3.13.19 \(\int \frac {1-10 x \log (x)+e^{e^x} (1+e^x x) \log (x)+\log (x) \log (\log (x))+(-e^{e^x} \log (x)+5 x \log (x)-\log (x) \log (\log (x))) \log (e^{e^x}-5 x+\log (\log (x)))}{e^{e^x} x^2 \log (x)-5 x^3 \log (x)+x^2 \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=22 \[ \frac {-1-x+\log \left (e^{e^x}-5 x+\log (\log (x))\right )}{x} \]

________________________________________________________________________________________

Rubi [F]  time = 5.42, 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 {1-10 x \log (x)+e^{e^x} \left (1+e^x x\right ) \log (x)+\log (x) \log (\log (x))+\left (-e^{e^x} \log (x)+5 x \log (x)-\log (x) \log (\log (x))\right ) \log \left (e^{e^x}-5 x+\log (\log (x))\right )}{e^{e^x} x^2 \log (x)-5 x^3 \log (x)+x^2 \log (x) \log (\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 - 10*x*Log[x] + E^E^x*(1 + E^x*x)*Log[x] + Log[x]*Log[Log[x]] + (-(E^E^x*Log[x]) + 5*x*Log[x] - Log[x]*
Log[Log[x]])*Log[E^E^x - 5*x + Log[Log[x]]])/(E^E^x*x^2*Log[x] - 5*x^3*Log[x] + x^2*Log[x]*Log[Log[x]]),x]

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.55, size = 23, normalized size = 1.05 \begin {gather*} -\frac {1}{x}+\frac {\log \left (e^{e^x}-5 x+\log (\log (x))\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 10*x*Log[x] + E^E^x*(1 + E^x*x)*Log[x] + Log[x]*Log[Log[x]] + (-(E^E^x*Log[x]) + 5*x*Log[x] - L
og[x]*Log[Log[x]])*Log[E^E^x - 5*x + Log[Log[x]]])/(E^E^x*x^2*Log[x] - 5*x^3*Log[x] + x^2*Log[x]*Log[Log[x]]),
x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 17, normalized size = 0.77 \begin {gather*} \frac {\log \left (-5 \, x + e^{\left (e^{x}\right )} + \log \left (\log \relax (x)\right )\right ) - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(x)*log(log(x))-log(x)*exp(exp(x))+5*x*log(x))*log(log(log(x))+exp(exp(x))-5*x)+log(x)*log(log
(x))+(exp(x)*x+1)*log(x)*exp(exp(x))-10*x*log(x)+1)/(x^2*log(x)*log(log(x))+x^2*log(x)*exp(exp(x))-5*x^3*log(x
)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.83, size = 33, normalized size = 1.50 \begin {gather*} \frac {\log \left (-{\left (5 \, x e^{x} - e^{x} \log \left (\log \relax (x)\right ) - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right ) - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(x)*log(log(x))-log(x)*exp(exp(x))+5*x*log(x))*log(log(log(x))+exp(exp(x))-5*x)+log(x)*log(log
(x))+(exp(x)*x+1)*log(x)*exp(exp(x))-10*x*log(x)+1)/(x^2*log(x)*log(log(x))+x^2*log(x)*exp(exp(x))-5*x^3*log(x
)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 22, normalized size = 1.00

method	result	size
risch	\(\frac {\ln \left (\ln \left (\ln \relax (x )\right )+{\mathrm e}^{{\mathrm e}^{x}}-5 x \right )}{x}-\frac {1}{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(x)*ln(ln(x))-ln(x)*exp(exp(x))+5*x*ln(x))*ln(ln(ln(x))+exp(exp(x))-5*x)+ln(x)*ln(ln(x))+(exp(x)*x+1)
*ln(x)*exp(exp(x))-10*x*ln(x)+1)/(x^2*ln(x)*ln(ln(x))+x^2*ln(x)*exp(exp(x))-5*x^3*ln(x)),x,method=_RETURNVERBO
SE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 17, normalized size = 0.77 \begin {gather*} \frac {\log \left (-5 \, x + e^{\left (e^{x}\right )} + \log \left (\log \relax (x)\right )\right ) - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(x)*log(log(x))-log(x)*exp(exp(x))+5*x*log(x))*log(log(log(x))+exp(exp(x))-5*x)+log(x)*log(log
(x))+(exp(x)*x+1)*log(x)*exp(exp(x))-10*x*log(x)+1)/(x^2*log(x)*log(log(x))+x^2*log(x)*exp(exp(x))-5*x^3*log(x
)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.11, size = 17, normalized size = 0.77 \begin {gather*} \frac {\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}-5\,x+\ln \left (\ln \relax (x)\right )\right )-1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x))*log(x) - 10*x*log(x) - log(exp(exp(x)) - 5*x + log(log(x)))*(exp(exp(x))*log(x) + log(log(x))
*log(x) - 5*x*log(x)) + exp(exp(x))*log(x)*(x*exp(x) + 1) + 1)/(x^2*exp(exp(x))*log(x) - 5*x^3*log(x) + x^2*lo
g(log(x))*log(x)),x)

[Out]

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

________________________________________________________________________________________

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(((-ln(x)*ln(ln(x))-ln(x)*exp(exp(x))+5*x*ln(x))*ln(ln(ln(x))+exp(exp(x))-5*x)+ln(x)*ln(ln(x))+(exp(x
)*x+1)*ln(x)*exp(exp(x))-10*x*ln(x)+1)/(x**2*ln(x)*ln(ln(x))+x**2*ln(x)*exp(exp(x))-5*x**3*ln(x)),x)

[Out]

Timed out

________________________________________________________________________________________