3.44.18 \(\int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log (e^{-x} x) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log (e^{-x} x) (x^2+2 x \log (\log (3)))} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\left (5+x \left (-x+\log \left (e^{-x} x\right )\right ) (x+2 \log (\log (3)))\right )^2\right ) \]

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Rubi [F]  time = 1.67, 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 {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x - 8*x^2 + 2*(2 - 6*x)*Log[Log[3]] + Log[x/E^x]*(4*x + 4*Log[Log[3]]))/(5 - x^3 - 2*x^2*Log[Log[3]] +
Log[x/E^x]*(x^2 + 2*x*Log[Log[3]])),x]

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 0.28, size = 70, normalized size = 2.69 \begin {gather*} 2 \log \left (5-2 x^3+x^2 \log (x)+x^2 \left (x-\log (x)+\log \left (e^{-x} x\right )\right )-4 x^2 \log (\log (3))+2 x \log (x) \log (\log (3))+2 x \left (x-\log (x)+\log \left (e^{-x} x\right )\right ) \log (\log (3))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 8*x^2 + 2*(2 - 6*x)*Log[Log[3]] + Log[x/E^x]*(4*x + 4*Log[Log[3]]))/(5 - x^3 - 2*x^2*Log[Log[
3]] + Log[x/E^x]*(x^2 + 2*x*Log[Log[3]])),x]

[Out]

2*Log[5 - 2*x^3 + x^2*Log[x] + x^2*(x - Log[x] + Log[x/E^x]) - 4*x^2*Log[Log[3]] + 2*x*Log[x]*Log[Log[3]] + 2*
x*(x - Log[x] + Log[x/E^x])*Log[Log[3]]]

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fricas [B]  time = 0.61, size = 63, normalized size = 2.42 \begin {gather*} 2 \, \log \left (x^{2} + 2 \, x \log \left (\log \relax (3)\right )\right ) + 2 \, \log \left (-\frac {x^{3} + 2 \, x^{2} \log \left (\log \relax (3)\right ) - {\left (x^{2} + 2 \, x \log \left (\log \relax (3)\right )\right )} \log \left (x e^{\left (-x\right )}\right ) - 5}{x^{2} + 2 \, x \log \left (\log \relax (3)\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(log(3))+4*x)*log(x/exp(x))+2*(-6*x+2)*log(log(3))-8*x^2+2*x)/((2*log(log(3))*x+x^2)*log(x/ex
p(x))-2*x^2*log(log(3))-x^3+5),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 33, normalized size = 1.27 \begin {gather*} 2 \, \log \left (2 \, x^{3} - x^{2} \log \relax (x) + 4 \, x^{2} \log \left (\log \relax (3)\right ) - 2 \, x \log \relax (x) \log \left (\log \relax (3)\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(log(3))+4*x)*log(x/exp(x))+2*(-6*x+2)*log(log(3))-8*x^2+2*x)/((2*log(log(3))*x+x^2)*log(x/ex
p(x))-2*x^2*log(log(3))-x^3+5),x, algorithm="giac")

[Out]

2*log(2*x^3 - x^2*log(x) + 4*x^2*log(log(3)) - 2*x*log(x)*log(log(3)) - 5)

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maple [A]  time = 0.19, size = 42, normalized size = 1.62

method	result	size
norman	\(2 \ln \left (-2 \ln \left (x \,{\mathrm e}^{-x}\right ) \ln \left (\ln \relax (3)\right ) x -\ln \left (x \,{\mathrm e}^{-x}\right ) x^{2}+2 x^{2} \ln \left (\ln \relax (3)\right )+x^{3}-5\right )\)	\(42\)
risch	\(2 \ln \left (2 \ln \left (\ln \relax (3)\right ) x +x^{2}\right )+2 \ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {-630-126 x^{2} \ln \relax (x )+126 x^{3}+252 x^{2} \ln \left (\ln \relax (3)\right )+2 i x^{3}+\pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-4 i \ln \left (\ln \relax (3)\right ) x \ln \relax (x )-2 \pi \ln \left (\ln \relax (3)\right ) x \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+63 i \pi \,x^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+\pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+126 i \pi \ln \left (\ln \relax (3)\right ) x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )-252 \ln \left (\ln \relax (3)\right ) x \ln \relax (x )-10 i+4 i \ln \left (\ln \relax (3)\right ) x^{2}-\pi \,x^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}-2 \pi \ln \left (\ln \relax (3)\right ) x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+63 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )-126 i \pi \ln \left (\ln \relax (3)\right ) x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-126 i \pi \ln \left (\ln \relax (3)\right ) x \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-63 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-63 i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+126 i \pi \ln \left (\ln \relax (3)\right ) x \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}-\pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+2 \pi \ln \left (\ln \relax (3)\right ) x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+2 \pi \ln \left (\ln \relax (3)\right ) x \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-2 i x^{2} \ln \relax (x )}{2 x \left (2 i \ln \left (\ln \relax (3)\right )+i x +126 \ln \left (\ln \relax (3)\right )+63 x \right )}\right )\)	\(494\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*ln(ln(3))+4*x)*ln(x/exp(x))+2*(-6*x+2)*ln(ln(3))-8*x^2+2*x)/((2*ln(ln(3))*x+x^2)*ln(x/exp(x))-2*x^2*ln
(ln(3))-x^3+5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.47, size = 61, normalized size = 2.35 \begin {gather*} 2 \, \log \left (x + 2 \, \log \left (\log \relax (3)\right )\right ) + 2 \, \log \relax (x) + 2 \, \log \left (-\frac {2 \, x^{3} + 4 \, x^{2} \log \left (\log \relax (3)\right ) - {\left (x^{2} + 2 \, x \log \left (\log \relax (3)\right )\right )} \log \relax (x) - 5}{x^{2} + 2 \, x \log \left (\log \relax (3)\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(log(3))+4*x)*log(x/exp(x))+2*(-6*x+2)*log(log(3))-8*x^2+2*x)/((2*log(log(3))*x+x^2)*log(x/ex
p(x))-2*x^2*log(log(3))-x^3+5),x, algorithm="maxima")

[Out]

2*log(x + 2*log(log(3))) + 2*log(x) + 2*log(-(2*x^3 + 4*x^2*log(log(3)) - (x^2 + 2*x*log(log(3)))*log(x) - 5)/
(x^2 + 2*x*log(log(3))))

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mupad [B]  time = 3.80, size = 59, normalized size = 2.27 \begin {gather*} 2\,\ln \left (\frac {x^2\,\left (4\,\ln \left (\ln \relax (3)\right )-\ln \relax (x)\right )+2\,x^3-2\,x\,\ln \left (\ln \relax (3)\right )\,\ln \relax (x)-5}{x^2+2\,\ln \left (\ln \relax (3)\right )\,x}\right )+2\,\ln \left (x^2+2\,\ln \left (\ln \relax (3)\right )\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(x*exp(-x))*(4*x + 4*log(log(3))) - 2*log(log(3))*(6*x - 2) - 8*x^2)/(2*x^2*log(log(3)) - log(x
*exp(-x))*(2*x*log(log(3)) + x^2) + x^3 - 5),x)

[Out]

2*log((x^2*(4*log(log(3)) - log(x)) + 2*x^3 - 2*x*log(log(3))*log(x) - 5)/(2*x*log(log(3)) + x^2)) + 2*log(2*x
*log(log(3)) + x^2)

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sympy [B]  time = 0.61, size = 51, normalized size = 1.96 \begin {gather*} 2 \log {\left (x^{2} + 2 x \log {\left (\log {\relax (3 )} \right )} \right )} + 2 \log {\left (\log {\left (x e^{- x} \right )} + \frac {- x^{3} - 2 x^{2} \log {\left (\log {\relax (3 )} \right )} + 5}{x^{2} + 2 x \log {\left (\log {\relax (3 )} \right )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*ln(ln(3))+4*x)*ln(x/exp(x))+2*(-6*x+2)*ln(ln(3))-8*x**2+2*x)/((2*ln(ln(3))*x+x**2)*ln(x/exp(x))-
2*x**2*ln(ln(3))-x**3+5),x)

[Out]

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

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