3.58.88 \(\int \frac {-8-4 x+24 x^2-32 x^3+(8+2 x-8 x^2+8 x^3) \log ^2(\frac {1}{4} (4 x+x^2-4 x^3+4 x^4))}{(4 x+x^2-4 x^3+4 x^4) \log (e^{\frac {1}{\log (\frac {1}{4} (4 x+x^2-4 x^3+4 x^4))}} x) \log ^2(\frac {1}{4} (4 x+x^2-4 x^3+4 x^4))} \, dx\)

Optimal. Leaf size=27 \[ \log \left (\log ^2\left (e^{\frac {1}{\log \left (x+\frac {1}{4} (1-2 x)^2 x^2\right )}} x\right )\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 32, normalized size of antiderivative = 1.19, number of steps used = 1, number of rules used = 1, integrand size = 130, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {6684} \begin {gather*} 2 \log \left (\log \left (x e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 - 4*x + 24*x^2 - 32*x^3 + (8 + 2*x - 8*x^2 + 8*x^3)*Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^2)/((4*x + x^2
- 4*x^3 + 4*x^4)*Log[E^Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^(-1)*x]*Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^2),x]

[Out]

2*Log[Log[E^Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^(-1)*x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 28, normalized size = 1.04 \begin {gather*} 2 \log \left (\log \left (e^{\frac {1}{\log \left (x+\frac {x^2}{4}-x^3+x^4\right )}} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 4*x + 24*x^2 - 32*x^3 + (8 + 2*x - 8*x^2 + 8*x^3)*Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^2)/((4*x
+ x^2 - 4*x^3 + 4*x^4)*Log[E^Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^(-1)*x]*Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^2),
x]

[Out]

2*Log[Log[E^Log[x + x^2/4 - x^3 + x^4]^(-1)*x]]

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fricas [A]  time = 0.55, size = 25, normalized size = 0.93 \begin {gather*} 2 \, \log \left (\log \left (x e^{\left (\frac {1}{\log \left (x^{4} - x^{3} + \frac {1}{4} \, x^{2} + x\right )}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3-8*x^2+2*x+8)*log(x^4-x^3+1/4*x^2+x)^2-32*x^3+24*x^2-4*x-8)/(4*x^4-4*x^3+x^2+4*x)/log(x^4-x^3
+1/4*x^2+x)^2/log(x*exp(1/log(x^4-x^3+1/4*x^2+x))),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 23.29, size = 64, normalized size = 2.37 \begin {gather*} 2 \, \log \left (2 \, \log \relax (2) \log \relax (x) - \log \left (4 \, x^{3} - 4 \, x^{2} + x + 4\right ) \log \relax (x) - \log \relax (x)^{2} - 1\right ) - 2 \, \log \left (2 \, \log \relax (2) - \log \left (4 \, x^{3} - 4 \, x^{2} + x + 4\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3-8*x^2+2*x+8)*log(x^4-x^3+1/4*x^2+x)^2-32*x^3+24*x^2-4*x-8)/(4*x^4-4*x^3+x^2+4*x)/log(x^4-x^3
+1/4*x^2+x)^2/log(x*exp(1/log(x^4-x^3+1/4*x^2+x))),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (8 x^{3}-8 x^{2}+2 x +8\right ) \ln \left (x^{4}-x^{3}+\frac {1}{4} x^{2}+x \right )^{2}-32 x^{3}+24 x^{2}-4 x -8}{\left (4 x^{4}-4 x^{3}+x^{2}+4 x \right ) \ln \left (x^{4}-x^{3}+\frac {1}{4} x^{2}+x \right )^{2} \ln \left (x \,{\mathrm e}^{\frac {1}{\ln \left (x^{4}-x^{3}+\frac {1}{4} x^{2}+x \right )}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^3-8*x^2+2*x+8)*ln(x^4-x^3+1/4*x^2+x)^2-32*x^3+24*x^2-4*x-8)/(4*x^4-4*x^3+x^2+4*x)/ln(x^4-x^3+1/4*x^2
+x)^2/ln(x*exp(1/ln(x^4-x^3+1/4*x^2+x))),x)

[Out]

int(((8*x^3-8*x^2+2*x+8)*ln(x^4-x^3+1/4*x^2+x)^2-32*x^3+24*x^2-4*x-8)/(4*x^4-4*x^3+x^2+4*x)/ln(x^4-x^3+1/4*x^2
+x)^2/ln(x*exp(1/ln(x^4-x^3+1/4*x^2+x))),x)

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maxima [A]  time = 0.54, size = 37, normalized size = 1.37 \begin {gather*} 2 \, \log \left (\log \relax (x) + \log \left (e^{\left (-\frac {1}{2 \, \log \relax (2) - \log \left (4 \, x^{3} - 4 \, x^{2} + x + 4\right ) - \log \relax (x)}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3-8*x^2+2*x+8)*log(x^4-x^3+1/4*x^2+x)^2-32*x^3+24*x^2-4*x-8)/(4*x^4-4*x^3+x^2+4*x)/log(x^4-x^3
+1/4*x^2+x)^2/log(x*exp(1/log(x^4-x^3+1/4*x^2+x))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.87, size = 25, normalized size = 0.93 \begin {gather*} 2\,\ln \left (\ln \left (x\,{\mathrm {e}}^{\frac {1}{\ln \left (x^4-x^3+\frac {x^2}{4}+x\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - 24*x^2 + 32*x^3 - log(x + x^2/4 - x^3 + x^4)^2*(2*x - 8*x^2 + 8*x^3 + 8) + 8)/(log(x + x^2/4 - x^3
+ x^4)^2*log(x*exp(1/log(x + x^2/4 - x^3 + x^4)))*(4*x + x^2 - 4*x^3 + 4*x^4)),x)

[Out]

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

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sympy [A]  time = 1.62, size = 24, normalized size = 0.89 \begin {gather*} 2 \log {\left (\log {\left (x e^{\frac {1}{\log {\left (x^{4} - x^{3} + \frac {x^{2}}{4} + x \right )}}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**3-8*x**2+2*x+8)*ln(x**4-x**3+1/4*x**2+x)**2-32*x**3+24*x**2-4*x-8)/(4*x**4-4*x**3+x**2+4*x)/l
n(x**4-x**3+1/4*x**2+x)**2/ln(x*exp(1/ln(x**4-x**3+1/4*x**2+x))),x)

[Out]

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

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