Integrand size = 46, antiderivative size = 31 \[ \int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2\left (\frac {9}{\log (5)}\right )-10 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx=x-\frac {2 \left (1+5 \left (2 x-\left (5+\log ^2\left (\frac {9}{\log (5)}\right )\right )^2\right )\right )}{\log (x)} \]
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2\left (\frac {9}{\log (5)}\right )-10 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx=x+\frac {248}{\log (x)}-\frac {20 x}{\log (x)}+\frac {100 \log ^2\left (\frac {9}{\log (5)}\right )}{\log (x)}+\frac {10 \log ^4\left (\frac {9}{\log (5)}\right )}{\log (x)} \]
Integrate[(-248 + 20*x - 20*x*Log[x] + x*Log[x]^2 - 100*Log[9/Log[5]]^2 - 10*Log[9/Log[5]]^4)/(x*Log[x]^2),x]
Time = 0.54 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, 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.
\(\displaystyle \int \frac {20 x+x \log ^2(x)-20 x \log (x)-248-10 \log ^4\left (\frac {9}{\log (5)}\right )-100 \log ^2\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {20 x+x \log ^2(x)-20 x \log (x)-248 \left (1+\frac {5}{124} \log ^2\left (\frac {9}{\log (5)}\right ) \left (10+\log ^2\left (\frac {9}{\log (5)}\right )\right )\right )}{x \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (10 x-124-5 \log ^4\left (\frac {9}{\log (5)}\right )-50 \log ^2\left (\frac {9}{\log (5)}\right )\right )}{x \log ^2(x)}-\frac {20}{\log (x)}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x+\frac {2 \left (124+5 \log ^4\left (\frac {9}{\log (5)}\right )+50 \log ^2\left (\frac {9}{\log (5)}\right )\right )}{\log (x)}-\frac {20 x}{\log (x)}\) |
Int[(-248 + 20*x - 20*x*Log[x] + x*Log[x]^2 - 100*Log[9/Log[5]]^2 - 10*Log [9/Log[5]]^4)/(x*Log[x]^2),x]
3.2.43.3.1 Defintions of rubi rules used
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19
method | result | size |
parallelrisch | \(\frac {10 \ln \left (\frac {9}{\ln \left (5\right )}\right )^{4}+100 \ln \left (\frac {9}{\ln \left (5\right )}\right )^{2}+x \ln \left (x \right )+248-20 x}{\ln \left (x \right )}\) | \(37\) |
risch | \(x +\frac {10 \ln \left (\ln \left (5\right )\right )^{4}-80 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )^{3}+240 \ln \left (3\right )^{2} \ln \left (\ln \left (5\right )\right )^{2}-320 \ln \left (3\right )^{3} \ln \left (\ln \left (5\right )\right )+160 \ln \left (3\right )^{4}+100 \ln \left (\ln \left (5\right )\right )^{2}-400 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )+400 \ln \left (3\right )^{2}-20 x +248}{\ln \left (x \right )}\) | \(76\) |
norman | \(\frac {x \ln \left (x \right )-20 x +248+10 \ln \left (\ln \left (5\right )\right )^{4}-80 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )^{3}+240 \ln \left (3\right )^{2} \ln \left (\ln \left (5\right )\right )^{2}-320 \ln \left (3\right )^{3} \ln \left (\ln \left (5\right )\right )+160 \ln \left (3\right )^{4}+100 \ln \left (\ln \left (5\right )\right )^{2}-400 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )+400 \ln \left (3\right )^{2}}{\ln \left (x \right )}\) | \(77\) |
default | \(\frac {160 \ln \left (3\right )^{4}}{\ln \left (x \right )}-\frac {320 \ln \left (3\right )^{3} \ln \left (\ln \left (5\right )\right )}{\ln \left (x \right )}+\frac {240 \ln \left (3\right )^{2} \ln \left (\ln \left (5\right )\right )^{2}}{\ln \left (x \right )}-\frac {80 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )^{3}}{\ln \left (x \right )}+\frac {10 \ln \left (\ln \left (5\right )\right )^{4}}{\ln \left (x \right )}+x +\frac {400 \ln \left (3\right )^{2}}{\ln \left (x \right )}-\frac {400 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )}{\ln \left (x \right )}+\frac {100 \ln \left (\ln \left (5\right )\right )^{2}}{\ln \left (x \right )}-\frac {20 x}{\ln \left (x \right )}+\frac {248}{\ln \left (x \right )}\) | \(110\) |
parts | \(\frac {160 \ln \left (3\right )^{4}}{\ln \left (x \right )}-\frac {320 \ln \left (3\right )^{3} \ln \left (\ln \left (5\right )\right )}{\ln \left (x \right )}+\frac {240 \ln \left (3\right )^{2} \ln \left (\ln \left (5\right )\right )^{2}}{\ln \left (x \right )}-\frac {80 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )^{3}}{\ln \left (x \right )}+\frac {10 \ln \left (\ln \left (5\right )\right )^{4}}{\ln \left (x \right )}+x +\frac {400 \ln \left (3\right )^{2}}{\ln \left (x \right )}-\frac {400 \ln \left (3\right ) \ln \left (\ln \left (5\right )\right )}{\ln \left (x \right )}+\frac {100 \ln \left (\ln \left (5\right )\right )^{2}}{\ln \left (x \right )}-\frac {20 x}{\ln \left (x \right )}+\frac {248}{\ln \left (x \right )}\) | \(110\) |
int((x*ln(x)^2-20*x*ln(x)-10*ln(9/ln(5))^4-100*ln(9/ln(5))^2+20*x-248)/x/l n(x)^2,x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2\left (\frac {9}{\log (5)}\right )-10 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx=\frac {10 \, \log \left (\frac {9}{\log \left (5\right )}\right )^{4} + x \log \left (x\right ) + 100 \, \log \left (\frac {9}{\log \left (5\right )}\right )^{2} - 20 \, x + 248}{\log \left (x\right )} \]
integrate((x*log(x)^2-20*x*log(x)-10*log(9/log(5))^4-100*log(9/log(5))^2+2 0*x-248)/x/log(x)^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (22) = 44\).
Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.81 \[ \int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2\left (\frac {9}{\log (5)}\right )-10 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx=x + \frac {- 20 x - 400 \log {\left (3 \right )} \log {\left (\log {\left (5 \right )} \right )} - 320 \log {\left (3 \right )}^{3} \log {\left (\log {\left (5 \right )} \right )} - 80 \log {\left (3 \right )} \log {\left (\log {\left (5 \right )} \right )}^{3} + 10 \log {\left (\log {\left (5 \right )} \right )}^{4} + 100 \log {\left (\log {\left (5 \right )} \right )}^{2} + 240 \log {\left (3 \right )}^{2} \log {\left (\log {\left (5 \right )} \right )}^{2} + 160 \log {\left (3 \right )}^{4} + 248 + 400 \log {\left (3 \right )}^{2}}{\log {\left (x \right )}} \]
integrate((x*ln(x)**2-20*x*ln(x)-10*ln(9/ln(5))**4-100*ln(9/ln(5))**2+20*x -248)/x/ln(x)**2,x)
x + (-20*x - 400*log(3)*log(log(5)) - 320*log(3)**3*log(log(5)) - 80*log(3 )*log(log(5))**3 + 10*log(log(5))**4 + 100*log(log(5))**2 + 240*log(3)**2* log(log(5))**2 + 160*log(3)**4 + 248 + 400*log(3)**2)/log(x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2\left (\frac {9}{\log (5)}\right )-10 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx=\frac {10 \, \log \left (\frac {9}{\log \left (5\right )}\right )^{4}}{\log \left (x\right )} + x + \frac {100 \, \log \left (\frac {9}{\log \left (5\right )}\right )^{2}}{\log \left (x\right )} + \frac {248}{\log \left (x\right )} - 20 \, {\rm Ei}\left (\log \left (x\right )\right ) + 20 \, \Gamma \left (-1, -\log \left (x\right )\right ) \]
integrate((x*log(x)^2-20*x*log(x)-10*log(9/log(5))^4-100*log(9/log(5))^2+2 0*x-248)/x/log(x)^2,x, algorithm=\
10*log(9/log(5))^4/log(x) + x + 100*log(9/log(5))^2/log(x) + 248/log(x) - 20*Ei(log(x)) + 20*gamma(-1, -log(x))
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (28) = 56\).
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.42 \[ \int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2\left (\frac {9}{\log (5)}\right )-10 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx=x + \frac {2 \, {\left (80 \, \log \left (3\right )^{4} - 160 \, \log \left (3\right )^{3} \log \left (\log \left (5\right )\right ) + 120 \, \log \left (3\right )^{2} \log \left (\log \left (5\right )\right )^{2} - 40 \, \log \left (3\right ) \log \left (\log \left (5\right )\right )^{3} + 5 \, \log \left (\log \left (5\right )\right )^{4} + 200 \, \log \left (3\right )^{2} - 200 \, \log \left (3\right ) \log \left (\log \left (5\right )\right ) + 50 \, \log \left (\log \left (5\right )\right )^{2} - 10 \, x + 124\right )}}{\log \left (x\right )} \]
integrate((x*log(x)^2-20*x*log(x)-10*log(9/log(5))^4-100*log(9/log(5))^2+2 0*x-248)/x/log(x)^2,x, algorithm=\
x + 2*(80*log(3)^4 - 160*log(3)^3*log(log(5)) + 120*log(3)^2*log(log(5))^2 - 40*log(3)*log(log(5))^3 + 5*log(log(5))^4 + 200*log(3)^2 - 200*log(3)*l og(log(5)) + 50*log(log(5))^2 - 10*x + 124)/log(x)
Time = 11.64 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-248+20 x-20 x \log (x)+x \log ^2(x)-100 \log ^2\left (\frac {9}{\log (5)}\right )-10 \log ^4\left (\frac {9}{\log (5)}\right )}{x \log ^2(x)} \, dx=x+\frac {100\,{\ln \left (\frac {9}{\ln \left (5\right )}\right )}^2-20\,x+10\,{\ln \left (\frac {9}{\ln \left (5\right )}\right )}^4+248}{\ln \left (x\right )} \]