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

Optimal. Leaf size=26 \[ e^{3-x} x \left (-e^x+x\right )+5 x \log (\log (3 \log (x))) \]

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Rubi [F]  time = 0.62, 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 {5+\left (-e^3+e^{3-x} \left (2 x-x^2\right )\right ) \log (x) \log (3 \log (x))+5 \log (x) \log (3 \log (x)) \log (\log (3 \log (x)))}{\log (x) \log (3 \log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 24, normalized size = 0.92 \begin {gather*} x \left (e^3 \left (-1+e^{-x} x\right )+5 \log (\log (3 \log (x)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*(E^3*(-1 + x/E^x) + 5*Log[Log[3*Log[x]]])

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fricas [A]  time = 0.66, size = 25, normalized size = 0.96 \begin {gather*} x^{2} e^{\left (-x + 3\right )} - x e^{3} + 5 \, x \log \left (\log \left (3 \, \log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(-x + 3) - x*e^3 + 5*x*log(log(3*log(x)))

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giac [A]  time = 0.39, size = 25, normalized size = 0.96 \begin {gather*} x^{2} e^{\left (-x + 3\right )} - x e^{3} + 5 \, x \log \left (\log \left (3 \, \log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(-x + 3) - x*e^3 + 5*x*log(log(3*log(x)))

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maple [A]  time = 0.12, size = 25, normalized size = 0.96

method	result	size
risch	\({\mathrm e}^{3-x} \left (x -{\mathrm e}^{x}\right ) x +5 \ln \left (\ln \left (3 \ln \relax (x )\right )\right ) x\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*ln(x)*ln(3*ln(x))*ln(ln(3*ln(x)))+(-exp(3-x)*exp(x)+(-x^2+2*x)*exp(3-x))*ln(x)*ln(3*ln(x))+5)/ln(x)/ln(
3*ln(x)),x,method=_RETURNVERBOSE)

[Out]

exp(3-x)*(x-exp(x))*x+5*ln(ln(3*ln(x)))*x

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maxima [B]  time = 0.60, size = 50, normalized size = 1.92 \begin {gather*} -x e^{3} + {\left (x^{2} e^{3} + 2 \, x e^{3} + 2 \, e^{3}\right )} e^{\left (-x\right )} - 2 \, {\left (x e^{3} + e^{3}\right )} e^{\left (-x\right )} + 5 \, x \log \left (\log \relax (3) + \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*e^3 + (x^2*e^3 + 2*x*e^3 + 2*e^3)*e^(-x) - 2*(x*e^3 + e^3)*e^(-x) + 5*x*log(log(3) + log(log(x)))

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mupad [B]  time = 0.91, size = 25, normalized size = 0.96 \begin {gather*} x^2\,{\mathrm {e}}^{3-x}-x\,{\mathrm {e}}^3+5\,x\,\ln \left (\ln \left (3\,\ln \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(log(3*log(x)))*log(3*log(x))*log(x) + log(3*log(x))*log(x)*(exp(3 - x)*(2*x - x^2) - exp(3 - x)*exp
(x)) + 5)/(log(3*log(x))*log(x)),x)

[Out]

x^2*exp(3 - x) - x*exp(3) + 5*x*log(log(3*log(x)))

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sympy [A]  time = 1.12, size = 26, normalized size = 1.00 \begin {gather*} x^{2} e^{3} e^{- x} + 5 x \log {\left (\log {\left (3 \log {\relax (x )} \right )} \right )} - x e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*exp(3)*exp(-x) + 5*x*log(log(3*log(x))) - x*exp(3)

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