3.12.19 \(\int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+(32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} (8 x^2+4 e^x x^2) \log (x)}{x^2}) \log (\log (x))+(\frac {e^{2 x} (4 x^2+8 x^3) \log (x)}{x^2}+\frac {e^{e^x+2 x} (x^2+2 x^3+e^x x^3) \log (x)}{x^2}) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=24 \[ 5+e^{2 x} \left (4+e^{e^x}\right ) \left (x+\frac {4}{\log (\log (x))}\right ) \]

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Rubi [F]  time = 6.99, 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 {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-16*E^(2*x))/x - (4*E^(E^x + 2*x))/x + (32*E^(2*x)*Log[x] + (E^(E^x + 2*x)*(8*x^2 + 4*E^x*x^2)*Log[x])/x
^2)*Log[Log[x]] + ((E^(2*x)*(4*x^2 + 8*x^3)*Log[x])/x^2 + (E^(E^x + 2*x)*(x^2 + 2*x^3 + E^x*x^3)*Log[x])/x^2)*
Log[Log[x]]^2)/(Log[x]*Log[Log[x]]^2),x]

[Out]

-E^E^x + E^(E^x + x) + (4*E^(2*x)*(4*x*Log[x]*Log[Log[x]] + x^2*Log[x]*Log[Log[x]]^2))/(x*Log[x]*Log[Log[x]]^2
) + 2*Defer[Int][E^(E^x + 2*x)*x, x] + Defer[Int][E^(E^x + 3*x)*x, x] - 4*Defer[Int][E^(E^x + 2*x)/(x*Log[x]*L
og[Log[x]]^2), x] + 8*Defer[Int][E^(E^x + 2*x)/Log[Log[x]], x] + 4*Defer[Int][E^(E^x + 3*x)/Log[Log[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (-4 \left (4+e^{e^x}\right )+x \log (x) \log (\log (x)) \left (4 \left (8+2 e^{e^x}+e^{e^x+x}\right )+\left (4+8 x+e^{e^x+x} x+e^{e^x} (1+2 x)\right ) \log (\log (x))\right )\right )}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\int \left (\frac {e^{e^x+3 x} (4+x \log (\log (x)))}{\log (\log (x))}+\frac {e^{2 x} \left (4+e^{e^x}\right ) \left (-4+8 x \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))+2 x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}\right ) \, dx\\ &=\int \frac {e^{e^x+3 x} (4+x \log (\log (x)))}{\log (\log (x))} \, dx+\int \frac {e^{2 x} \left (4+e^{e^x}\right ) \left (-4+8 x \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))+2 x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\int \left (e^{e^x+3 x} x+\frac {4 e^{e^x+3 x}}{\log (\log (x))}\right ) \, dx+\int \frac {e^{2 x} \left (4+e^{e^x}\right ) (-4+x \log (x) \log (\log (x)) (8+(1+2 x) \log (\log (x))))}{x \log (x) \log ^2(\log (x))} \, dx\\ &=4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+\int e^{e^x+3 x} x \, dx+\int \left (\frac {4 e^{2 x} \left (-4+8 x \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))+2 x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}+\frac {e^{e^x+2 x} \left (-4+8 x \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))+2 x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}\right ) \, dx\\ &=4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+4 \int \frac {e^{2 x} \left (-4+8 x \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))+2 x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))} \, dx+\int e^{e^x+3 x} x \, dx+\int \frac {e^{e^x+2 x} \left (-4+8 x \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))+2 x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {4 e^{2 x} \left (4 x \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}+4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+\int e^{e^x+3 x} x \, dx+\int \frac {e^{e^x+2 x} (-4+x \log (x) \log (\log (x)) (8+(1+2 x) \log (\log (x))))}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {4 e^{2 x} \left (4 x \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}+4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+\int e^{e^x+3 x} x \, dx+\int \left (e^{e^x+2 x}+2 e^{e^x+2 x} x-\frac {4 e^{e^x+2 x}}{x \log (x) \log ^2(\log (x))}+\frac {8 e^{e^x+2 x}}{\log (\log (x))}\right ) \, dx\\ &=\frac {4 e^{2 x} \left (4 x \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}+2 \int e^{e^x+2 x} x \, dx-4 \int \frac {e^{e^x+2 x}}{x \log (x) \log ^2(\log (x))} \, dx+4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+8 \int \frac {e^{e^x+2 x}}{\log (\log (x))} \, dx+\int e^{e^x+2 x} \, dx+\int e^{e^x+3 x} x \, dx\\ &=\frac {4 e^{2 x} \left (4 x \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}+2 \int e^{e^x+2 x} x \, dx-4 \int \frac {e^{e^x+2 x}}{x \log (x) \log ^2(\log (x))} \, dx+4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+8 \int \frac {e^{e^x+2 x}}{\log (\log (x))} \, dx+\int e^{e^x+3 x} x \, dx+\operatorname {Subst}\left (\int e^x x \, dx,x,e^x\right )\\ &=e^{e^x+x}+\frac {4 e^{2 x} \left (4 x \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}+2 \int e^{e^x+2 x} x \, dx-4 \int \frac {e^{e^x+2 x}}{x \log (x) \log ^2(\log (x))} \, dx+4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+8 \int \frac {e^{e^x+2 x}}{\log (\log (x))} \, dx+\int e^{e^x+3 x} x \, dx-\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=-e^{e^x}+e^{e^x+x}+\frac {4 e^{2 x} \left (4 x \log (x) \log (\log (x))+x^2 \log (x) \log ^2(\log (x))\right )}{x \log (x) \log ^2(\log (x))}+2 \int e^{e^x+2 x} x \, dx-4 \int \frac {e^{e^x+2 x}}{x \log (x) \log ^2(\log (x))} \, dx+4 \int \frac {e^{e^x+3 x}}{\log (\log (x))} \, dx+8 \int \frac {e^{e^x+2 x}}{\log (\log (x))} \, dx+\int e^{e^x+3 x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 25, normalized size = 1.04 \begin {gather*} \frac {e^{2 x} \left (4+e^{e^x}\right ) (4+x \log (\log (x)))}{\log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-16*E^(2*x))/x - (4*E^(E^x + 2*x))/x + (32*E^(2*x)*Log[x] + (E^(E^x + 2*x)*(8*x^2 + 4*E^x*x^2)*Log
[x])/x^2)*Log[Log[x]] + ((E^(2*x)*(4*x^2 + 8*x^3)*Log[x])/x^2 + (E^(E^x + 2*x)*(x^2 + 2*x^3 + E^x*x^3)*Log[x])
/x^2)*Log[Log[x]]^2)/(Log[x]*Log[Log[x]]^2),x]

[Out]

(E^(2*x)*(4 + E^E^x)*(4 + x*Log[Log[x]]))/Log[Log[x]]

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fricas [A]  time = 0.99, size = 56, normalized size = 2.33 \begin {gather*} \frac {4 \, x^{2} e^{\left (2 \, x + e^{x} - 2 \, \log \relax (x)\right )} + {\left (x^{3} e^{\left (2 \, x + e^{x} - 2 \, \log \relax (x)\right )} + 4 \, x e^{\left (2 \, x\right )}\right )} \log \left (\log \relax (x)\right ) + 16 \, e^{\left (2 \, x\right )}}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x^3+2*x^3+x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+(8*x^3+4*x^2)*log(x)*exp(x-log(x))^2)*lo
g(log(x))^2+((4*exp(x)*x^2+8*x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+32*x^2*log(x)*exp(x-log(x))^2)*log(log(x)
)-4*x*exp(x-log(x))^2*exp(exp(x))-16*x*exp(x-log(x))^2)/log(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

(4*x^2*e^(2*x + e^x - 2*log(x)) + (x^3*e^(2*x + e^x - 2*log(x)) + 4*x*e^(2*x))*log(log(x)) + 16*e^(2*x))/log(l
og(x))

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giac [A]  time = 0.39, size = 44, normalized size = 1.83 \begin {gather*} \frac {4 \, x e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right ) + x e^{\left (2 \, x + e^{x}\right )} \log \left (\log \relax (x)\right ) + 16 \, e^{\left (2 \, x\right )} + 4 \, e^{\left (2 \, x + e^{x}\right )}}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x^3+2*x^3+x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+(8*x^3+4*x^2)*log(x)*exp(x-log(x))^2)*lo
g(log(x))^2+((4*exp(x)*x^2+8*x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+32*x^2*log(x)*exp(x-log(x))^2)*log(log(x)
)-4*x*exp(x-log(x))^2*exp(exp(x))-16*x*exp(x-log(x))^2)/log(x)/log(log(x))^2,x, algorithm="giac")

[Out]

(4*x*e^(2*x)*log(log(x)) + x*e^(2*x + e^x)*log(log(x)) + 16*e^(2*x) + 4*e^(2*x + e^x))/log(log(x))

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maple [A]  time = 0.08, size = 34, normalized size = 1.42

method	result	size
risch	\(x \,{\mathrm e}^{{\mathrm e}^{x}+2 x}+4 x \,{\mathrm e}^{2 x}+\frac {4 \,{\mathrm e}^{2 x} \left ({\mathrm e}^{{\mathrm e}^{x}}+4\right )}{\ln \left (\ln \relax (x )\right )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(x)*x^3+2*x^3+x^2)*ln(x)*exp(x-ln(x))^2*exp(exp(x))+(8*x^3+4*x^2)*ln(x)*exp(x-ln(x))^2)*ln(ln(x))^2+
((4*exp(x)*x^2+8*x^2)*ln(x)*exp(x-ln(x))^2*exp(exp(x))+32*x^2*ln(x)*exp(x-ln(x))^2)*ln(ln(x))-4*x*exp(x-ln(x))
^2*exp(exp(x))-16*x*exp(x-ln(x))^2)/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(exp(x)+2*x)+4*x*exp(2*x)+4*exp(2*x)*(exp(exp(x))+4)/ln(ln(x))

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maxima [B]  time = 0.52, size = 66, normalized size = 2.75 \begin {gather*} 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + {\left (e^{x} - 1\right )} e^{\left (e^{x}\right )} + \frac {{\left ({\left (x e^{\left (2 \, x\right )} - e^{x} + 1\right )} \log \left (\log \relax (x)\right ) + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (e^{x}\right )} + 16 \, e^{\left (2 \, x\right )}}{\log \left (\log \relax (x)\right )} + 2 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x^3+2*x^3+x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+(8*x^3+4*x^2)*log(x)*exp(x-log(x))^2)*lo
g(log(x))^2+((4*exp(x)*x^2+8*x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+32*x^2*log(x)*exp(x-log(x))^2)*log(log(x)
)-4*x*exp(x-log(x))^2*exp(exp(x))-16*x*exp(x-log(x))^2)/log(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

2*(2*x - 1)*e^(2*x) + (e^x - 1)*e^(e^x) + (((x*e^(2*x) - e^x + 1)*log(log(x)) + 4*e^(2*x))*e^(e^x) + 16*e^(2*x
))/log(log(x)) + 2*e^(2*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\left ({\mathrm {e}}^{2\,x-2\,\ln \relax (x)}\,\ln \relax (x)\,\left (8\,x^3+4\,x^2\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x-2\,\ln \relax (x)}\,\ln \relax (x)\,\left (x^3\,{\mathrm {e}}^x+x^2+2\,x^3\right )\right )\,{\ln \left (\ln \relax (x)\right )}^2+\left (32\,x^2\,{\mathrm {e}}^{2\,x-2\,\ln \relax (x)}\,\ln \relax (x)+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x-2\,\ln \relax (x)}\,\ln \relax (x)\,\left (4\,x^2\,{\mathrm {e}}^x+8\,x^2\right )\right )\,\ln \left (\ln \relax (x)\right )-16\,x\,{\mathrm {e}}^{2\,x-2\,\ln \relax (x)}-4\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x-2\,\ln \relax (x)}}{{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x))*(32*x^2*exp(2*x - 2*log(x))*log(x) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(4*x^2*exp(x) + 8
*x^2)) + log(log(x))^2*(exp(2*x - 2*log(x))*log(x)*(4*x^2 + 8*x^3) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(x
^3*exp(x) + x^2 + 2*x^3)) - 16*x*exp(2*x - 2*log(x)) - 4*x*exp(exp(x))*exp(2*x - 2*log(x)))/(log(log(x))^2*log
(x)),x)

[Out]

int((log(log(x))*(32*x^2*exp(2*x - 2*log(x))*log(x) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(4*x^2*exp(x) + 8
*x^2)) + log(log(x))^2*(exp(2*x - 2*log(x))*log(x)*(4*x^2 + 8*x^3) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(x
^3*exp(x) + x^2 + 2*x^3)) - 16*x*exp(2*x - 2*log(x)) - 4*x*exp(exp(x))*exp(2*x - 2*log(x)))/(log(log(x))^2*log
(x)), x)

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sympy [B]  time = 0.61, size = 48, normalized size = 2.00 \begin {gather*} \frac {\left (4 x \log {\left (\log {\relax (x )} \right )} + 16\right ) e^{2 x}}{\log {\left (\log {\relax (x )} \right )}} + \frac {\left (x e^{2 x} \log {\left (\log {\relax (x )} \right )} + 4 e^{2 x}\right ) e^{e^{x}}}{\log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x**3+2*x**3+x**2)*ln(x)*exp(x-ln(x))**2*exp(exp(x))+(8*x**3+4*x**2)*ln(x)*exp(x-ln(x))**2)
*ln(ln(x))**2+((4*exp(x)*x**2+8*x**2)*ln(x)*exp(x-ln(x))**2*exp(exp(x))+32*x**2*ln(x)*exp(x-ln(x))**2)*ln(ln(x
))-4*x*exp(x-ln(x))**2*exp(exp(x))-16*x*exp(x-ln(x))**2)/ln(x)/ln(ln(x))**2,x)

[Out]

(4*x*log(log(x)) + 16)*exp(2*x)/log(log(x)) + (x*exp(2*x)*log(log(x)) + 4*exp(2*x))*exp(exp(x))/log(log(x))

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