Integrand size = 53, antiderivative size = 27 \[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=8+2 x-x (4+x)-\left (x+\frac {2 x}{3 \log (\log (x))}\right )^2 \] Output:
2*x-(x+2/3*x/ln(ln(x)))^2+8-(4+x)*x
Time = 0.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=-2 x-2 x^2-\frac {4 x^2}{9 \log ^2(\log (x))}-\frac {4 x^2}{3 \log (\log (x))} \] Input:
Integrate[(8*x + (12*x - 8*x*Log[x])*Log[Log[x]] - 24*x*Log[x]*Log[Log[x]] ^2 + (-18 - 36*x)*Log[x]*Log[Log[x]]^3)/(9*Log[x]*Log[Log[x]]^3),x]
Output:
-2*x - 2*x^2 - (4*x^2)/(9*Log[Log[x]]^2) - (4*x^2)/(3*Log[Log[x]])
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.
\(\displaystyle \int \frac {8 x+(-36 x-18) \log (x) \log ^3(\log (x))-24 x \log (x) \log ^2(\log (x))+(12 x-8 x \log (x)) \log (\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {2 \left (-9 (2 x+1) \log (x) \log ^3(\log (x))-12 x \log (x) \log ^2(\log (x))+2 (3 x-2 x \log (x)) \log (\log (x))+4 x\right )}{\log (x) \log ^3(\log (x))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{9} \int \frac {-9 (2 x+1) \log (x) \log ^3(\log (x))-12 x \log (x) \log ^2(\log (x))+2 (3 x-2 x \log (x)) \log (\log (x))+4 x}{\log (x) \log ^3(\log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{9} \int \left (-\frac {12 x}{\log (\log (x))}+\frac {4 x}{\log (x) \log ^3(\log (x))}-9 (2 x+1)-\frac {2 (2 x \log (x)-3 x)}{\log (x) \log ^2(\log (x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{9} \left (4 \int \frac {x}{\log (x) \log ^3(\log (x))}dx-4 \int \frac {x}{\log ^2(\log (x))}dx+6 \int \frac {x}{\log (x) \log ^2(\log (x))}dx-12 \int \frac {x}{\log (\log (x))}dx-\frac {9}{4} (2 x+1)^2\right )\) |
Input:
Int[(8*x + (12*x - 8*x*Log[x])*Log[Log[x]] - 24*x*Log[x]*Log[Log[x]]^2 + ( -18 - 36*x)*Log[x]*Log[Log[x]]^3)/(9*Log[x]*Log[Log[x]]^3),x]
Output:
$Aborted
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-2 x^{2}-2 x -\frac {4 x^{2} \left (3 \ln \left (\ln \left (x \right )\right )+1\right )}{9 \ln \left (\ln \left (x \right )\right )^{2}}\) | \(27\) |
parallelrisch | \(\frac {-18 x^{2} \ln \left (\ln \left (x \right )\right )^{2}-12 x^{2} \ln \left (\ln \left (x \right )\right )-18 x \ln \left (\ln \left (x \right )\right )^{2}-4 x^{2}}{9 \ln \left (\ln \left (x \right )\right )^{2}}\) | \(40\) |
Input:
int(1/9*((-36*x-18)*ln(x)*ln(ln(x))^3-24*x*ln(x)*ln(ln(x))^2+(-8*x*ln(x)+1 2*x)*ln(ln(x))+8*x)/ln(x)/ln(ln(x))^3,x,method=_RETURNVERBOSE)
Output:
-2*x^2-2*x-4/9*x^2*(3*ln(ln(x))+1)/ln(ln(x))^2
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=-\frac {2 \, {\left (6 \, x^{2} \log \left (\log \left (x\right )\right ) + 9 \, {\left (x^{2} + x\right )} \log \left (\log \left (x\right )\right )^{2} + 2 \, x^{2}\right )}}{9 \, \log \left (\log \left (x\right )\right )^{2}} \] Input:
integrate(1/9*((-36*x-18)*log(x)*log(log(x))^3-24*x*log(x)*log(log(x))^2+( -8*x*log(x)+12*x)*log(log(x))+8*x)/log(x)/log(log(x))^3,x, algorithm="fric as")
Output:
-2/9*(6*x^2*log(log(x)) + 9*(x^2 + x)*log(log(x))^2 + 2*x^2)/log(log(x))^2
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=- 2 x^{2} - 2 x + \frac {- 12 x^{2} \log {\left (\log {\left (x \right )} \right )} - 4 x^{2}}{9 \log {\left (\log {\left (x \right )} \right )}^{2}} \] Input:
integrate(1/9*((-36*x-18)*ln(x)*ln(ln(x))**3-24*x*ln(x)*ln(ln(x))**2+(-8*x *ln(x)+12*x)*ln(ln(x))+8*x)/ln(x)/ln(ln(x))**3,x)
Output:
-2*x**2 - 2*x + (-12*x**2*log(log(x)) - 4*x**2)/(9*log(log(x))**2)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=-2 \, x^{2} - 2 \, x - \frac {4 \, {\left (3 \, x^{2} \log \left (\log \left (x\right )\right ) + x^{2}\right )}}{9 \, \log \left (\log \left (x\right )\right )^{2}} \] Input:
integrate(1/9*((-36*x-18)*log(x)*log(log(x))^3-24*x*log(x)*log(log(x))^2+( -8*x*log(x)+12*x)*log(log(x))+8*x)/log(x)/log(log(x))^3,x, algorithm="maxi ma")
Output:
-2*x^2 - 2*x - 4/9*(3*x^2*log(log(x)) + x^2)/log(log(x))^2
\[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=\int { -\frac {2 \, {\left (9 \, {\left (2 \, x + 1\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right )^{3} + 12 \, x \log \left (x\right ) \log \left (\log \left (x\right )\right )^{2} + 2 \, {\left (2 \, x \log \left (x\right ) - 3 \, x\right )} \log \left (\log \left (x\right )\right ) - 4 \, x\right )}}{9 \, \log \left (x\right ) \log \left (\log \left (x\right )\right )^{3}} \,d x } \] Input:
integrate(1/9*((-36*x-18)*log(x)*log(log(x))^3-24*x*log(x)*log(log(x))^2+( -8*x*log(x)+12*x)*log(log(x))+8*x)/log(x)/log(log(x))^3,x, algorithm="giac ")
Output:
integrate(-2/9*(9*(2*x + 1)*log(x)*log(log(x))^3 + 12*x*log(x)*log(log(x)) ^2 + 2*(2*x*log(x) - 3*x)*log(log(x)) - 4*x)/(log(x)*log(log(x))^3), x)
Time = 2.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=-2\,x-2\,x^2-\frac {4\,x^2}{3\,\ln \left (\ln \left (x\right )\right )}-\frac {4\,x^2}{9\,{\ln \left (\ln \left (x\right )\right )}^2} \] Input:
int(((8*x)/9 + (log(log(x))*(12*x - 8*x*log(x)))/9 - (log(log(x))^3*log(x) *(36*x + 18))/9 - (8*x*log(log(x))^2*log(x))/3)/(log(log(x))^3*log(x)),x)
Output:
- 2*x - 2*x^2 - (4*x^2)/(3*log(log(x))) - (4*x^2)/(9*log(log(x))^2)
Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {8 x+(12 x-8 x \log (x)) \log (\log (x))-24 x \log (x) \log ^2(\log (x))+(-18-36 x) \log (x) \log ^3(\log (x))}{9 \log (x) \log ^3(\log (x))} \, dx=\frac {2 x \left (-9 \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2} x -9 \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2}-6 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x -2 x \right )}{9 \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2}} \] Input:
int(1/9*((-36*x-18)*log(x)*log(log(x))^3-24*x*log(x)*log(log(x))^2+(-8*x*l og(x)+12*x)*log(log(x))+8*x)/log(x)/log(log(x))^3,x)
Output:
(2*x*( - 9*log(log(x))**2*x - 9*log(log(x))**2 - 6*log(log(x))*x - 2*x))/( 9*log(log(x))**2)