3.102.48 \(\int \frac {e^{-\frac {-5 x+e^5 x+(-5+e^5) \log (x)+\log (\frac {x}{\log (x)})}{x+\log (x)}} (e^{\frac {-5 x+e^5 x+(-5+e^5) \log (x)+\log (\frac {x}{\log (x)})}{x+\log (x)}} (x^2 \log (x)+2 x \log ^2(x)+\log ^3(x))+e^{e^{-\frac {-5 x+e^5 x+(-5+e^5) \log (x)+\log (\frac {x}{\log (x)})}{x+\log (x)}} (-2 e^{\frac {-5 x+e^5 x+(-5+e^5) \log (x)+\log (\frac {x}{\log (x)})}{x+\log (x)}}+x)} (x+(1-x+x^2) \log (x)+(-1+2 x) \log ^2(x)+\log ^3(x)+(1+x) \log (x) \log (\frac {x}{\log (x)})))}{x^2 \log (x)+2 x \log ^2(x)+\log ^3(x)} \, dx\)

Optimal. Leaf size=32 \[ e^{-2+e^{5-e^5-\frac {\log \left (\frac {x}{\log (x)}\right )}{x+\log (x)}} x}+x \]

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Rubi [F]  time = 172.06, 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 (-\frac {-5 x+e^5 x+\left (-5+e^5\right ) \log (x)+\log \left (\frac {x}{\log (x)}\right )}{x+\log (x)}\right ) \left (\exp \left (\frac {-5 x+e^5 x+\left (-5+e^5\right ) \log (x)+\log \left (\frac {x}{\log (x)}\right )}{x+\log (x)}\right ) \left (x^2 \log (x)+2 x \log ^2(x)+\log ^3(x)\right )+\exp \left (\exp \left (-\frac {-5 x+e^5 x+\left (-5+e^5\right ) \log (x)+\log \left (\frac {x}{\log (x)}\right )}{x+\log (x)}\right ) \left (-2 \exp \left (\frac {-5 x+e^5 x+\left (-5+e^5\right ) \log (x)+\log \left (\frac {x}{\log (x)}\right )}{x+\log (x)}\right )+x\right )\right ) \left (x+\left (1-x+x^2\right ) \log (x)+(-1+2 x) \log ^2(x)+\log ^3(x)+(1+x) \log (x) \log \left (\frac {x}{\log (x)}\right )\right )\right )}{x^2 \log (x)+2 x \log ^2(x)+\log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-5*x + E^5*x + (-5 + E^5)*Log[x] + Log[x/Log[x]])/(x + Log[x]))*(x^2*Log[x] + 2*x*Log[x]^2 + Log[x]^3
) + E^((-2*E^((-5*x + E^5*x + (-5 + E^5)*Log[x] + Log[x/Log[x]])/(x + Log[x])) + x)/E^((-5*x + E^5*x + (-5 + E
^5)*Log[x] + Log[x/Log[x]])/(x + Log[x])))*(x + (1 - x + x^2)*Log[x] + (-1 + 2*x)*Log[x]^2 + Log[x]^3 + (1 + x
)*Log[x]*Log[x/Log[x]]))/(E^((-5*x + E^5*x + (-5 + E^5)*Log[x] + Log[x/Log[x]])/(x + Log[x]))*(x^2*Log[x] + 2*
x*Log[x]^2 + Log[x]^3)),x]

[Out]

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

Rubi steps

Aborted

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Mathematica [A]  time = 2.14, size = 56, normalized size = 1.75 \begin {gather*} e^{-2+e^{-\frac {\left (-5+e^5\right ) x}{x+\log (x)}+\frac {\left (5-e^5\right ) \log (x)}{x+\log (x)}} x \left (\frac {x}{\log (x)}\right )^{-\frac {1}{x+\log (x)}}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-5*x + E^5*x + (-5 + E^5)*Log[x] + Log[x/Log[x]])/(x + Log[x]))*(x^2*Log[x] + 2*x*Log[x]^2 + Lo
g[x]^3) + E^((-2*E^((-5*x + E^5*x + (-5 + E^5)*Log[x] + Log[x/Log[x]])/(x + Log[x])) + x)/E^((-5*x + E^5*x + (
-5 + E^5)*Log[x] + Log[x/Log[x]])/(x + Log[x])))*(x + (1 - x + x^2)*Log[x] + (-1 + 2*x)*Log[x]^2 + Log[x]^3 +
(1 + x)*Log[x]*Log[x/Log[x]]))/(E^((-5*x + E^5*x + (-5 + E^5)*Log[x] + Log[x/Log[x]])/(x + Log[x]))*(x^2*Log[x
] + 2*x*Log[x]^2 + Log[x]^3)),x]

[Out]

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

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fricas [B]  time = 0.54, size = 69, normalized size = 2.16 \begin {gather*} x + e^{\left ({\left (x - 2 \, e^{\left (\frac {x e^{5} + {\left (e^{5} - 5\right )} \log \relax (x) - 5 \, x + \log \left (\frac {x}{\log \relax (x)}\right )}{x + \log \relax (x)}\right )}\right )} e^{\left (-\frac {x e^{5} + {\left (e^{5} - 5\right )} \log \relax (x) - 5 \, x + \log \left (\frac {x}{\log \relax (x)}\right )}{x + \log \relax (x)}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(x)*log(x/log(x))+log(x)^3+(2*x-1)*log(x)^2+(x^2-x+1)*log(x)+x)*exp((-2*exp((log(x/log(x)
)+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x)))+x)/exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))
))+(log(x)^3+2*x*log(x)^2+x^2*log(x))*exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))))/(log(x)^
3+2*x*log(x)^2+x^2*log(x))/exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))),x, algorithm="fricas
")

[Out]

x + e^((x - 2*e^((x*e^5 + (e^5 - 5)*log(x) - 5*x + log(x/log(x)))/(x + log(x))))*e^(-(x*e^5 + (e^5 - 5)*log(x)
- 5*x + log(x/log(x)))/(x + 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((((x+1)*log(x)*log(x/log(x))+log(x)^3+(2*x-1)*log(x)^2+(x^2-x+1)*log(x)+x)*exp((-2*exp((log(x/log(x)
)+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x)))+x)/exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))
))+(log(x)^3+2*x*log(x)^2+x^2*log(x))*exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))))/(log(x)^
3+2*x*log(x)^2+x^2*log(x))/exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))),x, algorithm="giac")

[Out]

undef

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maple [C]  time = 0.45, size = 155, normalized size = 4.84

method	result	size
risch	\(x +{\mathrm e}^{\left (x \ln \relax (x )^{\frac {1}{x +\ln \relax (x )}} x^{\frac {4}{x +\ln \relax (x )}} {\mathrm e}^{-\frac {-i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-i \pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+2 x \,{\mathrm e}^{5}-10 x}{2 \left (x +\ln \relax (x )\right )}}-2 x^{\frac {{\mathrm e}^{5}}{x +\ln \relax (x )}}\right ) x^{-\frac {{\mathrm e}^{5}}{x +\ln \relax (x )}}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*ln(x)*ln(x/ln(x))+ln(x)^3+(2*x-1)*ln(x)^2+(x^2-x+1)*ln(x)+x)*exp((-2*exp((ln(x/ln(x))+(exp(5)-5)*l
n(x)+x*exp(5)-5*x)/(x+ln(x)))+x)/exp((ln(x/ln(x))+(exp(5)-5)*ln(x)+x*exp(5)-5*x)/(x+ln(x))))+(ln(x)^3+2*x*ln(x
)^2+x^2*ln(x))*exp((ln(x/ln(x))+(exp(5)-5)*ln(x)+x*exp(5)-5*x)/(x+ln(x))))/(ln(x)^3+2*x*ln(x)^2+x^2*ln(x))/exp
((ln(x/ln(x))+(exp(5)-5)*ln(x)+x*exp(5)-5*x)/(x+ln(x))),x,method=_RETURNVERBOSE)

[Out]

x+exp((x*ln(x)^(1/(x+ln(x)))*(x^(1/(x+ln(x))))^4*exp(-1/2*(-I*Pi*csgn(I*x/ln(x))^3+I*Pi*csgn(I*x/ln(x))^2*csgn
(I*x)+I*Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))-I*Pi*csgn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))+2*x*exp(5)-10*x)/(x+ln
(x)))-2*x^(1/(x+ln(x))*exp(5)))/(x^(1/(x+ln(x))*exp(5))))

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maxima [A]  time = 1.30, size = 38, normalized size = 1.19 \begin {gather*} {\left (x e^{2} + e^{\left (x e^{\left (-\frac {\log \relax (x)}{x + \log \relax (x)} + \frac {\log \left (\log \relax (x)\right )}{x + \log \relax (x)} - e^{5} + 5\right )}\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*log(x)*log(x/log(x))+log(x)^3+(2*x-1)*log(x)^2+(x^2-x+1)*log(x)+x)*exp((-2*exp((log(x/log(x)
)+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x)))+x)/exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))
))+(log(x)^3+2*x*log(x)^2+x^2*log(x))*exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))))/(log(x)^
3+2*x*log(x)^2+x^2*log(x))/exp((log(x/log(x))+(exp(5)-5)*log(x)+x*exp(5)-5*x)/(x+log(x))),x, algorithm="maxima
")

[Out]

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

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mupad [B]  time = 7.93, size = 68, normalized size = 2.12 \begin {gather*} x+{\mathrm {e}}^{\frac {x\,x^{\frac {5}{x+\ln \relax (x)}}\,{\mathrm {e}}^{\frac {5\,x}{x+\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{x+\ln \relax (x)}}}{x^{\frac {{\mathrm {e}}^5}{x+\ln \relax (x)}}\,{\left (\frac {x}{\ln \relax (x)}\right )}^{\frac {1}{x+\ln \relax (x)}}}}\,{\mathrm {e}}^{-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(log(x/log(x)) - 5*x + log(x)*(exp(5) - 5) + x*exp(5))/(x + log(x)))*(exp(exp(-(log(x/log(x)) - 5*x
+ log(x)*(exp(5) - 5) + x*exp(5))/(x + log(x)))*(x - 2*exp((log(x/log(x)) - 5*x + log(x)*(exp(5) - 5) + x*exp(
5))/(x + log(x)))))*(x + log(x)^3 + log(x)*(x^2 - x + 1) + log(x)^2*(2*x - 1) + log(x/log(x))*log(x)*(x + 1))
+ exp((log(x/log(x)) - 5*x + log(x)*(exp(5) - 5) + x*exp(5))/(x + log(x)))*(2*x*log(x)^2 + x^2*log(x) + log(x)
^3)))/(2*x*log(x)^2 + x^2*log(x) + log(x)^3),x)

[Out]

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

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sympy [B]  time = 106.05, size = 66, normalized size = 2.06 \begin {gather*} x + e^{\left (x - 2 e^{\frac {- 5 x + x e^{5} + \left (-5 + e^{5}\right ) \log {\relax (x )} + \log {\left (\frac {x}{\log {\relax (x )}} \right )}}{x + \log {\relax (x )}}}\right ) e^{- \frac {- 5 x + x e^{5} + \left (-5 + e^{5}\right ) \log {\relax (x )} + \log {\left (\frac {x}{\log {\relax (x )}} \right )}}{x + \log {\relax (x )}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*ln(x)*ln(x/ln(x))+ln(x)**3+(2*x-1)*ln(x)**2+(x**2-x+1)*ln(x)+x)*exp((-2*exp((ln(x/ln(x))+(ex
p(5)-5)*ln(x)+x*exp(5)-5*x)/(x+ln(x)))+x)/exp((ln(x/ln(x))+(exp(5)-5)*ln(x)+x*exp(5)-5*x)/(x+ln(x))))+(ln(x)**
3+2*x*ln(x)**2+x**2*ln(x))*exp((ln(x/ln(x))+(exp(5)-5)*ln(x)+x*exp(5)-5*x)/(x+ln(x))))/(ln(x)**3+2*x*ln(x)**2+
x**2*ln(x))/exp((ln(x/ln(x))+(exp(5)-5)*ln(x)+x*exp(5)-5*x)/(x+ln(x))),x)

[Out]

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

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