3.24.18 \(\int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} (10-2 e^x+2 x+(10 x-2 e^x x+2 x^2) \log (x)+e^{\frac {2 e^{-x}}{\log (x)}} (e^x x-e^{2 x} x) \log ^2(x))}{x \log ^2(x)} \, dx\) [2318]

Optimal. Leaf size=24 \[ e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \left (5-e^x+x\right ) \]

[Out]

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

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Rubi [F]
time = 7.09, 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^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} \left (10-2 e^x+2 x+\left (10 x-2 e^x x+2 x^2\right ) \log (x)+e^{\frac {2 e^{-x}}{\log (x)}} \left (e^x x-e^{2 x} x\right ) \log ^2(x)\right )}{x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^(-2/(E^x*Log[x])) - x - 2/(E^x*Log[x]))*(10 - 2*E^x + 2*x + (10*x - 2*E^x*x + 2*x^2)*Log[x] + E^(2/(
E^x*Log[x]))*(E^x*x - E^(2*x)*x)*Log[x]^2))/(x*Log[x]^2),x]

[Out]

Defer[Int][E^E^(-2/(E^x*Log[x])), x] - Defer[Int][E^(E^(-2/(E^x*Log[x])) + x), x] + 2*Defer[Int][E^(E^(-2/(E^x
*Log[x])) - x - 2/(E^x*Log[x]))/Log[x]^2, x] - 2*Defer[Int][E^(E^(-2/(E^x*Log[x])) - 2/(E^x*Log[x]))/(x*Log[x]
^2), x] + 10*Defer[Int][E^(E^(-2/(E^x*Log[x])) - x - 2/(E^x*Log[x]))/(x*Log[x]^2), x] - 2*Defer[Int][E^(E^(-2/
(E^x*Log[x])) - 2/(E^x*Log[x]))/Log[x], x] + 10*Defer[Int][E^(E^(-2/(E^x*Log[x])) - x - 2/(E^x*Log[x]))/Log[x]
, x] + 2*Defer[Int][(E^(E^(-2/(E^x*Log[x])) - x - 2/(E^x*Log[x]))*x)/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \left (-1+e^x\right )+\frac {2 e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} \left (5-e^x+x\right ) (1+x \log (x))}{x \log ^2(x)}\right ) \, dx\\ &=2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} \left (5-e^x+x\right ) (1+x \log (x))}{x \log ^2(x)} \, dx-\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \left (-1+e^x\right ) \, dx\\ &=2 \int \left (-\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}} (1+x \log (x))}{x \log ^2(x)}+\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} (5+x) (1+x \log (x))}{x \log ^2(x)}\right ) \, dx-\int \left (-e^{e^{-\frac {2 e^{-x}}{\log (x)}}}+e^{e^{-\frac {2 e^{-x}}{\log (x)}}+x}\right ) \, dx\\ &=-\left (2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}} (1+x \log (x))}{x \log ^2(x)} \, dx\right )+2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} (5+x) (1+x \log (x))}{x \log ^2(x)} \, dx+\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \, dx-\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}+x} \, dx\\ &=-\left (2 \int \left (\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{x \log ^2(x)}+\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{\log (x)}\right ) \, dx\right )+2 \int \left (\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} (5+x)}{x \log ^2(x)}+\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} (5+x)}{\log (x)}\right ) \, dx+\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \, dx-\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}+x} \, dx\\ &=-\left (2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{x \log ^2(x)} \, dx\right )+2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} (5+x)}{x \log ^2(x)} \, dx-2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{\log (x)} \, dx+2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} (5+x)}{\log (x)} \, dx+\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \, dx-\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}+x} \, dx\\ &=2 \int \left (\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}}}{\log ^2(x)}+\frac {5 e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}}}{x \log ^2(x)}\right ) \, dx+2 \int \left (\frac {5 e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}}}{\log (x)}+\frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} x}{\log (x)}\right ) \, dx-2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{x \log ^2(x)} \, dx-2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{\log (x)} \, dx+\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \, dx-\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}+x} \, dx\\ &=2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}}}{\log ^2(x)} \, dx-2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{x \log ^2(x)} \, dx-2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-\frac {2 e^{-x}}{\log (x)}}}{\log (x)} \, dx+2 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}} x}{\log (x)} \, dx+10 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}}}{x \log ^2(x)} \, dx+10 \int \frac {e^{e^{-\frac {2 e^{-x}}{\log (x)}}-x-\frac {2 e^{-x}}{\log (x)}}}{\log (x)} \, dx+\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \, dx-\int e^{e^{-\frac {2 e^{-x}}{\log (x)}}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.09, size = 25, normalized size = 1.04 \begin {gather*} -e^{e^{-\frac {2 e^{-x}}{\log (x)}}} \left (-5+e^x-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(-2/(E^x*Log[x])) - x - 2/(E^x*Log[x]))*(10 - 2*E^x + 2*x + (10*x - 2*E^x*x + 2*x^2)*Log[x] +
E^(2/(E^x*Log[x]))*(E^x*x - E^(2*x)*x)*Log[x]^2))/(x*Log[x]^2),x]

[Out]

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

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Maple [A]
time = 0.08, size = 21, normalized size = 0.88

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{-\frac {2 \,{\mathrm e}^{-x}}{\ln \left (x \right )}}} \left (x -{\mathrm e}^{x}+5\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(x)^2+exp(x)*x)*ln(x)^2*exp(1/exp(x)/ln(x))^2+(-2*exp(x)*x+2*x^2+10*x)*ln(x)-2*exp(x)+2*x+10)*exp(
1/exp(1/exp(x)/ln(x))^2)/x/exp(x)/ln(x)^2/exp(1/exp(x)/ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^2+exp(x)*x)*log(x)^2*exp(1/exp(x)/log(x))^2+(-2*exp(x)*x+2*x^2+10*x)*log(x)-2*exp(x)+2*x
+10)*exp(1/exp(1/exp(x)/log(x))^2)/x/exp(x)/log(x)^2/exp(1/exp(x)/log(x))^2,x, algorithm="maxima")

[Out]

-integrate(((x*e^(2*x) - x*e^x)*e^(2*e^(-x)/log(x))*log(x)^2 - 2*(x^2 - x*e^x + 5*x)*log(x) - 2*x + 2*e^x - 10
)*e^(-x - 2*e^(-x)/log(x) + e^(-2*e^(-x)/log(x)))/(x*log(x)^2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (20) = 40\).
time = 0.38, size = 74, normalized size = 3.08 \begin {gather*} {\left ({\left (x + 5\right )} e^{x} - e^{\left (2 \, x\right )}\right )} e^{\left (-\frac {{\left ({\left (x e^{x} \log \left (x\right ) + 2\right )} e^{\left (\frac {2 \, e^{\left (-x\right )}}{\log \left (x\right )}\right )} - e^{x} \log \left (x\right )\right )} e^{\left (-x - \frac {2 \, e^{\left (-x\right )}}{\log \left (x\right )}\right )}}{\log \left (x\right )} + \frac {2 \, e^{\left (-x\right )}}{\log \left (x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^2+exp(x)*x)*log(x)^2*exp(1/exp(x)/log(x))^2+(-2*exp(x)*x+2*x^2+10*x)*log(x)-2*exp(x)+2*x
+10)*exp(1/exp(1/exp(x)/log(x))^2)/x/exp(x)/log(x)^2/exp(1/exp(x)/log(x))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
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*exp(x)**2+exp(x)*x)*ln(x)**2*exp(1/exp(x)/ln(x))**2+(-2*exp(x)*x+2*x**2+10*x)*ln(x)-2*exp(x)+2*
x+10)*exp(1/exp(1/exp(x)/ln(x))**2)/x/exp(x)/ln(x)**2/exp(1/exp(x)/ln(x))**2,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^2+exp(x)*x)*log(x)^2*exp(1/exp(x)/log(x))^2+(-2*exp(x)*x+2*x^2+10*x)*log(x)-2*exp(x)+2*x
+10)*exp(1/exp(1/exp(x)/log(x))^2)/x/exp(x)/log(x)^2/exp(1/exp(x)/log(x))^2,x, algorithm="giac")

[Out]

undef

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{-x}}{\ln \left (x\right )}}\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{-x}}{\ln \left (x\right )}}}\,\left (2\,x-2\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (10\,x-2\,x\,{\mathrm {e}}^x+2\,x^2\right )-{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-x}}{\ln \left (x\right )}}\,{\ln \left (x\right )}^2\,\left (x\,{\mathrm {e}}^{2\,x}-x\,{\mathrm {e}}^x\right )+10\right )}{x\,{\ln \left (x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*exp(-(2*exp(-x))/log(x))*exp(exp(-(2*exp(-x))/log(x)))*(2*x - 2*exp(x) + log(x)*(10*x - 2*x*exp(x
) + 2*x^2) - exp((2*exp(-x))/log(x))*log(x)^2*(x*exp(2*x) - x*exp(x)) + 10))/(x*log(x)^2),x)

[Out]

int((exp(-x)*exp(-(2*exp(-x))/log(x))*exp(exp(-(2*exp(-x))/log(x)))*(2*x - 2*exp(x) + log(x)*(10*x - 2*x*exp(x
) + 2*x^2) - exp((2*exp(-x))/log(x))*log(x)^2*(x*exp(2*x) - x*exp(x)) + 10))/(x*log(x)^2), x)

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