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

Optimal. Leaf size=20 \[ \log ^{e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \]

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Rubi [F]  time = 9.71, 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^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Log[x]^(-1 + E^(-((5 - E^x + x + Log[4])/x)))*(x + (5 + E^x*(-1 + x) + Log[4])*Log[x]*Log[Log[x]]))/(E^((
5 - E^x + x + Log[4])/x)*x^2),x]

[Out]

Defer[Int][(E^(-1 + E^x/x - (5*(1 + (2*Log[2])/5))/x)*Log[x]^(-1 + E^(-((5 - E^x + x + Log[4])/x))))/x, x] + (
5 + Log[4])*Defer[Int][(E^(-1 + E^x/x - (5*(1 + (2*Log[2])/5))/x)*Log[x]^E^(-1 + E^x/x - (5*(1 + (2*Log[2])/5)
)/x)*Log[Log[x]])/x^2, x] - Defer[Int][(E^(-1 + E^x/x + x - (5*(1 + (2*Log[2])/5))/x)*Log[x]^E^(-1 + E^x/x - (
5*(1 + (2*Log[2])/5))/x)*Log[Log[x]])/x^2, x] + Defer[Int][(E^(-1 + E^x/x + x - (5*(1 + (2*Log[2])/5))/x)*Log[
x]^E^(-1 + E^x/x - (5*(1 + (2*Log[2])/5))/x)*Log[Log[x]])/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx\\ &=\int \left (\frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} (-1+x) \log ^{e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \log (\log (x))}{x^2}+\frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+5 \left (1+\frac {2 \log (2)}{5}\right ) \log (x) \log (\log (x))\right )}{x^2}\right ) \, dx\\ &=\int \frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} (-1+x) \log ^{e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \log (\log (x))}{x^2} \, dx+\int \frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+5 \left (1+\frac {2 \log (2)}{5}\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx\\ &=\int \frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} (-1+x) \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x^2} \, dx+\int \frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) (x+(5+\log (4)) \log (x) \log (\log (x)))}{x^2} \, dx\\ &=\int \left (-\frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x^2}+\frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x}\right ) \, dx+\int \left (\frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x)}{x}+\frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} (5+\log (4)) \log ^{e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \log (\log (x))}{x^2}\right ) \, dx\\ &=(5+\log (4)) \int \frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \log (\log (x))}{x^2} \, dx+\int \frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x)}{x} \, dx-\int \frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x^2} \, dx+\int \frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x} \, dx\\ &=(5+\log (4)) \int \frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x^2} \, dx+\int \frac {e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x)}{x} \, dx-\int \frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x^2} \, dx+\int \frac {e^{-1+\frac {e^x}{x}+x-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}} \log ^{e^{-1+\frac {e^x}{x}-\frac {5 \left (1+\frac {2 \log (2)}{5}\right )}{x}}}(x) \log (\log (x))}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.00, size = 25, normalized size = 1.25 \begin {gather*} \log ^{4^{-1/x} e^{\frac {-5+e^x-x}{x}}}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[x]^(-1 + E^(-((5 - E^x + x + Log[4])/x)))*(x + (5 + E^x*(-1 + x) + Log[4])*Log[x]*Log[Log[x]]))
/(E^((5 - E^x + x + Log[4])/x)*x^2),x]

[Out]

Log[x]^(E^((-5 + E^x - x)/x)/4^x^(-1))

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fricas [A]  time = 0.81, size = 20, normalized size = 1.00 \begin {gather*} \log \relax (x)^{e^{\left (-\frac {x - e^{x} + 2 \, \log \relax (2) + 5}{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 20, normalized size = 1.00

method	result	size
risch	\(\ln \relax (x )^{{\mathrm e}^{\frac {{\mathrm e}^{x}-2 \ln \relax (2)-5-x}{x}}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.77, size = 24, normalized size = 1.20 \begin {gather*} \log \relax (x)^{e^{\left (\frac {e^{x}}{x} - \frac {2 \, \log \relax (2)}{x} - \frac {5}{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.40, size = 28, normalized size = 1.40 \begin {gather*} {\ln \relax (x)}^{\frac {{\mathrm {e}}^{-1}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{-\frac {5}{x}}}{2^{2/x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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