Integrand size = 79, antiderivative size = 23 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=4 \left (x-2 \left (x+\log (5)+(\log (\log (-5+x))-\log (\log (x)))^2\right )\right ) \] Output:
-4*x-8*ln(5)-8*(ln(ln(-5+x))-ln(ln(x)))^2
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=-4 x-8 \log ^2(\log (-5+x))+16 \log (\log (-5+x)) \log (\log (x))-8 \log ^2(\log (x)) \] Input:
Integrate[((20*x - 4*x^2)*Log[-5 + x]*Log[x] + ((-80 + 16*x)*Log[-5 + x] - 16*x*Log[x])*Log[Log[-5 + x]] + ((80 - 16*x)*Log[-5 + x] + 16*x*Log[x])*L og[Log[x]])/((-5*x + x^2)*Log[-5 + x]*Log[x]),x]
Output:
-4*x - 8*Log[Log[-5 + x]]^2 + 16*Log[Log[-5 + x]]*Log[Log[x]] - 8*Log[Log[ x]]^2
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.
\(\displaystyle \int \frac {\left (20 x-4 x^2\right ) \log (x-5) \log (x)+((16 x-80) \log (x-5)-16 x \log (x)) \log (\log (x-5))+((80-16 x) \log (x-5)+16 x \log (x)) \log (\log (x))}{\left (x^2-5 x\right ) \log (x-5) \log (x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (20 x-4 x^2\right ) \log (x-5) \log (x)+((16 x-80) \log (x-5)-16 x \log (x)) \log (\log (x-5))+((80-16 x) \log (x-5)+16 x \log (x)) \log (\log (x))}{(x-5) x \log (x-5) \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {4 \left (x^2 \log (x-5) \log (x)-5 x \log (x-5) \log (x)-4 x \log (x-5) \log (\log (x-5))+4 x \log (x) \log (\log (x-5))+20 \log (x-5) \log (\log (x-5))\right )}{(x-5) x \log (x-5) \log (x)}-\frac {16 (x \log (x-5)-5 \log (x-5)-x \log (x)) \log (\log (x))}{(x-5) x \log (x-5) \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 16 \int \frac {\log (\log (x-5))}{x \log (x)}dx+16 \int \frac {\log (\log (x))}{(x-5) \log (x-5)}dx-4 x-8 \log ^2(\log (x-5))-8 \log ^2(\log (x))\) |
Input:
Int[((20*x - 4*x^2)*Log[-5 + x]*Log[x] + ((-80 + 16*x)*Log[-5 + x] - 16*x* Log[x])*Log[Log[-5 + x]] + ((80 - 16*x)*Log[-5 + x] + 16*x*Log[x])*Log[Log [x]])/((-5*x + x^2)*Log[-5 + x]*Log[x]),x]
Output:
$Aborted
Time = 7.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35
method | result | size |
risch | \(-8 \ln \left (\ln \left (x \right )\right )^{2}+16 \ln \left (\ln \left (x \right )\right ) \ln \left (\ln \left (-5+x \right )\right )-8 \ln \left (\ln \left (-5+x \right )\right )^{2}-4 x\) | \(31\) |
Input:
int(((16*x*ln(x)+(-16*x+80)*ln(-5+x))*ln(ln(x))+(-16*x*ln(x)+(16*x-80)*ln( -5+x))*ln(ln(-5+x))+(-4*x^2+20*x)*ln(-5+x)*ln(x))/(x^2-5*x)/ln(-5+x)/ln(x) ,x,method=_RETURNVERBOSE)
Output:
-8*ln(ln(x))^2+16*ln(ln(x))*ln(ln(-5+x))-8*ln(ln(-5+x))^2-4*x
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=-8 \, \log \left (\log \left (x - 5\right )\right )^{2} + 16 \, \log \left (\log \left (x - 5\right )\right ) \log \left (\log \left (x\right )\right ) - 8 \, \log \left (\log \left (x\right )\right )^{2} - 4 \, x \] Input:
integrate(((16*x*log(x)+(-16*x+80)*log(-5+x))*log(log(x))+(-16*x*log(x)+(1 6*x-80)*log(-5+x))*log(log(-5+x))+(-4*x^2+20*x)*log(-5+x)*log(x))/(x^2-5*x )/log(-5+x)/log(x),x, algorithm="fricas")
Output:
-8*log(log(x - 5))^2 + 16*log(log(x - 5))*log(log(x)) - 8*log(log(x))^2 - 4*x
Time = 0.93 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=- 4 x - 8 \log {\left (\log {\left (x \right )} \right )}^{2} + 16 \log {\left (\log {\left (x \right )} \right )} \log {\left (\log {\left (x - 5 \right )} \right )} - 8 \log {\left (\log {\left (x - 5 \right )} \right )}^{2} \] Input:
integrate(((16*x*ln(x)+(-16*x+80)*ln(-5+x))*ln(ln(x))+(-16*x*ln(x)+(16*x-8 0)*ln(-5+x))*ln(ln(-5+x))+(-4*x**2+20*x)*ln(-5+x)*ln(x))/(x**2-5*x)/ln(-5+ x)/ln(x),x)
Output:
-4*x - 8*log(log(x))**2 + 16*log(log(x))*log(log(x - 5)) - 8*log(log(x - 5 ))**2
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=-8 \, \log \left (\log \left (x - 5\right )\right )^{2} + 16 \, \log \left (\log \left (x - 5\right )\right ) \log \left (\log \left (x\right )\right ) - 8 \, \log \left (\log \left (x\right )\right )^{2} - 4 \, x \] Input:
integrate(((16*x*log(x)+(-16*x+80)*log(-5+x))*log(log(x))+(-16*x*log(x)+(1 6*x-80)*log(-5+x))*log(log(-5+x))+(-4*x^2+20*x)*log(-5+x)*log(x))/(x^2-5*x )/log(-5+x)/log(x),x, algorithm="maxima")
Output:
-8*log(log(x - 5))^2 + 16*log(log(x - 5))*log(log(x)) - 8*log(log(x))^2 - 4*x
Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=-8 \, \log \left (\log \left (x - 5\right )\right )^{2} + 16 \, \log \left (\log \left (x - 5\right )\right ) \log \left (\log \left (x\right )\right ) - 8 \, \log \left (\log \left (x\right )\right )^{2} - 4 \, x \] Input:
integrate(((16*x*log(x)+(-16*x+80)*log(-5+x))*log(log(x))+(-16*x*log(x)+(1 6*x-80)*log(-5+x))*log(log(-5+x))+(-4*x^2+20*x)*log(-5+x)*log(x))/(x^2-5*x )/log(-5+x)/log(x),x, algorithm="giac")
Output:
-8*log(log(x - 5))^2 + 16*log(log(x - 5))*log(log(x)) - 8*log(log(x))^2 - 4*x
Time = 3.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=-8\,{\ln \left (\ln \left (x\right )\right )}^2+16\,\ln \left (\ln \left (x\right )\right )\,\ln \left (\ln \left (x-5\right )\right )-8\,{\ln \left (\ln \left (x-5\right )\right )}^2-4\,x \] Input:
int(-(log(log(x))*(16*x*log(x) - log(x - 5)*(16*x - 80)) - log(log(x - 5)) *(16*x*log(x) - log(x - 5)*(16*x - 80)) + log(x - 5)*log(x)*(20*x - 4*x^2) )/(log(x - 5)*log(x)*(5*x - x^2)),x)
Output:
16*log(log(x))*log(log(x - 5)) - 8*log(log(x - 5))^2 - 4*x - 8*log(log(x)) ^2
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx=-8 \mathrm {log}\left (\mathrm {log}\left (-5+x \right )\right )^{2}+16 \,\mathrm {log}\left (\mathrm {log}\left (-5+x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right )-8 \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2}-4 x \] Input:
int(((16*x*log(x)+(-16*x+80)*log(-5+x))*log(log(x))+(-16*x*log(x)+(16*x-80 )*log(-5+x))*log(log(-5+x))+(-4*x^2+20*x)*log(-5+x)*log(x))/(x^2-5*x)/log( -5+x)/log(x),x)
Output:
4*( - 2*log(log(x - 5))**2 + 4*log(log(x - 5))*log(log(x)) - 2*log(log(x)) **2 - x)