3.22.31 \(\int \frac {-e^x x+\log (\frac {1}{4} (8+e^x)) (8+e^x+(-16-2 e^x) \log (x)) \log (\log (\frac {1}{4} (8+e^x))) \log (\log (\log (\frac {1}{4} (8+e^x))))+(16+2 e^x) \log (\frac {1}{4} (8+e^x)) \log (\log (\frac {1}{4} (8+e^x))) \log (\log (\log (\frac {1}{4} (8+e^x)))) \log (\log (\log (\log (\frac {1}{4} (8+e^x)))))}{(80 x^3+10 e^x x^3) \log (\frac {1}{4} (8+e^x)) \log (\log (\frac {1}{4} (8+e^x))) \log (\log (\log (\frac {1}{4} (8+e^x))))} \, dx\) [2131]

Optimal. Leaf size=25 \[ \frac {\log (x)-\log \left (\log \left (\log \left (\log \left (2+\frac {e^x}{4}\right )\right )\right )\right )}{10 x^2} \]

[Out]

1/10*(ln(x)-ln(ln(ln(ln(1/4*exp(x)+2)))))/x^2

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Rubi [F]
time = 2.62, 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^x x+\log \left (\frac {1}{4} \left (8+e^x\right )\right ) \left (8+e^x+\left (-16-2 e^x\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )+\left (16+2 e^x\right ) \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right ) \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{\left (80 x^3+10 e^x x^3\right ) \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^x*x) + Log[(8 + E^x)/4]*(8 + E^x + (-16 - 2*E^x)*Log[x])*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 + E^x)/
4]]] + (16 + 2*E^x)*Log[(8 + E^x)/4]*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 + E^x)/4]]]*Log[Log[Log[Log[(8 + E^x
)/4]]]])/((80*x^3 + 10*E^x*x^3)*Log[(8 + E^x)/4]*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 + E^x)/4]]]),x]

[Out]

Log[x]/(10*x^2) - Defer[Int][1/(x^2*Log[(8 + E^x)/4]*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 + E^x)/4]]]), x]/10
+ (4*Defer[Int][1/((8 + E^x)*x^2*Log[(8 + E^x)/4]*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 + E^x)/4]]]), x])/5 + D
efer[Int][Log[Log[Log[Log[(8 + E^x)/4]]]]/x^3, x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-2 \log (x)-\frac {e^x x}{\left (8+e^x\right ) \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}+2 \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{10 x^3} \, dx\\ &=\frac {1}{10} \int \frac {1-2 \log (x)-\frac {e^x x}{\left (8+e^x\right ) \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}+2 \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3} \, dx\\ &=\frac {1}{10} \int \left (\frac {8}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}+\frac {-x+\log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )-2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log (x) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )+2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right ) \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}\right ) \, dx\\ &=\frac {1}{10} \int \frac {-x+\log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )-2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log (x) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )+2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right ) \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ &=\frac {1}{10} \int \frac {1-2 \log (x)-\frac {x}{\log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}+2 \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3} \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ &=\frac {1}{10} \int \left (\frac {-x+\log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )-2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log (x) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}{x^3 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}+\frac {2 \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3}\right ) \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ &=\frac {1}{10} \int \frac {-x+\log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )-2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log (x) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}{x^3 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx+\frac {1}{5} \int \frac {\log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3} \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ &=\frac {1}{10} \int \frac {1-2 \log (x)-\frac {x}{\log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}}{x^3} \, dx+\frac {1}{5} \int \frac {\log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3} \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ &=\frac {1}{10} \int \left (\frac {1-2 \log (x)}{x^3}-\frac {1}{x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )}\right ) \, dx+\frac {1}{5} \int \frac {\log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3} \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ &=\frac {1}{10} \int \frac {1-2 \log (x)}{x^3} \, dx-\frac {1}{10} \int \frac {1}{x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx+\frac {1}{5} \int \frac {\log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3} \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ &=\frac {\log (x)}{10 x^2}-\frac {1}{10} \int \frac {1}{x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx+\frac {1}{5} \int \frac {\log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^3} \, dx+\frac {4}{5} \int \frac {1}{\left (8+e^x\right ) x^2 \log \left (\frac {1}{4} \left (8+e^x\right )\right ) \log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.26, size = 29, normalized size = 1.16 \begin {gather*} \frac {1}{10} \left (\frac {\log (x)}{x^2}-\frac {\log \left (\log \left (\log \left (\log \left (\frac {1}{4} \left (8+e^x\right )\right )\right )\right )\right )}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^x*x) + Log[(8 + E^x)/4]*(8 + E^x + (-16 - 2*E^x)*Log[x])*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 +
E^x)/4]]] + (16 + 2*E^x)*Log[(8 + E^x)/4]*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 + E^x)/4]]]*Log[Log[Log[Log[(8
+ E^x)/4]]]])/((80*x^3 + 10*E^x*x^3)*Log[(8 + E^x)/4]*Log[Log[(8 + E^x)/4]]*Log[Log[Log[(8 + E^x)/4]]]),x]

[Out]

(Log[x]/x^2 - Log[Log[Log[Log[(8 + E^x)/4]]]]/x^2)/10

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Maple [A]
time = 0.07, size = 24, normalized size = 0.96

method	result	size
risch	\(-\frac {\ln \left (\ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{x}}{4}+2\right )\right )\right )\right )}{10 x^{2}}+\frac {\ln \left (x \right )}{10 x^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)+16)*ln(1/4*exp(x)+2)*ln(ln(1/4*exp(x)+2))*ln(ln(ln(1/4*exp(x)+2)))*ln(ln(ln(ln(1/4*exp(x)+2))))
+((-2*exp(x)-16)*ln(x)+exp(x)+8)*ln(1/4*exp(x)+2)*ln(ln(1/4*exp(x)+2))*ln(ln(ln(1/4*exp(x)+2)))-exp(x)*x)/(10*
exp(x)*x^3+80*x^3)/ln(1/4*exp(x)+2)/ln(ln(1/4*exp(x)+2))/ln(ln(ln(1/4*exp(x)+2))),x,method=_RETURNVERBOSE)

[Out]

-1/10/x^2*ln(ln(ln(ln(1/4*exp(x)+2))))+1/10*ln(x)/x^2

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Maxima [A]
time = 0.62, size = 23, normalized size = 0.92 \begin {gather*} \frac {\log \left (x\right ) - \log \left (\log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (e^{x} + 8\right )\right )\right )\right )}{10 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)+16)*log(1/4*exp(x)+2)*log(log(1/4*exp(x)+2))*log(log(log(1/4*exp(x)+2)))*log(log(log(log(
1/4*exp(x)+2))))+((-2*exp(x)-16)*log(x)+exp(x)+8)*log(1/4*exp(x)+2)*log(log(1/4*exp(x)+2))*log(log(log(1/4*exp
(x)+2)))-exp(x)*x)/(10*exp(x)*x^3+80*x^3)/log(1/4*exp(x)+2)/log(log(1/4*exp(x)+2))/log(log(log(1/4*exp(x)+2)))
,x, algorithm="maxima")

[Out]

1/10*(log(x) - log(log(log(-2*log(2) + log(e^x + 8)))))/x^2

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Fricas [A]
time = 0.39, size = 20, normalized size = 0.80 \begin {gather*} \frac {\log \left (x\right ) - \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \, e^{x} + 2\right )\right )\right )\right )}{10 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)+16)*log(1/4*exp(x)+2)*log(log(1/4*exp(x)+2))*log(log(log(1/4*exp(x)+2)))*log(log(log(log(
1/4*exp(x)+2))))+((-2*exp(x)-16)*log(x)+exp(x)+8)*log(1/4*exp(x)+2)*log(log(1/4*exp(x)+2))*log(log(log(1/4*exp
(x)+2)))-exp(x)*x)/(10*exp(x)*x^3+80*x^3)/log(1/4*exp(x)+2)/log(log(1/4*exp(x)+2))/log(log(log(1/4*exp(x)+2)))
,x, algorithm="fricas")

[Out]

1/10*(log(x) - log(log(log(log(1/4*e^x + 2)))))/x^2

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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(((2*exp(x)+16)*ln(1/4*exp(x)+2)*ln(ln(1/4*exp(x)+2))*ln(ln(ln(1/4*exp(x)+2)))*ln(ln(ln(ln(1/4*exp(x)
+2))))+((-2*exp(x)-16)*ln(x)+exp(x)+8)*ln(1/4*exp(x)+2)*ln(ln(1/4*exp(x)+2))*ln(ln(ln(1/4*exp(x)+2)))-exp(x)*x
)/(10*exp(x)*x**3+80*x**3)/ln(1/4*exp(x)+2)/ln(ln(1/4*exp(x)+2))/ln(ln(ln(1/4*exp(x)+2))),x)

[Out]

Timed out

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Giac [A]
time = 1.28, size = 20, normalized size = 0.80 \begin {gather*} \frac {\log \left (x\right ) - \log \left (\log \left (\log \left (\log \left (\frac {1}{4} \, e^{x} + 2\right )\right )\right )\right )}{10 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)+16)*log(1/4*exp(x)+2)*log(log(1/4*exp(x)+2))*log(log(log(1/4*exp(x)+2)))*log(log(log(log(
1/4*exp(x)+2))))+((-2*exp(x)-16)*log(x)+exp(x)+8)*log(1/4*exp(x)+2)*log(log(1/4*exp(x)+2))*log(log(log(1/4*exp
(x)+2)))-exp(x)*x)/(10*exp(x)*x^3+80*x^3)/log(1/4*exp(x)+2)/log(log(1/4*exp(x)+2))/log(log(log(1/4*exp(x)+2)))
,x, algorithm="giac")

[Out]

1/10*(log(x) - log(log(log(log(1/4*e^x + 2)))))/x^2

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Mupad [B]
time = 5.13, size = 20, normalized size = 0.80 \begin {gather*} -\frac {\ln \left (\ln \left (\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{4}+2\right )\right )\right )\right )-\ln \left (x\right )}{10\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(x)/4 + 2)*log(log(exp(x)/4 + 2))*log(log(log(exp(x)/4 + 2)))*(exp(x) - log(x)*(2*exp(x) + 16) + 8
) - x*exp(x) + log(exp(x)/4 + 2)*log(log(log(log(exp(x)/4 + 2))))*log(log(exp(x)/4 + 2))*log(log(log(exp(x)/4
+ 2)))*(2*exp(x) + 16))/(log(exp(x)/4 + 2)*log(log(exp(x)/4 + 2))*log(log(log(exp(x)/4 + 2)))*(10*x^3*exp(x) +
80*x^3)),x)

[Out]

-(log(log(log(log(exp(x)/4 + 2)))) - log(x))/(10*x^2)

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