Integrand size = 132, antiderivative size = 27 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right ) \] Output:
16*ln(2*x-ln(ln(5/(4*ln(2)+4*x^2+8*x+16))))
Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right ) \] Input:
Integrate[(-32 - 32*x + (-128 - 64*x - 32*x^2 - 32*Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])])/((-8*x - 4*x^2 - 2*x^3 - 2*x*Log[2])*Log[5/(16 + 8* x + 4*x^2 + 4*Log[2])] + (4 + 2*x + x^2 + Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])]*Log[Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])]]),x]
Output:
16*Log[2*x - Log[Log[5/(4*(4 + 2*x + x^2 + Log[2]))]]]
Time = 0.60 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {7292, 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.
\(\displaystyle \int \frac {\left (-32 x^2-64 x-128-32 \log (2)\right ) \log \left (\frac {5}{4 x^2+8 x+16+4 \log (2)}\right )-32 x-32}{\left (x^2+2 x+4+\log (2)\right ) \log \left (\log \left (\frac {5}{4 x^2+8 x+16+4 \log (2)}\right )\right ) \log \left (\frac {5}{4 x^2+8 x+16+4 \log (2)}\right )+\left (-2 x^3-4 x^2-8 x-2 x \log (2)\right ) \log \left (\frac {5}{4 x^2+8 x+16+4 \log (2)}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (-32 x^2-64 x-128-32 \log (2)\right ) \log \left (\frac {5}{4 x^2+8 x+16+4 \log (2)}\right )+32 x+32}{\left (x^2+2 x+4+\log (2)\right ) \log \left (\frac {5}{4 \left (x^2+2 x+4+\log (2)\right )}\right ) \left (2 x-\log \left (\log \left (\frac {5}{4 \left (x^2+2 x+4+\log (2)\right )}\right )\right )\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle 16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (x^2+2 x+4+\log (2)\right )}\right )\right )\right )\) |
Input:
Int[(-32 - 32*x + (-128 - 64*x - 32*x^2 - 32*Log[2])*Log[5/(16 + 8*x + 4*x ^2 + 4*Log[2])])/((-8*x - 4*x^2 - 2*x^3 - 2*x*Log[2])*Log[5/(16 + 8*x + 4* x^2 + 4*Log[2])] + (4 + 2*x + x^2 + Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*Lo g[2])]*Log[Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])]]),x]
Output:
16*Log[2*x - Log[Log[5/(4*(4 + 2*x + x^2 + Log[2]))]]]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 1.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(16 \ln \left (x -\frac {\ln \left (\ln \left (\frac {5}{4 \left (\ln \left (2\right )+x^{2}+2 x +4\right )}\right )\right )}{2}\right )\) | \(24\) |
Input:
int(((-32*ln(2)-32*x^2-64*x-128)*ln(5/(4*ln(2)+4*x^2+8*x+16))-32*x-32)/((l n(2)+x^2+2*x+4)*ln(5/(4*ln(2)+4*x^2+8*x+16))*ln(ln(5/(4*ln(2)+4*x^2+8*x+16 )))+(-2*x*ln(2)-2*x^3-4*x^2-8*x)*ln(5/(4*ln(2)+4*x^2+8*x+16))),x,method=_R ETURNVERBOSE)
Output:
16*ln(x-1/2*ln(ln(5/4/(ln(2)+x^2+2*x+4))))
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \, \log \left (-2 \, x + \log \left (\log \left (\frac {5}{4 \, {\left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )}}\right )\right )\right ) \] Input:
integrate(((-32*log(2)-32*x^2-64*x-128)*log(5/(4*log(2)+4*x^2+8*x+16))-32* x-32)/((log(2)+x^2+2*x+4)*log(5/(4*log(2)+4*x^2+8*x+16))*log(log(5/(4*log( 2)+4*x^2+8*x+16)))+(-2*x*log(2)-2*x^3-4*x^2-8*x)*log(5/(4*log(2)+4*x^2+8*x +16))),x, algorithm="fricas")
Output:
16*log(-2*x + log(log(5/4/(x^2 + 2*x + log(2) + 4))))
Time = 0.38 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \log {\left (- 2 x + \log {\left (\log {\left (\frac {5}{4 x^{2} + 8 x + 4 \log {\left (2 \right )} + 16} \right )} \right )} \right )} \] Input:
integrate(((-32*ln(2)-32*x**2-64*x-128)*ln(5/(4*ln(2)+4*x**2+8*x+16))-32*x -32)/((ln(2)+x**2+2*x+4)*ln(5/(4*ln(2)+4*x**2+8*x+16))*ln(ln(5/(4*ln(2)+4* x**2+8*x+16)))+(-2*x*ln(2)-2*x**3-4*x**2-8*x)*ln(5/(4*ln(2)+4*x**2+8*x+16) )),x)
Output:
16*log(-2*x + log(log(5/(4*x**2 + 8*x + 4*log(2) + 16))))
Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \, \log \left (-2 \, x + \log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) - \log \left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )\right )\right ) \] Input:
integrate(((-32*log(2)-32*x^2-64*x-128)*log(5/(4*log(2)+4*x^2+8*x+16))-32* x-32)/((log(2)+x^2+2*x+4)*log(5/(4*log(2)+4*x^2+8*x+16))*log(log(5/(4*log( 2)+4*x^2+8*x+16)))+(-2*x*log(2)-2*x^3-4*x^2-8*x)*log(5/(4*log(2)+4*x^2+8*x +16))),x, algorithm="maxima")
Output:
16*log(-2*x + log(log(5) - 2*log(2) - log(x^2 + 2*x + log(2) + 4)))
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \, \log \left (2 \, x - \log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) - \log \left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )\right )\right ) \] Input:
integrate(((-32*log(2)-32*x^2-64*x-128)*log(5/(4*log(2)+4*x^2+8*x+16))-32* x-32)/((log(2)+x^2+2*x+4)*log(5/(4*log(2)+4*x^2+8*x+16))*log(log(5/(4*log( 2)+4*x^2+8*x+16)))+(-2*x*log(2)-2*x^3-4*x^2-8*x)*log(5/(4*log(2)+4*x^2+8*x +16))),x, algorithm="giac")
Output:
16*log(2*x - log(log(5) - 2*log(2) - log(x^2 + 2*x + log(2) + 4)))
Timed out. \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=\text {Hanged} \] Input:
int((32*x + log(5/(8*x + 4*log(2) + 4*x^2 + 16))*(64*x + 32*log(2) + 32*x^ 2 + 128) + 32)/(log(5/(8*x + 4*log(2) + 4*x^2 + 16))*(8*x + 2*x*log(2) + 4 *x^2 + 2*x^3) - log(5/(8*x + 4*log(2) + 4*x^2 + 16))*log(log(5/(8*x + 4*lo g(2) + 4*x^2 + 16)))*(2*x + log(2) + x^2 + 4)),x)
Output:
\text{Hanged}
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {5}{4 \,\mathrm {log}\left (2\right )+4 x^{2}+8 x +16}\right )\right )-2 x \right ) \] Input:
int(((-32*log(2)-32*x^2-64*x-128)*log(5/(4*log(2)+4*x^2+8*x+16))-32*x-32)/ ((log(2)+x^2+2*x+4)*log(5/(4*log(2)+4*x^2+8*x+16))*log(log(5/(4*log(2)+4*x ^2+8*x+16)))+(-2*x*log(2)-2*x^3-4*x^2-8*x)*log(5/(4*log(2)+4*x^2+8*x+16))) ,x)
Output:
16*log(log(log(5/(4*log(2) + 4*x**2 + 8*x + 16))) - 2*x)