3.24.68 \(\int \frac {-4 x^2+2 x^5-4 x^3 \log (2)+2 x \log ^2(2)+(-4 x^2-4 x^5+(4+8 x^3) \log (2)-4 x \log ^2(2)) \log (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}) \log (\log (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}))}{(x^3+x^6+(-x-2 x^4) \log (2)+x^2 \log ^2(2)) \log (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}) \log (\log (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}))} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\frac {48 \log ^2\left (\log \left (x+\frac {1}{x^2-\log (2)}\right )\right )}{x^4}\right ) \]

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Rubi [F]  time = 3.10, 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 x^2+2 x^5-4 x^3 \log (2)+2 x \log ^2(2)+\left (-4 x^2-4 x^5+\left (4+8 x^3\right ) \log (2)-4 x \log ^2(2)\right ) \log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right ) \log \left (\log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right )\right )}{\left (x^3+x^6+\left (-x-2 x^4\right ) \log (2)+x^2 \log ^2(2)\right ) \log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right ) \log \left (\log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x^2 + 2*x^5 - 4*x^3*Log[2] + 2*x*Log[2]^2 + (-4*x^2 - 4*x^5 + (4 + 8*x^3)*Log[2] - 4*x*Log[2]^2)*Log[(
-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]*Log[Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]])/((x^3 + x^6 + (-x - 2*x
^4)*Log[2] + x^2*Log[2]^2)*Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]*Log[Log[(-1 - x^3 + x*Log[2])/(-x^2 + Lo
g[2])]]),x]

[Out]

-4*Log[x] + 2*Defer[Int][1/((-x + Sqrt[Log[2]])*Log[(1 + x^3 - x*Log[2])/(x^2 - Log[2])]*Log[Log[(1 + x^3 - x*
Log[2])/(x^2 - Log[2])]]), x] - 2*Defer[Int][1/((x + Sqrt[Log[2]])*Log[(1 + x^3 - x*Log[2])/(x^2 - Log[2])]*Lo
g[Log[(1 + x^3 - x*Log[2])/(x^2 - Log[2])]]), x] + 6*Defer[Int][x^2/((1 + x^3 - x*Log[2])*Log[(1 + x^3 - x*Log
[2])/(x^2 - Log[2])]*Log[Log[(1 + x^3 - x*Log[2])/(x^2 - Log[2])]]), x] + 2*Log[2]*Defer[Int][1/((-1 - x^3 + x
*Log[2])*Log[(1 + x^3 - x*Log[2])/(x^2 - Log[2])]*Log[Log[(1 + x^3 - x*Log[2])/(x^2 - Log[2])]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {4}{x}+\frac {2 \left (-2 x+x^4+\log ^2(2)-x^2 \log (4)\right )}{\left (x^2-\log (2)\right ) \left (1+x^3-x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )}\right ) \, dx\\ &=-4 \log (x)+2 \int \frac {-2 x+x^4+\log ^2(2)-x^2 \log (4)}{\left (x^2-\log (2)\right ) \left (1+x^3-x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )} \, dx\\ &=-4 \log (x)+2 \int \left (-\frac {2 x}{\left (x^2-\log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )}+\frac {3 x^2-\log (2)}{\left (1+x^3-x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )}\right ) \, dx\\ &=-4 \log (x)+2 \int \frac {3 x^2-\log (2)}{\left (1+x^3-x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )} \, dx-4 \int \frac {x}{\left (x^2-\log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )} \, dx\\ &=-4 \log (x)+2 \int \left (\frac {3 x^2}{\left (1+x^3-x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )}+\frac {\log (2)}{\left (-1-x^3+x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )}\right ) \, dx-4 \int \left (-\frac {1}{2 \left (-x+\sqrt {\log (2)}\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )}+\frac {1}{2 \left (x+\sqrt {\log (2)}\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )}\right ) \, dx\\ &=-4 \log (x)+2 \int \frac {1}{\left (-x+\sqrt {\log (2)}\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )} \, dx-2 \int \frac {1}{\left (x+\sqrt {\log (2)}\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )} \, dx+6 \int \frac {x^2}{\left (1+x^3-x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )} \, dx+(2 \log (2)) \int \frac {1}{\left (-1-x^3+x \log (2)\right ) \log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right ) \log \left (\log \left (\frac {1+x^3-x \log (2)}{x^2-\log (2)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.20, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-4 x^2+2 x^5-4 x^3 \log (2)+2 x \log ^2(2)+\left (-4 x^2-4 x^5+\left (4+8 x^3\right ) \log (2)-4 x \log ^2(2)\right ) \log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right ) \log \left (\log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right )\right )}{\left (x^3+x^6+\left (-x-2 x^4\right ) \log (2)+x^2 \log ^2(2)\right ) \log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right ) \log \left (\log \left (\frac {-1-x^3+x \log (2)}{-x^2+\log (2)}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-4*x^2 + 2*x^5 - 4*x^3*Log[2] + 2*x*Log[2]^2 + (-4*x^2 - 4*x^5 + (4 + 8*x^3)*Log[2] - 4*x*Log[2]^2)
*Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]*Log[Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]])/((x^3 + x^6 + (-x
- 2*x^4)*Log[2] + x^2*Log[2]^2)*Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]*Log[Log[(-1 - x^3 + x*Log[2])/(-x^
2 + Log[2])]]),x]

[Out]

Integrate[(-4*x^2 + 2*x^5 - 4*x^3*Log[2] + 2*x*Log[2]^2 + (-4*x^2 - 4*x^5 + (4 + 8*x^3)*Log[2] - 4*x*Log[2]^2)
*Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]*Log[Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]])/((x^3 + x^6 + (-x
- 2*x^4)*Log[2] + x^2*Log[2]^2)*Log[(-1 - x^3 + x*Log[2])/(-x^2 + Log[2])]*Log[Log[(-1 - x^3 + x*Log[2])/(-x^
2 + Log[2])]]), x]

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fricas [A]  time = 0.51, size = 31, normalized size = 1.41 \begin {gather*} -4 \, \log \relax (x) + 2 \, \log \left (\log \left (\log \left (\frac {x^{3} - x \log \relax (2) + 1}{x^{2} - \log \relax (2)}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2)^2+(8*x^3+4)*log(2)-4*x^5-4*x^2)*log((x*log(2)-x^3-1)/(log(2)-x^2))*log(log((x*log(2)-x
^3-1)/(log(2)-x^2)))+2*x*log(2)^2-4*x^3*log(2)+2*x^5-4*x^2)/(x^2*log(2)^2+(-2*x^4-x)*log(2)+x^6+x^3)/log((x*lo
g(2)-x^3-1)/(log(2)-x^2))/log(log((x*log(2)-x^3-1)/(log(2)-x^2))),x, algorithm="fricas")

[Out]

-4*log(x) + 2*log(log(log((x^3 - x*log(2) + 1)/(x^2 - log(2)))))

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giac [C]  time = 0.91, size = 35, normalized size = 1.59 \begin {gather*} -4 \, \log \relax (x) + 2 \, \log \left (\log \left (2 i \, \pi + \log \left (x^{3} - x \log \relax (2) + 1\right ) - \log \left (x^{2} - \log \relax (2)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2)^2+(8*x^3+4)*log(2)-4*x^5-4*x^2)*log((x*log(2)-x^3-1)/(log(2)-x^2))*log(log((x*log(2)-x
^3-1)/(log(2)-x^2)))+2*x*log(2)^2-4*x^3*log(2)+2*x^5-4*x^2)/(x^2*log(2)^2+(-2*x^4-x)*log(2)+x^6+x^3)/log((x*lo
g(2)-x^3-1)/(log(2)-x^2))/log(log((x*log(2)-x^3-1)/(log(2)-x^2))),x, algorithm="giac")

[Out]

-4*log(x) + 2*log(log(2*I*pi + log(x^3 - x*log(2) + 1) - log(x^2 - log(2))))

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-4 x \ln \relax (2)^{2}+\left (8 x^{3}+4\right ) \ln \relax (2)-4 x^{5}-4 x^{2}\right ) \ln \left (\frac {x \ln \relax (2)-x^{3}-1}{\ln \relax (2)-x^{2}}\right ) \ln \left (\ln \left (\frac {x \ln \relax (2)-x^{3}-1}{\ln \relax (2)-x^{2}}\right )\right )+2 x \ln \relax (2)^{2}-4 x^{3} \ln \relax (2)+2 x^{5}-4 x^{2}}{\left (x^{2} \ln \relax (2)^{2}+\left (-2 x^{4}-x \right ) \ln \relax (2)+x^{6}+x^{3}\right ) \ln \left (\frac {x \ln \relax (2)-x^{3}-1}{\ln \relax (2)-x^{2}}\right ) \ln \left (\ln \left (\frac {x \ln \relax (2)-x^{3}-1}{\ln \relax (2)-x^{2}}\right )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*ln(2)^2+(8*x^3+4)*ln(2)-4*x^5-4*x^2)*ln((x*ln(2)-x^3-1)/(ln(2)-x^2))*ln(ln((x*ln(2)-x^3-1)/(ln(2)-x
^2)))+2*x*ln(2)^2-4*x^3*ln(2)+2*x^5-4*x^2)/(x^2*ln(2)^2+(-2*x^4-x)*ln(2)+x^6+x^3)/ln((x*ln(2)-x^3-1)/(ln(2)-x^
2))/ln(ln((x*ln(2)-x^3-1)/(ln(2)-x^2))),x)

[Out]

int(((-4*x*ln(2)^2+(8*x^3+4)*ln(2)-4*x^5-4*x^2)*ln((x*ln(2)-x^3-1)/(ln(2)-x^2))*ln(ln((x*ln(2)-x^3-1)/(ln(2)-x
^2)))+2*x*ln(2)^2-4*x^3*ln(2)+2*x^5-4*x^2)/(x^2*ln(2)^2+(-2*x^4-x)*ln(2)+x^6+x^3)/ln((x*ln(2)-x^3-1)/(ln(2)-x^
2))/ln(ln((x*ln(2)-x^3-1)/(ln(2)-x^2))),x)

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maxima [A]  time = 0.67, size = 32, normalized size = 1.45 \begin {gather*} -4 \, \log \relax (x) + 2 \, \log \left (\log \left (\log \left (x^{3} - x \log \relax (2) + 1\right ) - \log \left (x^{2} - \log \relax (2)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2)^2+(8*x^3+4)*log(2)-4*x^5-4*x^2)*log((x*log(2)-x^3-1)/(log(2)-x^2))*log(log((x*log(2)-x
^3-1)/(log(2)-x^2)))+2*x*log(2)^2-4*x^3*log(2)+2*x^5-4*x^2)/(x^2*log(2)^2+(-2*x^4-x)*log(2)+x^6+x^3)/log((x*lo
g(2)-x^3-1)/(log(2)-x^2))/log(log((x*log(2)-x^3-1)/(log(2)-x^2))),x, algorithm="maxima")

[Out]

-4*log(x) + 2*log(log(log(x^3 - x*log(2) + 1) - log(x^2 - log(2))))

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mupad [B]  time = 3.99, size = 32, normalized size = 1.45 \begin {gather*} 2\,\ln \left (\ln \left (\ln \left (-\frac {x^3-\ln \relax (2)\,x+1}{\ln \relax (2)-x^2}\right )\right )\right )-4\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x^3*log(2) - 2*x*log(2)^2 + 4*x^2 - 2*x^5 + log(log(-(x^3 - x*log(2) + 1)/(log(2) - x^2)))*log(-(x^3 -
x*log(2) + 1)/(log(2) - x^2))*(4*x*log(2)^2 - log(2)*(8*x^3 + 4) + 4*x^2 + 4*x^5))/(log(log(-(x^3 - x*log(2)
+ 1)/(log(2) - x^2)))*log(-(x^3 - x*log(2) + 1)/(log(2) - x^2))*(x^2*log(2)^2 + x^3 + x^6 - log(2)*(x + 2*x^4)
)),x)

[Out]

2*log(log(log(-(x^3 - x*log(2) + 1)/(log(2) - x^2)))) - 4*log(x)

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sympy [A]  time = 1.12, size = 27, normalized size = 1.23 \begin {gather*} - 4 \log {\relax (x )} + 2 \log {\left (\log {\left (\log {\left (\frac {- x^{3} + x \log {\relax (2 )} - 1}{- x^{2} + \log {\relax (2 )}} \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*ln(2)**2+(8*x**3+4)*ln(2)-4*x**5-4*x**2)*ln((x*ln(2)-x**3-1)/(ln(2)-x**2))*ln(ln((x*ln(2)-x**
3-1)/(ln(2)-x**2)))+2*x*ln(2)**2-4*x**3*ln(2)+2*x**5-4*x**2)/(x**2*ln(2)**2+(-2*x**4-x)*ln(2)+x**6+x**3)/ln((x
*ln(2)-x**3-1)/(ln(2)-x**2))/ln(ln((x*ln(2)-x**3-1)/(ln(2)-x**2))),x)

[Out]

-4*log(x) + 2*log(log(log((-x**3 + x*log(2) - 1)/(-x**2 + log(2)))))

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