Integrand size = 100, antiderivative size = 19 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=25+4 x \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right ) \] Output:
4*ln(ln(ln(x))-5+1/25/x^2+x)*x+25
Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=4 x \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right ) \] Input:
Integrate[(100*x^2 + (-8 + 100*x^3)*Log[x] + ((4 - 500*x^2 + 100*x^3)*Log[ x] + 100*x^2*Log[x]*Log[Log[x]])*Log[(1 - 125*x^2 + 25*x^3 + 25*x^2*Log[Lo g[x]])/(25*x^2)])/((1 - 125*x^2 + 25*x^3)*Log[x] + 25*x^2*Log[x]*Log[Log[x ]]),x]
Output:
4*x*Log[-5 + 1/(25*x^2) + x + 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.
\(\displaystyle \int \frac {\left (100 x^3-8\right ) \log (x)+100 x^2+\left (100 x^2 \log (x) \log (\log (x))+\left (100 x^3-500 x^2+4\right ) \log (x)\right ) \log \left (\frac {25 x^3-125 x^2+25 x^2 \log (\log (x))+1}{25 x^2}\right )}{25 x^2 \log (x) \log (\log (x))+\left (25 x^3-125 x^2+1\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (100 x^3-8\right ) \log (x)+100 x^2+\left (100 x^2 \log (x) \log (\log (x))+\left (100 x^3-500 x^2+4\right ) \log (x)\right ) \log \left (\frac {25 x^3-125 x^2+25 x^2 \log (\log (x))+1}{25 x^2}\right )}{\log (x) \left (25 x^3-125 x^2+25 x^2 \log (\log (x))+1\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (4 \log \left (\frac {1}{25 x^2}+x+\log (\log (x))-5\right )+\frac {4 \left (25 x^3 \log (x)+25 x^2-2 \log (x)\right )}{\log (x) \left (25 x^3-125 x^2+25 x^2 \log (\log (x))+1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \log \left (x+\log (\log (x))-5+\frac {1}{25 x^2}\right )dx-8 \int \frac {1}{25 x^3+25 \log (\log (x)) x^2-125 x^2+1}dx+100 \int \frac {x^3}{25 x^3+25 \log (\log (x)) x^2-125 x^2+1}dx+100 \int \frac {x^2}{\log (x) \left (25 x^3+25 \log (\log (x)) x^2-125 x^2+1\right )}dx\) |
Input:
Int[(100*x^2 + (-8 + 100*x^3)*Log[x] + ((4 - 500*x^2 + 100*x^3)*Log[x] + 1 00*x^2*Log[x]*Log[Log[x]])*Log[(1 - 125*x^2 + 25*x^3 + 25*x^2*Log[Log[x]]) /(25*x^2)])/((1 - 125*x^2 + 25*x^3)*Log[x] + 25*x^2*Log[x]*Log[Log[x]]),x]
Output:
$Aborted
Time = 4.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58
method | result | size |
parallelrisch | \(4 x \ln \left (\frac {25 x^{2} \ln \left (\ln \left (x \right )\right )+25 x^{3}-125 x^{2}+1}{25 x^{2}}\right )\) | \(30\) |
risch | \(4 x \ln \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \left (x \right )\right )-5\right ) x^{2}\right )+2 i \pi x \,\operatorname {csgn}\left (i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \left (x \right )\right )-5\right ) x^{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \left (x \right )\right )-5\right ) x^{2}\right )}{x^{2}}\right )}^{2}-2 i \pi x \,\operatorname {csgn}\left (i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \left (x \right )\right )-5\right ) x^{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \left (x \right )\right )-5\right ) x^{2}\right )}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )-2 i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \left (x \right )\right )-5\right ) x^{2}\right )}{x^{2}}\right )}^{3}+2 i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \left (x \right )\right )-5\right ) x^{2}\right )}{x^{2}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+2 i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-4 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+2 i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}-8 x \ln \left (x \right )\) | \(237\) |
Input:
int(((100*x^2*ln(x)*ln(ln(x))+(100*x^3-500*x^2+4)*ln(x))*ln(1/25*(25*x^2*l n(ln(x))+25*x^3-125*x^2+1)/x^2)+(100*x^3-8)*ln(x)+100*x^2)/(25*x^2*ln(x)*l n(ln(x))+(25*x^3-125*x^2+1)*ln(x)),x,method=_RETURNVERBOSE)
Output:
4*x*ln(1/25*(25*x^2*ln(ln(x))+25*x^3-125*x^2+1)/x^2)
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=4 \, x \log \left (\frac {25 \, x^{3} + 25 \, x^{2} \log \left (\log \left (x\right )\right ) - 125 \, x^{2} + 1}{25 \, x^{2}}\right ) \] Input:
integrate(((100*x^2*log(x)*log(log(x))+(100*x^3-500*x^2+4)*log(x))*log(1/2 5*(25*x^2*log(log(x))+25*x^3-125*x^2+1)/x^2)+(100*x^3-8)*log(x)+100*x^2)/( 25*x^2*log(x)*log(log(x))+(25*x^3-125*x^2+1)*log(x)),x, algorithm="fricas" )
Output:
4*x*log(1/25*(25*x^3 + 25*x^2*log(log(x)) - 125*x^2 + 1)/x^2)
Time = 0.74 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=4 x \log {\left (\frac {x^{3} + x^{2} \log {\left (\log {\left (x \right )} \right )} - 5 x^{2} + \frac {1}{25}}{x^{2}} \right )} \] Input:
integrate(((100*x**2*ln(x)*ln(ln(x))+(100*x**3-500*x**2+4)*ln(x))*ln(1/25* (25*x**2*ln(ln(x))+25*x**3-125*x**2+1)/x**2)+(100*x**3-8)*ln(x)+100*x**2)/ (25*x**2*ln(x)*ln(ln(x))+(25*x**3-125*x**2+1)*ln(x)),x)
Output:
4*x*log((x**3 + x**2*log(log(x)) - 5*x**2 + 1/25)/x**2)
Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).
Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=-8 \, x \log \left (5\right ) + 4 \, x \log \left (25 \, x^{3} + 25 \, x^{2} \log \left (\log \left (x\right )\right ) - 125 \, x^{2} + 1\right ) - 8 \, x \log \left (x\right ) \] Input:
integrate(((100*x^2*log(x)*log(log(x))+(100*x^3-500*x^2+4)*log(x))*log(1/2 5*(25*x^2*log(log(x))+25*x^3-125*x^2+1)/x^2)+(100*x^3-8)*log(x)+100*x^2)/( 25*x^2*log(x)*log(log(x))+(25*x^3-125*x^2+1)*log(x)),x, algorithm="maxima" )
Output:
-8*x*log(5) + 4*x*log(25*x^3 + 25*x^2*log(log(x)) - 125*x^2 + 1) - 8*x*log (x)
Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).
Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=-8 \, x \log \left (5\right ) + 4 \, x \log \left (25 \, x^{3} + 25 \, x^{2} \log \left (\log \left (x\right )\right ) - 125 \, x^{2} + 1\right ) - 8 \, x \log \left (x\right ) \] Input:
integrate(((100*x^2*log(x)*log(log(x))+(100*x^3-500*x^2+4)*log(x))*log(1/2 5*(25*x^2*log(log(x))+25*x^3-125*x^2+1)/x^2)+(100*x^3-8)*log(x)+100*x^2)/( 25*x^2*log(x)*log(log(x))+(25*x^3-125*x^2+1)*log(x)),x, algorithm="giac")
Output:
-8*x*log(5) + 4*x*log(25*x^3 + 25*x^2*log(log(x)) - 125*x^2 + 1) - 8*x*log (x)
Time = 2.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=4\,x\,\ln \left (\frac {x^2\,\ln \left (\ln \left (x\right )\right )-5\,x^2+x^3+\frac {1}{25}}{x^2}\right ) \] Input:
int((log((x^2*log(log(x)) - 5*x^2 + x^3 + 1/25)/x^2)*(log(x)*(100*x^3 - 50 0*x^2 + 4) + 100*x^2*log(log(x))*log(x)) + 100*x^2 + log(x)*(100*x^3 - 8)) /(log(x)*(25*x^3 - 125*x^2 + 1) + 25*x^2*log(log(x))*log(x)),x)
Output:
4*x*log((x^2*log(log(x)) - 5*x^2 + x^3 + 1/25)/x^2)
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx=4 \,\mathrm {log}\left (\frac {25 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x^{2}+25 x^{3}-125 x^{2}+1}{25 x^{2}}\right ) x \] Input:
int(((100*x^2*log(x)*log(log(x))+(100*x^3-500*x^2+4)*log(x))*log(1/25*(25* x^2*log(log(x))+25*x^3-125*x^2+1)/x^2)+(100*x^3-8)*log(x)+100*x^2)/(25*x^2 *log(x)*log(log(x))+(25*x^3-125*x^2+1)*log(x)),x)
Output:
4*log((25*log(log(x))*x**2 + 25*x**3 - 125*x**2 + 1)/(25*x**2))*x