1)*exp(exp(exp(1)))-4*x**2*exp(x+exp(1))*ln(x)-4*x)*ln(1/4*exp(exp(exp(1))
)+x)+(x**2*exp(x+exp(1))-x)*ln(x)**2*exp(exp(exp(1)))+(4*x**3*exp(x+exp(1))-4*x**2)*ln(x)**2+4*x*ln(x))/(x*ln(
x)**2*exp(exp(exp(1)))+4*x**2*ln(x)**2)/exp(exp(x+exp(1))),x)

[Out]

(-x*log(x) + log(x + exp(exp(E))/4))*exp(-exp(x + E))/log(x)

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