Integrand size = 132, antiderivative size = 24 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 x \log \left (-\log \left (\frac {8 x}{5-x}+\log \left (\frac {x}{e}\right )\right )\right ) \] Output:
3*x*ln(-ln(ln(x/exp(1))+8*x/(5-x)))
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 x \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right ) \] Input:
Integrate[(75 + 90*x + 3*x^2 + (120*x - 24*x^2 + (75 - 30*x + 3*x^2)*Log[x /E])*Log[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]*Log[-Log[(-8*x + (-5 + x)*Lo g[x/E])/(-5 + x)]])/((40*x - 8*x^2 + (25 - 10*x + x^2)*Log[x/E])*Log[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]),x]
Output:
3*x*Log[-Log[(5 - 9*x + (-5 + x)*Log[x])/(-5 + x)]]
Time = 1.86 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, 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.
\(\displaystyle \int \frac {3 x^2+\left (-24 x^2+\left (3 x^2-30 x+75\right ) \log \left (\frac {x}{e}\right )+120 x\right ) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right ) \log \left (-\log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )\right )+90 x+75}{\left (-8 x^2+\left (x^2-10 x+25\right ) \log \left (\frac {x}{e}\right )+40 x\right ) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-3 x^2-\left (-24 x^2+\left (3 x^2-30 x+75\right ) \log \left (\frac {x}{e}\right )+120 x\right ) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right ) \log \left (-\log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )\right )-90 x-75}{(5-x) (-9 x+x \log (x)-5 \log (x)+5) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 \left (x^2+30 x+25\right )}{(x-5) (-9 x+x \log (x)-5 \log (x)+5) \log \left (\frac {-9 x+(x-5) \log (x)+5}{x-5}\right )}+3 \log \left (-\log \left (\frac {-9 x+(x-5) \log (x)+5}{x-5}\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 x \log \left (-\log \left (-\frac {-9 x-(5-x) \log (x)+5}{5-x}\right )\right )\) |
Input:
Int[(75 + 90*x + 3*x^2 + (120*x - 24*x^2 + (75 - 30*x + 3*x^2)*Log[x/E])*L og[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]*Log[-Log[(-8*x + (-5 + x)*Log[x/E] )/(-5 + x)]])/((40*x - 8*x^2 + (25 - 10*x + x^2)*Log[x/E])*Log[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]),x]
Output:
3*x*Log[-Log[-((5 - 9*x - (5 - x)*Log[x])/(5 - x))]]
Time = 17.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(3 x \ln \left (-\ln \left (\frac {\left (-5+x \right ) \ln \left (x \,{\mathrm e}^{-1}\right )-8 x}{-5+x}\right )\right )\) | \(29\) |
risch | \(3 \ln \left (\ln \left (-5+x \right )-\ln \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )+\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )}{-5+x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )}{-5+x}\right )+\operatorname {csgn}\left (\frac {i}{-5+x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )}{-5+x}\right )+\operatorname {csgn}\left (i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )\right )\right )}{2}\right ) x\) | \(148\) |
Input:
int((((3*x^2-30*x+75)*ln(x/exp(1))-24*x^2+120*x)*ln(((-5+x)*ln(x/exp(1))-8 *x)/(-5+x))*ln(-ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)))+3*x^2+90*x+75)/((x^2 -10*x+25)*ln(x/exp(1))-8*x^2+40*x)/ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)),x, method=_RETURNVERBOSE)
Output:
3*x*ln(-ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)))
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 \, x \log \left (-\log \left (\frac {{\left (x - 5\right )} \log \left (x e^{\left (-1\right )}\right ) - 8 \, x}{x - 5}\right )\right ) \] Input:
integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/ exp(1))-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90 *x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))- 8*x)/(-5+x)),x, algorithm="fricas")
Output:
3*x*log(-log(((x - 5)*log(x*e^(-1)) - 8*x)/(x - 5)))
Exception generated. \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((3*x**2-30*x+75)*ln(x/exp(1))-24*x**2+120*x)*ln(((-5+x)*ln(x/e xp(1))-8*x)/(-5+x))*ln(-ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)))+3*x**2+90*x+ 75)/((x**2-10*x+25)*ln(x/exp(1))-8*x**2+40*x)/ln(((-5+x)*ln(x/exp(1))-8*x) /(-5+x)),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 \, x \log \left (-\log \left ({\left (x - 5\right )} \log \left (x\right ) - 9 \, x + 5\right ) + \log \left (x - 5\right )\right ) \] Input:
integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/ exp(1))-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90 *x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))- 8*x)/(-5+x)),x, algorithm="maxima")
Output:
3*x*log(-log((x - 5)*log(x) - 9*x + 5) + log(x - 5))
Time = 0.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 \, x \log \left (-\log \left (x \log \left (x\right ) - 9 \, x - 5 \, \log \left (x\right ) + 5\right ) + \log \left (x - 5\right )\right ) \] Input:
integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/ exp(1))-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90 *x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))- 8*x)/(-5+x)),x, algorithm="giac")
Output:
3*x*log(-log(x*log(x) - 9*x - 5*log(x) + 5) + log(x - 5))
Time = 5.81 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3\,x\,\ln \left (-\ln \left (-\frac {8\,x-\ln \left (x\,{\mathrm {e}}^{-1}\right )\,\left (x-5\right )}{x-5}\right )\right ) \] Input:
int((90*x + 3*x^2 + log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5))*log(-log( -(8*x - log(x*exp(-1))*(x - 5))/(x - 5)))*(120*x + log(x*exp(-1))*(3*x^2 - 30*x + 75) - 24*x^2) + 75)/(log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5))* (40*x + log(x*exp(-1))*(x^2 - 10*x + 25) - 8*x^2)),x)
Output:
3*x*log(-log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5)))
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 4.00 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=90 \,\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x}{e}\right ) x -5 \,\mathrm {log}\left (\frac {x}{e}\right )-8 x}{-5+x}\right )\right )+3 \,\mathrm {log}\left (-\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x}{e}\right ) x -5 \,\mathrm {log}\left (\frac {x}{e}\right )-8 x}{-5+x}\right )\right ) x -90 \,\mathrm {log}\left (-\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x}{e}\right ) x -5 \,\mathrm {log}\left (\frac {x}{e}\right )-8 x}{-5+x}\right )\right ) \] Input:
int((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/exp(1) )-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90*x+75) /((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))-8*x)/( -5+x)),x)
Output:
3*(30*log(log((log(x/e)*x - 5*log(x/e) - 8*x)/(x - 5))) + log( - log((log( x/e)*x - 5*log(x/e) - 8*x)/(x - 5)))*x - 30*log( - log((log(x/e)*x - 5*log (x/e) - 8*x)/(x - 5))))