3.12.23 \(\int \frac {4+(-4 x+2 e^5 x^3-6 e^8 x^5) \log (x)+(-2 e^5 x^2+6 e^8 x^4) \log (x) \log (\log (x))+(2 x \log (x)-2 \log (x) \log (\log (x))) \log (x^2-2 x \log (\log (x))+\log ^2(\log (x)))}{(-e^{10} x^5+2 e^{13} x^7-e^{16} x^9) \log (x)+(e^{10} x^4-2 e^{13} x^6+e^{16} x^8) \log (x) \log (\log (x))+((2 e^5 x^3-2 e^8 x^5) \log (x)+(-2 e^5 x^2+2 e^8 x^4) \log (x) \log (\log (x))) \log (x^2-2 x \log (\log (x))+\log ^2(\log (x)))+(-x \log (x)+\log (x) \log (\log (x))) \log ^2(x^2-2 x \log (\log (x))+\log ^2(\log (x)))} \, dx\)

Optimal. Leaf size=34 \[ \frac {2 x}{x^2 \left (e^5-e^8 x^2\right )-\log \left ((x-\log (\log (x)))^2\right )} \]

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Rubi [F]  time = 3.42, 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 {4+\left (-4 x+2 e^5 x^3-6 e^8 x^5\right ) \log (x)+\left (-2 e^5 x^2+6 e^8 x^4\right ) \log (x) \log (\log (x))+(2 x \log (x)-2 \log (x) \log (\log (x))) \log \left (x^2-2 x \log (\log (x))+\log ^2(\log (x))\right )}{\left (-e^{10} x^5+2 e^{13} x^7-e^{16} x^9\right ) \log (x)+\left (e^{10} x^4-2 e^{13} x^6+e^{16} x^8\right ) \log (x) \log (\log (x))+\left (\left (2 e^5 x^3-2 e^8 x^5\right ) \log (x)+\left (-2 e^5 x^2+2 e^8 x^4\right ) \log (x) \log (\log (x))\right ) \log \left (x^2-2 x \log (\log (x))+\log ^2(\log (x))\right )+(-x \log (x)+\log (x) \log (\log (x))) \log ^2\left (x^2-2 x \log (\log (x))+\log ^2(\log (x))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4 + (-4*x + 2*E^5*x^3 - 6*E^8*x^5)*Log[x] + (-2*E^5*x^2 + 6*E^8*x^4)*Log[x]*Log[Log[x]] + (2*x*Log[x] - 2
*Log[x]*Log[Log[x]])*Log[x^2 - 2*x*Log[Log[x]] + Log[Log[x]]^2])/((-(E^10*x^5) + 2*E^13*x^7 - E^16*x^9)*Log[x]
+ (E^10*x^4 - 2*E^13*x^6 + E^16*x^8)*Log[x]*Log[Log[x]] + ((2*E^5*x^3 - 2*E^8*x^5)*Log[x] + (-2*E^5*x^2 + 2*E
^8*x^4)*Log[x]*Log[Log[x]])*Log[x^2 - 2*x*Log[Log[x]] + Log[Log[x]]^2] + (-(x*Log[x]) + Log[x]*Log[Log[x]])*Lo
g[x^2 - 2*x*Log[Log[x]] + Log[Log[x]]^2]^2),x]

[Out]

4*Defer[Int][x/((x - Log[Log[x]])*(-(E^5*x^2) + E^8*x^4 + Log[(x - Log[Log[x]])^2])^2), x] - 4*E^5*Defer[Int][
x^3/((x - Log[Log[x]])*(-(E^5*x^2) + E^8*x^4 + Log[(x - Log[Log[x]])^2])^2), x] + 8*E^8*Defer[Int][x^5/((x - L
og[Log[x]])*(-(E^5*x^2) + E^8*x^4 + Log[(x - Log[Log[x]])^2])^2), x] - 4*Defer[Int][1/(Log[x]*(x - Log[Log[x]]
)*(-(E^5*x^2) + E^8*x^4 + Log[(x - Log[Log[x]])^2])^2), x] + 4*E^5*Defer[Int][(x^2*Log[Log[x]])/((x - Log[Log[
x]])*(-(E^5*x^2) + E^8*x^4 + Log[(x - Log[Log[x]])^2])^2), x] - 8*E^8*Defer[Int][(x^4*Log[Log[x]])/((x - Log[L
og[x]])*(-(E^5*x^2) + E^8*x^4 + Log[(x - Log[Log[x]])^2])^2), x] - 2*Defer[Int][(-(E^5*x^2) + E^8*x^4 + Log[(x
- Log[Log[x]])^2])^(-1), x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 31, normalized size = 0.91 \begin {gather*} -\frac {2 x}{-e^5 x^2+e^8 x^4+\log \left ((x-\log (\log (x)))^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + (-4*x + 2*E^5*x^3 - 6*E^8*x^5)*Log[x] + (-2*E^5*x^2 + 6*E^8*x^4)*Log[x]*Log[Log[x]] + (2*x*Log[
x] - 2*Log[x]*Log[Log[x]])*Log[x^2 - 2*x*Log[Log[x]] + Log[Log[x]]^2])/((-(E^10*x^5) + 2*E^13*x^7 - E^16*x^9)*
Log[x] + (E^10*x^4 - 2*E^13*x^6 + E^16*x^8)*Log[x]*Log[Log[x]] + ((2*E^5*x^3 - 2*E^8*x^5)*Log[x] + (-2*E^5*x^2
+ 2*E^8*x^4)*Log[x]*Log[Log[x]])*Log[x^2 - 2*x*Log[Log[x]] + Log[Log[x]]^2] + (-(x*Log[x]) + Log[x]*Log[Log[x
]])*Log[x^2 - 2*x*Log[Log[x]] + Log[Log[x]]^2]^2),x]

[Out]

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

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fricas [A]  time = 1.00, size = 35, normalized size = 1.03 \begin {gather*} -\frac {2 \, x}{x^{4} e^{8} - x^{2} e^{5} + \log \left (x^{2} - 2 \, x \log \left (\log \relax (x)\right ) + \log \left (\log \relax (x)\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)*log(log(x))+2*x*log(x))*log(log(log(x))^2-2*x*log(log(x))+x^2)+(-2*x^2*exp(5)+6*x^4*exp(
4)^2)*log(x)*log(log(x))+(2*x^3*exp(5)-6*x^5*exp(4)^2-4*x)*log(x)+4)/((log(x)*log(log(x))-x*log(x))*log(log(lo
g(x))^2-2*x*log(log(x))+x^2)^2+((-2*x^2*exp(5)+2*x^4*exp(4)^2)*log(x)*log(log(x))+(2*x^3*exp(5)-2*x^5*exp(4)^2
)*log(x))*log(log(log(x))^2-2*x*log(log(x))+x^2)+(x^4*exp(5)^2-2*x^6*exp(4)^2*exp(5)+x^8*exp(4)^4)*log(x)*log(
log(x))+(-x^5*exp(5)^2+2*x^7*exp(4)^2*exp(5)-x^9*exp(4)^4)*log(x)),x, algorithm="fricas")

[Out]

-2*x/(x^4*e^8 - x^2*e^5 + log(x^2 - 2*x*log(log(x)) + log(log(x))^2))

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giac [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(((-2*log(x)*log(log(x))+2*x*log(x))*log(log(log(x))^2-2*x*log(log(x))+x^2)+(-2*x^2*exp(5)+6*x^4*exp(
4)^2)*log(x)*log(log(x))+(2*x^3*exp(5)-6*x^5*exp(4)^2-4*x)*log(x)+4)/((log(x)*log(log(x))-x*log(x))*log(log(lo
g(x))^2-2*x*log(log(x))+x^2)^2+((-2*x^2*exp(5)+2*x^4*exp(4)^2)*log(x)*log(log(x))+(2*x^3*exp(5)-2*x^5*exp(4)^2
)*log(x))*log(log(log(x))^2-2*x*log(log(x))+x^2)+(x^4*exp(5)^2-2*x^6*exp(4)^2*exp(5)+x^8*exp(4)^4)*log(x)*log(
log(x))+(-x^5*exp(5)^2+2*x^7*exp(4)^2*exp(5)-x^9*exp(4)^4)*log(x)),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.13, size = 110, normalized size = 3.24

method	result	size
risch	\(-\frac {4 x}{2 x^{4} {\mathrm e}^{8}-2 x^{2} {\mathrm e}^{5}-i \pi \mathrm {csgn}\left (i \left (x -\ln \left (\ln \relax (x )\right )\right )\right )^{2} \mathrm {csgn}\left (i \left (x -\ln \left (\ln \relax (x )\right )\right )^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i \left (x -\ln \left (\ln \relax (x )\right )\right )\right ) \mathrm {csgn}\left (i \left (x -\ln \left (\ln \relax (x )\right )\right )^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i \left (x -\ln \left (\ln \relax (x )\right )\right )^{2}\right )^{3}+4 \ln \left (x -\ln \left (\ln \relax (x )\right )\right )}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*ln(x)*ln(ln(x))+2*x*ln(x))*ln(ln(ln(x))^2-2*x*ln(ln(x))+x^2)+(-2*x^2*exp(5)+6*x^4*exp(4)^2)*ln(x)*ln(
ln(x))+(2*x^3*exp(5)-6*x^5*exp(4)^2-4*x)*ln(x)+4)/((ln(x)*ln(ln(x))-x*ln(x))*ln(ln(ln(x))^2-2*x*ln(ln(x))+x^2)
^2+((-2*x^2*exp(5)+2*x^4*exp(4)^2)*ln(x)*ln(ln(x))+(2*x^3*exp(5)-2*x^5*exp(4)^2)*ln(x))*ln(ln(ln(x))^2-2*x*ln(
ln(x))+x^2)+(x^4*exp(5)^2-2*x^6*exp(4)^2*exp(5)+x^8*exp(4)^4)*ln(x)*ln(ln(x))+(-x^5*exp(5)^2+2*x^7*exp(4)^2*ex
p(5)-x^9*exp(4)^4)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-4*x/(2*x^4*exp(8)-2*x^2*exp(5)-I*Pi*csgn(I*(x-ln(ln(x))))^2*csgn(I*(x-ln(ln(x)))^2)+2*I*Pi*csgn(I*(x-ln(ln(x)
)))*csgn(I*(x-ln(ln(x)))^2)^2-I*Pi*csgn(I*(x-ln(ln(x)))^2)^3+4*ln(x-ln(ln(x))))

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maxima [A]  time = 0.61, size = 29, normalized size = 0.85 \begin {gather*} -\frac {2 \, x}{x^{4} e^{8} - x^{2} e^{5} + 2 \, \log \left (-x + \log \left (\log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)*log(log(x))+2*x*log(x))*log(log(log(x))^2-2*x*log(log(x))+x^2)+(-2*x^2*exp(5)+6*x^4*exp(
4)^2)*log(x)*log(log(x))+(2*x^3*exp(5)-6*x^5*exp(4)^2-4*x)*log(x)+4)/((log(x)*log(log(x))-x*log(x))*log(log(lo
g(x))^2-2*x*log(log(x))+x^2)^2+((-2*x^2*exp(5)+2*x^4*exp(4)^2)*log(x)*log(log(x))+(2*x^3*exp(5)-2*x^5*exp(4)^2
)*log(x))*log(log(log(x))^2-2*x*log(log(x))+x^2)+(x^4*exp(5)^2-2*x^6*exp(4)^2*exp(5)+x^8*exp(4)^4)*log(x)*log(
log(x))+(-x^5*exp(5)^2+2*x^7*exp(4)^2*exp(5)-x^9*exp(4)^4)*log(x)),x, algorithm="maxima")

[Out]

-2*x/(x^4*e^8 - x^2*e^5 + 2*log(-x + log(log(x))))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \left (x^2-2\,x\,\ln \left (\ln \relax (x)\right )+{\ln \left (\ln \relax (x)\right )}^2\right )\,\left (2\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)-2\,x\,\ln \relax (x)\right )+\ln \relax (x)\,\left (6\,{\mathrm {e}}^8\,x^5-2\,{\mathrm {e}}^5\,x^3+4\,x\right )+\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (2\,x^2\,{\mathrm {e}}^5-6\,x^4\,{\mathrm {e}}^8\right )-4}{\left (\ln \left (\ln \relax (x)\right )\,\ln \relax (x)-x\,\ln \relax (x)\right )\,{\ln \left (x^2-2\,x\,\ln \left (\ln \relax (x)\right )+{\ln \left (\ln \relax (x)\right )}^2\right )}^2+\left (\ln \relax (x)\,\left (2\,x^3\,{\mathrm {e}}^5-2\,x^5\,{\mathrm {e}}^8\right )-\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (2\,x^2\,{\mathrm {e}}^5-2\,x^4\,{\mathrm {e}}^8\right )\right )\,\ln \left (x^2-2\,x\,\ln \left (\ln \relax (x)\right )+{\ln \left (\ln \relax (x)\right )}^2\right )-\ln \relax (x)\,\left ({\mathrm {e}}^{16}\,x^9-2\,{\mathrm {e}}^{13}\,x^7+{\mathrm {e}}^{10}\,x^5\right )+\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left ({\mathrm {e}}^{16}\,x^8-2\,{\mathrm {e}}^{13}\,x^6+{\mathrm {e}}^{10}\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(log(x))^2 - 2*x*log(log(x)) + x^2)*(2*log(log(x))*log(x) - 2*x*log(x)) + log(x)*(4*x - 2*x^3*exp
(5) + 6*x^5*exp(8)) + log(log(x))*log(x)*(2*x^2*exp(5) - 6*x^4*exp(8)) - 4)/(log(log(log(x))^2 - 2*x*log(log(x
)) + x^2)*(log(x)*(2*x^3*exp(5) - 2*x^5*exp(8)) - log(log(x))*log(x)*(2*x^2*exp(5) - 2*x^4*exp(8))) - log(x)*(
x^5*exp(10) - 2*x^7*exp(13) + x^9*exp(16)) + log(log(log(x))^2 - 2*x*log(log(x)) + x^2)^2*(log(log(x))*log(x)
- x*log(x)) + log(log(x))*log(x)*(x^4*exp(10) - 2*x^6*exp(13) + x^8*exp(16))),x)

[Out]

int(-(log(log(log(x))^2 - 2*x*log(log(x)) + x^2)*(2*log(log(x))*log(x) - 2*x*log(x)) + log(x)*(4*x - 2*x^3*exp
(5) + 6*x^5*exp(8)) + log(log(x))*log(x)*(2*x^2*exp(5) - 6*x^4*exp(8)) - 4)/(log(log(log(x))^2 - 2*x*log(log(x
)) + x^2)*(log(x)*(2*x^3*exp(5) - 2*x^5*exp(8)) - log(log(x))*log(x)*(2*x^2*exp(5) - 2*x^4*exp(8))) - log(x)*(
x^5*exp(10) - 2*x^7*exp(13) + x^9*exp(16)) + log(log(log(x))^2 - 2*x*log(log(x)) + x^2)^2*(log(log(x))*log(x)
- x*log(x)) + log(log(x))*log(x)*(x^4*exp(10) - 2*x^6*exp(13) + x^8*exp(16))), x)

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sympy [A]  time = 0.63, size = 37, normalized size = 1.09 \begin {gather*} - \frac {2 x}{x^{4} e^{8} - x^{2} e^{5} + \log {\left (x^{2} - 2 x \log {\left (\log {\relax (x )} \right )} + \log {\left (\log {\relax (x )} \right )}^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*ln(x)*ln(ln(x))+2*x*ln(x))*ln(ln(ln(x))**2-2*x*ln(ln(x))+x**2)+(-2*x**2*exp(5)+6*x**4*exp(4)**2
)*ln(x)*ln(ln(x))+(2*x**3*exp(5)-6*x**5*exp(4)**2-4*x)*ln(x)+4)/((ln(x)*ln(ln(x))-x*ln(x))*ln(ln(ln(x))**2-2*x
*ln(ln(x))+x**2)**2+((-2*x**2*exp(5)+2*x**4*exp(4)**2)*ln(x)*ln(ln(x))+(2*x**3*exp(5)-2*x**5*exp(4)**2)*ln(x))
*ln(ln(ln(x))**2-2*x*ln(ln(x))+x**2)+(x**4*exp(5)**2-2*x**6*exp(4)**2*exp(5)+x**8*exp(4)**4)*ln(x)*ln(ln(x))+(
-x**5*exp(5)**2+2*x**7*exp(4)**2*exp(5)-x**9*exp(4)**4)*ln(x)),x)

[Out]

-2*x/(x**4*exp(8) - x**2*exp(5) + log(x**2 - 2*x*log(log(x)) + log(log(x))**2))

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