Integrand size = 130, antiderivative size = 27 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=\log \left (\log ^2\left (e^{\frac {1}{\log \left (x+\frac {1}{4} (1-2 x)^2 x^2\right )}} x\right )\right ) \] Output:
ln(ln(exp(1/ln(x+1/2*(1-2*x)*(-x+1/2)*x^2))*x)^2)
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=2 \log \left (\log \left (e^{\frac {1}{\log \left (x+\frac {x^2}{4}-x^3+x^4\right )}} x\right )\right ) \] Input:
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]
Output:
2*Log[Log[E^Log[x + x^2/4 - x^3 + x^4]^(-1)*x]]
Time = 0.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2026, 7235}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-32 x^3+24 x^2+\left (8 x^3-8 x^2+2 x+8\right ) \log ^2\left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )-4 x-8}{\left (4 x^4-4 x^3+x^2+4 x\right ) \log \left (x e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )}}\right ) \log ^2\left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-32 x^3+24 x^2+\left (8 x^3-8 x^2+2 x+8\right ) \log ^2\left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )-4 x-8}{x \left (4 x^3-4 x^2+x+4\right ) \log \left (x e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )}}\right ) \log ^2\left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle 2 \log \left (\log \left (x e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x^4-4 x^3+x^2+4 x\right )\right )}}\right )\right )\) |
Input:
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]
Output:
2*Log[Log[E^Log[(4*x + x^2 - 4*x^3 + 4*x^4)/4]^(-1)*x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 55.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(2 \ln \left (\ln \left (x \,{\mathrm e}^{\frac {1}{\ln \left (x^{4}-x^{3}+\frac {1}{4} x^{2}+x \right )}}\right )\right )\) | \(26\) |
Input:
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,method=_RETURNVERBOSE)
Output:
2*ln(ln(x*exp(1/ln(x^4-x^3+1/4*x^2+x))))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=2 \, \log \left (\log \left (x e^{\left (\frac {1}{\log \left (x^{4} - x^{3} + \frac {1}{4} \, x^{2} + x\right )}\right )}\right )\right ) \] Input:
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")
Output:
2*log(log(x*e^(1/log(x^4 - x^3 + 1/4*x^2 + x))))
Time = 0.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=2 \log {\left (\log {\left (x e^{\frac {1}{\log {\left (x^{4} - x^{3} + \frac {x^{2}}{4} + x \right )}}} \right )} \right )} \] Input:
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)/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)
Output:
2*log(log(x*exp(1/log(x**4 - x**3 + x**2/4 + x))))
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=2 \, \log \left (\log \left (x\right ) + \log \left (e^{\left (-\frac {1}{2 \, \log \left (2\right ) - \log \left (4 \, x^{3} - 4 \, x^{2} + x + 4\right ) - \log \left (x\right )}\right )}\right )\right ) \] Input:
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")
Output:
2*log(log(x) + log(e^(-1/(2*log(2) - log(4*x^3 - 4*x^2 + x + 4) - log(x))) ))
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (24) = 48\).
Time = 2.87 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.37 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=2 \, \log \left (2 \, \log \left (2\right ) \log \left (x\right ) - \log \left (4 \, x^{3} - 4 \, x^{2} + x + 4\right ) \log \left (x\right ) - \log \left (x\right )^{2} - 1\right ) - 2 \, \log \left (2 \, \log \left (2\right ) - \log \left (4 \, x^{3} - 4 \, x^{2} + x + 4\right ) - \log \left (x\right )\right ) \] Input:
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")
Output:
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))
Time = 4.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=2\,\ln \left (\ln \left (x\,{\mathrm {e}}^{\frac {1}{\ln \left (x^4-x^3+\frac {x^2}{4}+x\right )}}\right )\right ) \] Input:
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)
Output:
2*log(log(x*exp(1/log(x + x^2/4 - x^3 + x^4))))
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-8-4 x+24 x^2-32 x^3+\left (8+2 x-8 x^2+8 x^3\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}{\left (4 x+x^2-4 x^3+4 x^4\right ) \log \left (e^{\frac {1}{\log \left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )}} x\right ) \log ^2\left (\frac {1}{4} \left (4 x+x^2-4 x^3+4 x^4\right )\right )} \, dx=2 \,\mathrm {log}\left (\mathrm {log}\left (e^{\frac {1}{\mathrm {log}\left (x^{4}-x^{3}+\frac {1}{4} x^{2}+x \right )}} x \right )\right ) \] Input:
int(((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)
Output:
2*log(log(e**(1/log((4*x**4 - 4*x**3 + x**2 + 4*x)/4))*x))