3.84.20 \(\int \frac {(-4+x^2) \log (x) \log ^2(\log (x))+e^{\frac {4 x}{\log (\log (x))}} (-4 x+4 x \log (x) \log (\log (x))-\log (x) \log ^2(\log (x)))}{e^{\frac {8 x}{\log (\log (x))}} \log (x) \log ^2(\log (x))+e^{\frac {4 x}{\log (\log (x))}} (8-4 x+2 x^2) \log (x) \log ^2(\log (x))+(16-16 x+12 x^2-4 x^3+x^4) \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=25 \[ \frac {x}{-3-e^{\frac {4 x}{\log (\log (x))}}-(-1+x)^2} \]

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Rubi [F]  time = 6.28, 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 {\left (-4+x^2\right ) \log (x) \log ^2(\log (x))+e^{\frac {4 x}{\log (\log (x))}} \left (-4 x+4 x \log (x) \log (\log (x))-\log (x) \log ^2(\log (x))\right )}{e^{\frac {8 x}{\log (\log (x))}} \log (x) \log ^2(\log (x))+e^{\frac {4 x}{\log (\log (x))}} \left (8-4 x+2 x^2\right ) \log (x) \log ^2(\log (x))+\left (16-16 x+12 x^2-4 x^3+x^4\right ) \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-4 + x^2)*Log[x]*Log[Log[x]]^2 + E^((4*x)/Log[Log[x]])*(-4*x + 4*x*Log[x]*Log[Log[x]] - Log[x]*Log[Log[x
]]^2))/(E^((8*x)/Log[Log[x]])*Log[x]*Log[Log[x]]^2 + E^((4*x)/Log[Log[x]])*(8 - 4*x + 2*x^2)*Log[x]*Log[Log[x]
]^2 + (16 - 16*x + 12*x^2 - 4*x^3 + x^4)*Log[x]*Log[Log[x]]^2),x]

[Out]

-2*Defer[Int][x/(4 + E^((4*x)/Log[Log[x]]) - 2*x + x^2)^2, x] + 2*Defer[Int][x^2/(4 + E^((4*x)/Log[Log[x]]) -
2*x + x^2)^2, x] - Defer[Int][(4 + E^((4*x)/Log[Log[x]]) - 2*x + x^2)^(-1), x] + 16*Defer[Int][x/((4 + E^((4*x
)/Log[Log[x]]) - 2*x + x^2)^2*Log[x]*Log[Log[x]]^2), x] - 8*Defer[Int][x^2/((4 + E^((4*x)/Log[Log[x]]) - 2*x +
 x^2)^2*Log[x]*Log[Log[x]]^2), x] + 4*Defer[Int][x^3/((4 + E^((4*x)/Log[Log[x]]) - 2*x + x^2)^2*Log[x]*Log[Log
[x]]^2), x] - 4*Defer[Int][x/((4 + E^((4*x)/Log[Log[x]]) - 2*x + x^2)*Log[x]*Log[Log[x]]^2), x] - 16*Defer[Int
][x/((4 + E^((4*x)/Log[Log[x]]) - 2*x + x^2)^2*Log[Log[x]]), x] + 8*Defer[Int][x^2/((4 + E^((4*x)/Log[Log[x]])
 - 2*x + x^2)^2*Log[Log[x]]), x] - 4*Defer[Int][x^3/((4 + E^((4*x)/Log[Log[x]]) - 2*x + x^2)^2*Log[Log[x]]), x
] + 4*Defer[Int][x/((4 + E^((4*x)/Log[Log[x]]) - 2*x + x^2)*Log[Log[x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 e^{\frac {4 x}{\log (\log (x))}} x+\log (x) \log (\log (x)) \left (4 e^{\frac {4 x}{\log (\log (x))}} x+\left (-4-e^{\frac {4 x}{\log (\log (x))}}+x^2\right ) \log (\log (x))\right )}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))} \, dx\\ &=\int \left (\frac {-4 x+4 x \log (x) \log (\log (x))-\log (x) \log ^2(\log (x))}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (x) \log ^2(\log (x))}-\frac {2 x \left (-8+4 x-2 x^2+8 \log (x) \log (\log (x))-4 x \log (x) \log (\log (x))+2 x^2 \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))-x \log (x) \log ^2(\log (x))\right )}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))}\right ) \, dx\\ &=-\left (2 \int \frac {x \left (-8+4 x-2 x^2+8 \log (x) \log (\log (x))-4 x \log (x) \log (\log (x))+2 x^2 \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))-x \log (x) \log ^2(\log (x))\right )}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))} \, dx\right )+\int \frac {-4 x+4 x \log (x) \log (\log (x))-\log (x) \log ^2(\log (x))}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (x) \log ^2(\log (x))} \, dx\\ &=-\left (2 \int \frac {x \left (-2 \left (4-2 x+x^2\right )+\log (x) \log (\log (x)) \left (8-4 x+2 x^2+\log (\log (x))-x \log (\log (x))\right )\right )}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))} \, dx\right )+\int \left (-\frac {1}{4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2}-\frac {4 x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (x) \log ^2(\log (x))}+\frac {4 x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (\log (x))}\right ) \, dx\\ &=-\left (2 \int \left (\frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2}-\frac {x^2}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2}-\frac {8 x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))}+\frac {4 x^2}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))}-\frac {2 x^3}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))}+\frac {8 x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (\log (x))}-\frac {4 x^2}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (\log (x))}+\frac {2 x^3}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (\log (x))}\right ) \, dx\right )-4 \int \frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (x) \log ^2(\log (x))} \, dx+4 \int \frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (\log (x))} \, dx-\int \frac {1}{4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2} \, dx\\ &=-\left (2 \int \frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2} \, dx\right )+2 \int \frac {x^2}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2} \, dx+4 \int \frac {x^3}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))} \, dx-4 \int \frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (x) \log ^2(\log (x))} \, dx-4 \int \frac {x^3}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (\log (x))} \, dx+4 \int \frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right ) \log (\log (x))} \, dx-8 \int \frac {x^2}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))} \, dx+8 \int \frac {x^2}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (\log (x))} \, dx+16 \int \frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (x) \log ^2(\log (x))} \, dx-16 \int \frac {x}{\left (4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2\right )^2 \log (\log (x))} \, dx-\int \frac {1}{4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 23, normalized size = 0.92 \begin {gather*} -\frac {x}{4+e^{\frac {4 x}{\log (\log (x))}}-2 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-4 + x^2)*Log[x]*Log[Log[x]]^2 + E^((4*x)/Log[Log[x]])*(-4*x + 4*x*Log[x]*Log[Log[x]] - Log[x]*Log
[Log[x]]^2))/(E^((8*x)/Log[Log[x]])*Log[x]*Log[Log[x]]^2 + E^((4*x)/Log[Log[x]])*(8 - 4*x + 2*x^2)*Log[x]*Log[
Log[x]]^2 + (16 - 16*x + 12*x^2 - 4*x^3 + x^4)*Log[x]*Log[Log[x]]^2),x]

[Out]

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

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fricas [A]  time = 0.93, size = 22, normalized size = 0.88 \begin {gather*} -\frac {x}{x^{2} - 2 \, x + e^{\left (\frac {4 \, x}{\log \left (\log \relax (x)\right )}\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(x)*log(log(x))^2+4*x*log(x)*log(log(x))-4*x)*exp(2*x/log(log(x)))^2+(x^2-4)*log(x)*log(log(x)
)^2)/(log(x)*log(log(x))^2*exp(2*x/log(log(x)))^4+(2*x^2-4*x+8)*log(x)*log(log(x))^2*exp(2*x/log(log(x)))^2+(x
^4-4*x^3+12*x^2-16*x+16)*log(x)*log(log(x))^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(x)*log(log(x))^2+4*x*log(x)*log(log(x))-4*x)*exp(2*x/log(log(x)))^2+(x^2-4)*log(x)*log(log(x)
)^2)/(log(x)*log(log(x))^2*exp(2*x/log(log(x)))^4+(2*x^2-4*x+8)*log(x)*log(log(x))^2*exp(2*x/log(log(x)))^2+(x
^4-4*x^3+12*x^2-16*x+16)*log(x)*log(log(x))^2),x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.06, size = 23, normalized size = 0.92




method result size



risch \(-\frac {x}{{\mathrm e}^{\frac {4 x}{\ln \left (\ln \relax (x )\right )}}+x^{2}-2 x +4}\) \(23\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(x)*ln(ln(x))^2+4*x*ln(x)*ln(ln(x))-4*x)*exp(2*x/ln(ln(x)))^2+(x^2-4)*ln(x)*ln(ln(x))^2)/(ln(x)*ln(ln
(x))^2*exp(2*x/ln(ln(x)))^4+(2*x^2-4*x+8)*ln(x)*ln(ln(x))^2*exp(2*x/ln(ln(x)))^2+(x^4-4*x^3+12*x^2-16*x+16)*ln
(x)*ln(ln(x))^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.43, size = 22, normalized size = 0.88 \begin {gather*} -\frac {x}{x^{2} - 2 \, x + e^{\left (\frac {4 \, x}{\log \left (\log \relax (x)\right )}\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(x)*log(log(x))^2+4*x*log(x)*log(log(x))-4*x)*exp(2*x/log(log(x)))^2+(x^2-4)*log(x)*log(log(x)
)^2)/(log(x)*log(log(x))^2*exp(2*x/log(log(x)))^4+(2*x^2-4*x+8)*log(x)*log(log(x))^2*exp(2*x/log(log(x)))^2+(x
^4-4*x^3+12*x^2-16*x+16)*log(x)*log(log(x))^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.61, size = 209, normalized size = 8.36 \begin {gather*} -\frac {x\,\left ({\ln \left (\ln \relax (x)\right )}^4\,{\ln \relax (x)}^2+8\,{\ln \left (\ln \relax (x)\right )}^3\,{\ln \relax (x)}^2-8\,{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)\right )-x^2\,\left ({\ln \left (\ln \relax (x)\right )}^4\,{\ln \relax (x)}^2+4\,{\ln \left (\ln \relax (x)\right )}^3\,{\ln \relax (x)}^2-4\,{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)\right )+x^3\,\left (2\,{\ln \left (\ln \relax (x)\right )}^3\,{\ln \relax (x)}^2-2\,{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)\right )}{\left ({\mathrm {e}}^{\frac {4\,x}{\ln \left (\ln \relax (x)\right )}}-2\,x+x^2+4\right )\,\left (2\,x^2\,{\ln \left (\ln \relax (x)\right )}^3\,{\ln \relax (x)}^2-2\,x^2\,{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)-x\,{\ln \left (\ln \relax (x)\right )}^4\,{\ln \relax (x)}^2-4\,x\,{\ln \left (\ln \relax (x)\right )}^3\,{\ln \relax (x)}^2+4\,x\,{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)+{\ln \left (\ln \relax (x)\right )}^4\,{\ln \relax (x)}^2+8\,{\ln \left (\ln \relax (x)\right )}^3\,{\ln \relax (x)}^2-8\,{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((4*x)/log(log(x)))*(4*x + log(log(x))^2*log(x) - 4*x*log(log(x))*log(x)) - log(log(x))^2*log(x)*(x^2
 - 4))/(log(log(x))^2*log(x)*(12*x^2 - 16*x - 4*x^3 + x^4 + 16) + log(log(x))^2*exp((8*x)/log(log(x)))*log(x)
+ log(log(x))^2*exp((4*x)/log(log(x)))*log(x)*(2*x^2 - 4*x + 8)),x)

[Out]

-(x*(8*log(log(x))^3*log(x)^2 + log(log(x))^4*log(x)^2 - 8*log(log(x))^2*log(x)) - x^2*(4*log(log(x))^3*log(x)
^2 + log(log(x))^4*log(x)^2 - 4*log(log(x))^2*log(x)) + x^3*(2*log(log(x))^3*log(x)^2 - 2*log(log(x))^2*log(x)
))/((exp((4*x)/log(log(x))) - 2*x + x^2 + 4)*(8*log(log(x))^3*log(x)^2 + log(log(x))^4*log(x)^2 - 8*log(log(x)
)^2*log(x) + 2*x^2*log(log(x))^3*log(x)^2 + 4*x*log(log(x))^2*log(x) - 2*x^2*log(log(x))^2*log(x) - 4*x*log(lo
g(x))^3*log(x)^2 - x*log(log(x))^4*log(x)^2))

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sympy [A]  time = 0.48, size = 20, normalized size = 0.80 \begin {gather*} - \frac {x}{x^{2} - 2 x + e^{\frac {4 x}{\log {\left (\log {\relax (x )} \right )}}} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-ln(x)*ln(ln(x))**2+4*x*ln(x)*ln(ln(x))-4*x)*exp(2*x/ln(ln(x)))**2+(x**2-4)*ln(x)*ln(ln(x))**2)/(l
n(x)*ln(ln(x))**2*exp(2*x/ln(ln(x)))**4+(2*x**2-4*x+8)*ln(x)*ln(ln(x))**2*exp(2*x/ln(ln(x)))**2+(x**4-4*x**3+1
2*x**2-16*x+16)*ln(x)*ln(ln(x))**2),x)

[Out]

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

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