Integrand size = 55, antiderivative size = 21 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx=4-\frac {2 (-4+x)}{x^2 \log (5 x) \log (\log (x))} \] Output:
4-2/x^2*(-4+x)/ln(ln(x))/ln(5*x)
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx=\frac {2 (4-x)}{x^2 \log (5 x) \log (\log (x))} \] Input:
Integrate[((-8 + 2*x)*Log[5*x] + ((-8 + 2*x)*Log[x] + (-16 + 2*x)*Log[x]*L og[5*x])*Log[Log[x]])/(x^3*Log[x]*Log[5*x]^2*Log[Log[x]]^2),x]
Output:
(2*(4 - x))/(x^2*Log[5*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.
\(\displaystyle \int \frac {(2 x-8) \log (5 x)+((2 x-8) \log (x)+(2 x-16) \log (5 x) \log (x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 (x-4)}{x^3 \log (x) \log (5 x) \log ^2(\log (x))}+\frac {2 (x+x \log (5 x)-8 \log (5 x)-4)}{x^3 \log ^2(5 x) \log (\log (x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {1}{x^3 \log (x) \log (5 x) \log ^2(\log (x))}dx-8 \int \frac {1}{x^3 \log ^2(5 x) \log (\log (x))}dx-16 \int \frac {1}{x^3 \log (5 x) \log (\log (x))}dx+2 \int \frac {1}{x^2 \log (x) \log (5 x) \log ^2(\log (x))}dx+2 \int \frac {1}{x^2 \log ^2(5 x) \log (\log (x))}dx+2 \int \frac {1}{x^2 \log (5 x) \log (\log (x))}dx\) |
Input:
Int[((-8 + 2*x)*Log[5*x] + ((-8 + 2*x)*Log[x] + (-16 + 2*x)*Log[x]*Log[5*x ])*Log[Log[x]])/(x^3*Log[x]*Log[5*x]^2*Log[Log[x]]^2),x]
Output:
$Aborted
Time = 1.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05
method | result | size |
parallelrisch | \(\frac {-12 x +48}{6 x^{2} \ln \left (5 x \right ) \ln \left (\ln \left (x \right )\right )}\) | \(22\) |
risch | \(\frac {4 i \left (x -4\right )}{x^{2} \left (-2 i \ln \left (5\right )-2 i \ln \left (x \right )\right ) \ln \left (\ln \left (x \right )\right )}\) | \(28\) |
Input:
int((((2*x-16)*ln(x)*ln(5*x)+(2*x-8)*ln(x))*ln(ln(x))+(2*x-8)*ln(5*x))/x^3 /ln(x)/ln(5*x)^2/ln(ln(x))^2,x,method=_RETURNVERBOSE)
Output:
1/6/x^2*(-12*x+48)/ln(5*x)/ln(ln(x))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx=-\frac {2 \, {\left (x - 4\right )}}{{\left (x^{2} \log \left (5\right ) + x^{2} \log \left (x\right )\right )} \log \left (\log \left (x\right )\right )} \] Input:
integrate((((2*x-16)*log(x)*log(5*x)+(2*x-8)*log(x))*log(log(x))+(2*x-8)*l og(5*x))/x^3/log(x)/log(5*x)^2/log(log(x))^2,x, algorithm="fricas")
Output:
-2*(x - 4)/((x^2*log(5) + x^2*log(x))*log(log(x)))
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx=\frac {8 - 2 x}{\left (x^{2} \log {\left (x \right )} + x^{2} \log {\left (5 \right )}\right ) \log {\left (\log {\left (x \right )} \right )}} \] Input:
integrate((((2*x-16)*ln(x)*ln(5*x)+(2*x-8)*ln(x))*ln(ln(x))+(2*x-8)*ln(5*x ))/x**3/ln(x)/ln(5*x)**2/ln(ln(x))**2,x)
Output:
(8 - 2*x)/((x**2*log(x) + x**2*log(5))*log(log(x)))
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx=-\frac {2 \, {\left (x - 4\right )}}{{\left (x^{2} \log \left (5\right ) + x^{2} \log \left (x\right )\right )} \log \left (\log \left (x\right )\right )} \] Input:
integrate((((2*x-16)*log(x)*log(5*x)+(2*x-8)*log(x))*log(log(x))+(2*x-8)*l og(5*x))/x^3/log(x)/log(5*x)^2/log(log(x))^2,x, algorithm="maxima")
Output:
-2*(x - 4)/((x^2*log(5) + x^2*log(x))*log(log(x)))
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx=-\frac {2 \, {\left (x - 4\right )}}{x^{2} \log \left (5\right ) \log \left (\log \left (x\right )\right ) + x^{2} \log \left (x\right ) \log \left (\log \left (x\right )\right )} \] Input:
integrate((((2*x-16)*log(x)*log(5*x)+(2*x-8)*log(x))*log(log(x))+(2*x-8)*l og(5*x))/x^3/log(x)/log(5*x)^2/log(log(x))^2,x, algorithm="giac")
Output:
-2*(x - 4)/(x^2*log(5)*log(log(x)) + x^2*log(x)*log(log(x)))
Time = 4.26 (sec) , antiderivative size = 513, normalized size of antiderivative = 24.43 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx =\text {Too large to display} \] Input:
int((log(log(x))*(log(x)*(2*x - 8) + log(5*x)*log(x)*(2*x - 16)) + log(5*x )*(2*x - 8))/(x^3*log(5*x)^2*log(log(x))^2*log(x)),x)
Output:
((x - 8*log(5*x) + 8*log(x) - x*(log(5*x) - log(x))^2 + x*(log(5*x) - log( x))^3 + 16*(log(5*x) - log(x))^2 - 32*(log(5*x) - log(x))^3 + x*(log(5*x) - log(x)) - 4)/x^2 + (log(x)^2*(32*log(x) - 32*log(5*x) - x + x*(log(5*x) - log(x)) + 16))/x^2 - (log(x)*(32*log(x) - 32*log(5*x) - x - 2*x*(log(5*x ) - log(x))^2 + 64*(log(5*x) - log(x))^2 + 2*x*(log(5*x) - log(x)) + 8))/x ^2)/log(5*x) - ((2*(x - 4))/(x^2*log(5*x)) + (2*log(log(x))*log(x)*(x - 8* log(5*x) + x*log(x) + x*(log(5*x) - log(x)) - 4))/(x^2*log(5*x)^2))/log(lo g(x)) - (32*log(5*x) - 32*log(x) - 32*(log(5*x) - log(x))^2 + x*(2*log(x) - 2*log(5*x) + (log(5*x) - log(x))^2))/x^2 - ((4*log(x) - 4*log(5*x) + x*( log(5*x) - log(x))^2 + x*(log(5*x) - log(x))^3 - 8*(log(5*x) - log(x))^2 - 16*(log(5*x) - log(x))^3 + x*(log(5*x) - log(x)))/x^2 + (log(x)^2*(16*log (x) - 16*log(5*x) - x + x*(log(5*x) - log(x)) + 8))/x^2 - (log(x)*(x - 2*x *(log(5*x) - log(x))^2 + 32*(log(5*x) - log(x))^2 - 4))/x^2)/(2*log(x)*(lo g(5*x) - log(x)) + log(x)^2 + (log(5*x) - log(x))^2) + (log(x)*(32*log(5*x ) - 32*log(x) + x*(log(x) - log(5*x) + 1) - 16))/x^2
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx=\frac {-2 x +8}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (5 x \right ) x^{2}} \] Input:
int((((2*x-16)*log(x)*log(5*x)+(2*x-8)*log(x))*log(log(x))+(2*x-8)*log(5*x ))/x^3/log(x)/log(5*x)^2/log(log(x))^2,x)
Output:
(2*( - x + 4))/(log(log(x))*log(5*x)*x**2)