Integrand size = 66, antiderivative size = 35 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x-\frac {3 \left (2+x+\frac {1}{2} (i \pi +\log (-\log (\log (2))))^2\right )}{4+\log \left (4 x^2\right )} \] Output:
x-3*(x+2+1/2*ln(ln(ln(2)))^2)/(ln(4*x^2)+4)
Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x+\frac {3 \left (-4+\pi ^2-2 x-2 i \pi \log (-\log (\log (2)))-\log ^2(-\log (\log (2)))\right )}{2 \left (4+\log \left (4 x^2\right )\right )} \] Input:
Integrate[(12 + 10*x + 5*x*Log[4*x^2] + x*Log[4*x^2]^2 + 3*(I*Pi + Log[-Lo g[Log[2]]])^2)/(16*x + 8*x*Log[4*x^2] + x*Log[4*x^2]^2),x]
Output:
x + (3*(-4 + Pi^2 - 2*x - (2*I)*Pi*Log[-Log[Log[2]]] - Log[-Log[Log[2]]]^2 ))/(2*(4 + Log[4*x^2]))
Time = 0.72 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {x \log ^2\left (4 x^2\right )+5 x \log \left (4 x^2\right )+10 x+12+3 (\log (-\log (\log (2)))+i \pi )^2}{x \log ^2\left (4 x^2\right )+8 x \log \left (4 x^2\right )+16 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x \log ^2\left (4 x^2\right )+5 x \log \left (4 x^2\right )+10 x+12 \left (1+\frac {1}{4} (\log (-\log (\log (2)))+i \pi )^2\right )}{x \left (\log \left (4 x^2\right )+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 \left (2 x-\pi ^2+4+\log ^2(-\log (\log (2)))+2 i \pi \log (-\log (\log (2)))\right )}{x \left (\log \left (4 x^2\right )+4\right )^2}-\frac {3}{\log \left (4 x^2\right )+4}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 \left (4-\pi ^2+\log ^2(-\log (\log (2)))+2 i \pi \log (-\log (\log (2)))\right )}{2 \left (\log \left (4 x^2\right )+4\right )}-\frac {3 x}{\log \left (4 x^2\right )+4}+x\) |
Input:
Int[(12 + 10*x + 5*x*Log[4*x^2] + x*Log[4*x^2]^2 + 3*(I*Pi + Log[-Log[Log[ 2]]])^2)/(16*x + 8*x*Log[4*x^2] + x*Log[4*x^2]^2),x]
Output:
x - (3*x)/(4 + Log[4*x^2]) - (3*(4 - Pi^2 + (2*I)*Pi*Log[-Log[Log[2]]] + L og[-Log[Log[2]]]^2))/(2*(4 + Log[4*x^2]))
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74
method | result | size |
risch | \(x -\frac {3 \left (\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}+2 x +4\right )}{2 \left (\ln \left (4 x^{2}\right )+4\right )}\) | \(26\) |
norman | \(\frac {x +x \ln \left (4 x^{2}\right )-6-\frac {3 \ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{2}}{\ln \left (4 x^{2}\right )+4}\) | \(31\) |
parallelrisch | \(\frac {-12-3 \ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}+2 x \ln \left (4 x^{2}\right )+2 x}{2 \ln \left (4 x^{2}\right )+8}\) | \(35\) |
default | \(\frac {x \ln \left (x^{2}\right )+\left (1+2 \ln \left (2\right )\right ) x -6}{2 \ln \left (2\right )+\ln \left (x^{2}\right )+4}-\frac {3 \ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{2 \left (2 \ln \left (2\right )+\ln \left (x^{2}\right )+4\right )}\) | \(51\) |
Input:
int((3*ln(ln(ln(2)))^2+x*ln(4*x^2)^2+5*x*ln(4*x^2)+10*x+12)/(x*ln(4*x^2)^2 +8*x*ln(4*x^2)+16*x),x,method=_RETURNVERBOSE)
Output:
x-3/2*(ln(ln(ln(2)))^2+2*x+4)/(ln(4*x^2)+4)
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=\frac {2 \, x \log \left (4 \, x^{2}\right ) - 3 \, \log \left (\log \left (\log \left (2\right )\right )\right )^{2} + 2 \, x - 12}{2 \, {\left (\log \left (4 \, x^{2}\right ) + 4\right )}} \] Input:
integrate((3*log(log(log(2)))^2+x*log(4*x^2)^2+5*x*log(4*x^2)+10*x+12)/(x* log(4*x^2)^2+8*x*log(4*x^2)+16*x),x, algorithm="fricas")
Output:
1/2*(2*x*log(4*x^2) - 3*log(log(log(2)))^2 + 2*x - 12)/(log(4*x^2) + 4)
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x + \frac {- 6 x - 12 - 3 \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}^{2} + 3 \pi ^{2} - 6 i \pi \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}}{2 \log {\left (x^{2} \right )} + 4 \log {\left (2 \right )} + 8} \] Input:
integrate((3*ln(ln(ln(2)))**2+x*ln(4*x**2)**2+5*x*ln(4*x**2)+10*x+12)/(x*l n(4*x**2)**2+8*x*ln(4*x**2)+16*x),x)
Output:
x + (-6*x - 12 - 3*log(-log(log(2)))**2 + 3*pi**2 - 6*I*pi*log(-log(log(2) )))/(2*log(x**2) + 4*log(2) + 8)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (25) = 50\).
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.46 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=-\frac {3 \, \log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{4 \, {\left (\log \left (2\right ) + \log \left (x\right ) + 2\right )}} + \frac {x {\left (2 \, \log \left (2\right ) + 1\right )} + 2 \, x \log \left (x\right )}{2 \, {\left (\log \left (2\right ) + \log \left (x\right ) + 2\right )}} - \frac {3}{\log \left (2\right ) + \log \left (x\right ) + 2} \] Input:
integrate((3*log(log(log(2)))^2+x*log(4*x^2)^2+5*x*log(4*x^2)+10*x+12)/(x* log(4*x^2)^2+8*x*log(4*x^2)+16*x),x, algorithm="maxima")
Output:
-3/4*log(log(log(2)))^2/(log(2) + log(x) + 2) + 1/2*(x*(2*log(2) + 1) + 2* x*log(x))/(log(2) + log(x) + 2) - 3/(log(2) + log(x) + 2)
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=x - \frac {3 \, {\left (\log \left (\log \left (\log \left (2\right )\right )\right )^{2} + 2 \, x + 4\right )}}{2 \, {\left (2 \, \log \left (2\right ) + \log \left (x^{2}\right ) + 4\right )}} \] Input:
integrate((3*log(log(log(2)))^2+x*log(4*x^2)^2+5*x*log(4*x^2)+10*x+12)/(x* log(4*x^2)^2+8*x*log(4*x^2)+16*x),x, algorithm="giac")
Output:
x - 3/2*(log(log(log(2)))^2 + 2*x + 4)/(2*log(2) + log(x^2) + 4)
Time = 3.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=\frac {x\,\left (\ln \left (4\,x^2\right )+1\right )}{\ln \left (4\,x^2\right )+4}+\frac {\ln \left (4\,x^2\right )\,\left (\frac {3\,{\ln \left (\ln \left (\ln \left (2\right )\right )\right )}^2}{8}+\frac {3}{2}\right )}{\ln \left (4\,x^2\right )+4} \] Input:
int((10*x + 5*x*log(4*x^2) + 3*log(log(log(2)))^2 + x*log(4*x^2)^2 + 12)/( 16*x + 8*x*log(4*x^2) + x*log(4*x^2)^2),x)
Output:
(x*(log(4*x^2) + 1))/(log(4*x^2) + 4) + (log(4*x^2)*((3*log(log(log(2)))^2 )/8 + 3/2))/(log(4*x^2) + 4)
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {12+10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )} \, dx=\frac {3 \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )\right )\right )^{2} \mathrm {log}\left (4 x^{2}\right )+8 \,\mathrm {log}\left (4 x^{2}\right ) x +12 \,\mathrm {log}\left (4 x^{2}\right )+8 x}{8 \,\mathrm {log}\left (4 x^{2}\right )+32} \] Input:
int((3*log(log(log(2)))^2+x*log(4*x^2)^2+5*x*log(4*x^2)+10*x+12)/(x*log(4* x^2)^2+8*x*log(4*x^2)+16*x),x)
Output:
(3*log(log(log(2)))**2*log(4*x**2) + 8*log(4*x**2)*x + 12*log(4*x**2) + 8* x)/(8*(log(4*x**2) + 4))