3.43.2 \(\int \frac {20-2 e^x+(-20 e^x x+2 e^{2 x} x) \log (x) \log (\log (x))+(2-2 e^x x \log (x) \log (\log (x))) \log (\log (\log (x)))}{9 x \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{9} \left (-16+e^x+2 (3-x)+2 x-\log (\log (\log (x)))\right )^2 \]

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Rubi [A]  time = 0.63, antiderivative size = 17, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 3, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 6688, 6686} \begin {gather*} \frac {1}{9} \left (-e^x+\log (\log (\log (x)))+10\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 - 2*E^x + (-20*E^x*x + 2*E^(2*x)*x)*Log[x]*Log[Log[x]] + (2 - 2*E^x*x*Log[x]*Log[Log[x]])*Log[Log[Log[
x]]])/(9*x*Log[x]*Log[Log[x]]),x]

[Out]

(10 - E^x + Log[Log[Log[x]]])^2/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \frac {20-2 e^x+\left (-20 e^x x+2 e^{2 x} x\right ) \log (x) \log (\log (x))+\left (2-2 e^x x \log (x) \log (\log (x))\right ) \log (\log (\log (x)))}{x \log (x) \log (\log (x))} \, dx\\ &=\frac {1}{9} \int \frac {2 \left (1-e^x x \log (x) \log (\log (x))\right ) \left (10-e^x+\log (\log (\log (x)))\right )}{x \log (x) \log (\log (x))} \, dx\\ &=\frac {2}{9} \int \frac {\left (1-e^x x \log (x) \log (\log (x))\right ) \left (10-e^x+\log (\log (\log (x)))\right )}{x \log (x) \log (\log (x))} \, dx\\ &=\frac {1}{9} \left (10-e^x+\log (\log (\log (x)))\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 0.63 \begin {gather*} \frac {1}{9} \left (-10+e^x-\log (\log (\log (x)))\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 - 2*E^x + (-20*E^x*x + 2*E^(2*x)*x)*Log[x]*Log[Log[x]] + (2 - 2*E^x*x*Log[x]*Log[Log[x]])*Log[Lo
g[Log[x]]])/(9*x*Log[x]*Log[Log[x]]),x]

[Out]

(-10 + E^x - Log[Log[Log[x]]])^2/9

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fricas [B]  time = 0.69, size = 29, normalized size = 1.07 \begin {gather*} -\frac {2}{9} \, {\left (e^{x} - 10\right )} \log \left (\log \left (\log \relax (x)\right )\right ) + \frac {1}{9} \, \log \left (\log \left (\log \relax (x)\right )\right )^{2} + \frac {1}{9} \, e^{\left (2 \, x\right )} - \frac {20}{9} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-2*x*exp(x)*log(x)*log(log(x))+2)*log(log(log(x)))+(2*x*exp(x)^2-20*exp(x)*x)*log(x)*log(log(x
))-2*exp(x)+20)/x/log(x)/log(log(x)),x, algorithm="fricas")

[Out]

-2/9*(e^x - 10)*log(log(log(x))) + 1/9*log(log(log(x)))^2 + 1/9*e^(2*x) - 20/9*e^x

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giac [B]  time = 0.18, size = 33, normalized size = 1.22 \begin {gather*} -\frac {2}{9} \, e^{x} \log \left (\log \left (\log \relax (x)\right )\right ) + \frac {1}{9} \, \log \left (\log \left (\log \relax (x)\right )\right )^{2} + \frac {1}{9} \, e^{\left (2 \, x\right )} - \frac {20}{9} \, e^{x} + \frac {20}{9} \, \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-2*x*exp(x)*log(x)*log(log(x))+2)*log(log(log(x)))+(2*x*exp(x)^2-20*exp(x)*x)*log(x)*log(log(x
))-2*exp(x)+20)/x/log(x)/log(log(x)),x, algorithm="giac")

[Out]

-2/9*e^x*log(log(log(x))) + 1/9*log(log(log(x)))^2 + 1/9*e^(2*x) - 20/9*e^x + 20/9*log(log(log(x)))

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maple [A]  time = 0.04, size = 34, normalized size = 1.26

method	result	size
risch	\(\frac {\ln \left (\ln \left (\ln \relax (x )\right )\right )^{2}}{9}-\frac {2 \,{\mathrm e}^{x} \ln \left (\ln \left (\ln \relax (x )\right )\right )}{9}+\frac {{\mathrm e}^{2 x}}{9}-\frac {20 \,{\mathrm e}^{x}}{9}+\frac {20 \ln \left (\ln \left (\ln \relax (x )\right )\right )}{9}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((-2*x*exp(x)*ln(x)*ln(ln(x))+2)*ln(ln(ln(x)))+(2*x*exp(x)^2-20*exp(x)*x)*ln(x)*ln(ln(x))-2*exp(x)+20)
/x/ln(x)/ln(ln(x)),x,method=_RETURNVERBOSE)

[Out]

1/9*ln(ln(ln(x)))^2-2/9*exp(x)*ln(ln(ln(x)))+1/9*exp(2*x)-20/9*exp(x)+20/9*ln(ln(ln(x)))

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maxima [B]  time = 0.39, size = 33, normalized size = 1.22 \begin {gather*} -\frac {2}{9} \, e^{x} \log \left (\log \left (\log \relax (x)\right )\right ) + \frac {1}{9} \, \log \left (\log \left (\log \relax (x)\right )\right )^{2} + \frac {1}{9} \, e^{\left (2 \, x\right )} - \frac {20}{9} \, e^{x} + \frac {20}{9} \, \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-2*x*exp(x)*log(x)*log(log(x))+2)*log(log(log(x)))+(2*x*exp(x)^2-20*exp(x)*x)*log(x)*log(log(x
))-2*exp(x)+20)/x/log(x)/log(log(x)),x, algorithm="maxima")

[Out]

-2/9*e^x*log(log(log(x))) + 1/9*log(log(log(x)))^2 + 1/9*e^(2*x) - 20/9*e^x + 20/9*log(log(log(x)))

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mupad [B]  time = 3.04, size = 33, normalized size = 1.22 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}}{9}+\frac {20\,\ln \left (\ln \left (\ln \relax (x)\right )\right )}{9}-\frac {20\,{\mathrm {e}}^x}{9}+\frac {{\ln \left (\ln \left (\ln \relax (x)\right )\right )}^2}{9}-\frac {2\,{\mathrm {e}}^x\,\ln \left (\ln \left (\ln \relax (x)\right )\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*exp(x))/9 + (log(log(log(x)))*(2*x*log(log(x))*exp(x)*log(x) - 2))/9 - (log(log(x))*log(x)*(2*x*exp(2
*x) - 20*x*exp(x)))/9 - 20/9)/(x*log(log(x))*log(x)),x)

[Out]

exp(2*x)/9 + (20*log(log(log(x))))/9 - (20*exp(x))/9 + log(log(log(x)))^2/9 - (2*exp(x)*log(log(log(x))))/9

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sympy [B]  time = 18.86, size = 42, normalized size = 1.56 \begin {gather*} \frac {\left (- 18 \log {\left (\log {\left (\log {\relax (x )} \right )} \right )} - 180\right ) e^{x}}{81} + \frac {e^{2 x}}{9} + \frac {\log {\left (\log {\left (\log {\relax (x )} \right )} \right )}^{2}}{9} + \frac {20 \log {\left (\log {\left (\log {\relax (x )} \right )} \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-2*x*exp(x)*ln(x)*ln(ln(x))+2)*ln(ln(ln(x)))+(2*x*exp(x)**2-20*exp(x)*x)*ln(x)*ln(ln(x))-2*exp
(x)+20)/x/ln(x)/ln(ln(x)),x)

[Out]

(-18*log(log(log(x))) - 180)*exp(x)/81 + exp(2*x)/9 + log(log(log(x)))**2/9 + 20*log(log(log(x)))/9

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