Integrand size = 156, antiderivative size = 23 \[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x} \] Output:
ln(x*ln(-ln(x)+exp(-2+x)+9/4*x))/x
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x} \] Input:
Integrate[(4 - 9*x - 4*E^(-2 + x)*x + (-4*E^(-2 + x) - 9*x + 4*Log[x])*Log [(4*E^(-2 + x) + 9*x - 4*Log[x])/4] + (4*E^(-2 + x) + 9*x - 4*Log[x])*Log[ (4*E^(-2 + x) + 9*x - 4*Log[x])/4]*Log[x*Log[(4*E^(-2 + x) + 9*x - 4*Log[x ])/4]])/((-4*E^(-2 + x)*x^2 - 9*x^3 + 4*x^2*Log[x])*Log[(4*E^(-2 + x) + 9* x - 4*Log[x])/4]),x]
Output:
Log[x*Log[E^(-2 + x) + (9*x)/4 - Log[x]]]/x
Time = 6.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {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.
| \(\displaystyle \int \frac {-4 e^{x-2} x-9 x+\left (-9 x-4 e^{x-2}+4 \log (x)\right ) \log \left (\frac {1}{4} \left (9 x+4 e^{x-2}-4 \log (x)\right )\right )+\left (9 x+4 e^{x-2}-4 \log (x)\right ) \log \left (\frac {1}{4} \left (9 x+4 e^{x-2}-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (9 x+4 e^{x-2}-4 \log (x)\right )\right )\right )+4}{\left (-9 x^3-4 e^{x-2} x^2+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (9 x+4 e^{x-2}-4 \log (x)\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x+\log \left (\frac {9 x}{4}+e^{x-2}-\log (x)\right )-\log \left (\frac {9 x}{4}+e^{x-2}-\log (x)\right ) \log \left (x \log \left (\frac {9 x}{4}+e^{x-2}-\log (x)\right )\right )}{x^2 \log \left (\frac {9 x}{4}+e^{x-2}-\log (x)\right )}-\frac {e^2 \left (9 x^2-9 x-4 x \log (x)+4\right )}{x^2 \left (9 e^2 x+4 e^x-4 e^2 \log (x)\right ) \log \left (\frac {9 x}{4}+e^{x-2}-\log (x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\log \left (x \log \left (\frac {9 x}{4}+e^{x-2}-\log (x)\right )\right )}{x}\) |
Input:
Int[(4 - 9*x - 4*E^(-2 + x)*x + (-4*E^(-2 + x) - 9*x + 4*Log[x])*Log[(4*E^ (-2 + x) + 9*x - 4*Log[x])/4] + (4*E^(-2 + x) + 9*x - 4*Log[x])*Log[(4*E^( -2 + x) + 9*x - 4*Log[x])/4]*Log[x*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4]] )/((-4*E^(-2 + x)*x^2 - 9*x^3 + 4*x^2*Log[x])*Log[(4*E^(-2 + x) + 9*x - 4* Log[x])/4]),x]
Output:
Log[x*Log[E^(-2 + x) + (9*x)/4 - Log[x]]]/x
Time = 12.62 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
| method | result | size |
| parallelrisch | \(\frac {\ln \left (x \ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right )}{x}\) | \(21\) |
| risch | \(\frac {\ln \left (\ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right )}{x}+\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right ) {\operatorname {csgn}\left (i x \ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right ) \operatorname {csgn}\left (i x \ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right ) \operatorname {csgn}\left (i x \right )-i \pi {\operatorname {csgn}\left (i x \ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right )}^{3}+i \pi {\operatorname {csgn}\left (i x \ln \left (-\ln \left (x \right )+{\mathrm e}^{-2+x}+\frac {9 x}{4}\right )\right )}^{2} \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )}{2 x}\) | \(168\) |
Input:
int(((-4*ln(x)+4*exp(-2+x)+9*x)*ln(-ln(x)+exp(-2+x)+9/4*x)*ln(x*ln(-ln(x)+ exp(-2+x)+9/4*x))+(4*ln(x)-4*exp(-2+x)-9*x)*ln(-ln(x)+exp(-2+x)+9/4*x)-4*x *exp(-2+x)-9*x+4)/(4*x^2*ln(x)-4*x^2*exp(-2+x)-9*x^3)/ln(-ln(x)+exp(-2+x)+ 9/4*x),x,method=_RETURNVERBOSE)
Output:
ln(x*ln(-ln(x)+exp(-2+x)+9/4*x))/x
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\frac {\log \left (x \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \left (x\right )\right )\right )}{x} \] Input:
integrate(((-4*log(x)+4*exp(-2+x)+9*x)*log(-log(x)+exp(-2+x)+9/4*x)*log(x* log(-log(x)+exp(-2+x)+9/4*x))+(4*log(x)-4*exp(-2+x)-9*x)*log(-log(x)+exp(- 2+x)+9/4*x)-4*x*exp(-2+x)-9*x+4)/(4*x^2*log(x)-4*x^2*exp(-2+x)-9*x^3)/log( -log(x)+exp(-2+x)+9/4*x),x, algorithm="fricas")
Output:
log(x*log(9/4*x + e^(x - 2) - log(x)))/x
Time = 43.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\frac {\log {\left (x \log {\left (\frac {9 x}{4} + e^{x - 2} - \log {\left (x \right )} \right )} \right )}}{x} \] Input:
integrate(((-4*ln(x)+4*exp(-2+x)+9*x)*ln(-ln(x)+exp(-2+x)+9/4*x)*ln(x*ln(- ln(x)+exp(-2+x)+9/4*x))+(4*ln(x)-4*exp(-2+x)-9*x)*ln(-ln(x)+exp(-2+x)+9/4* x)-4*x*exp(-2+x)-9*x+4)/(4*x**2*ln(x)-4*x**2*exp(-2+x)-9*x**3)/ln(-ln(x)+e xp(-2+x)+9/4*x),x)
Output:
log(x*log(9*x/4 + exp(x - 2) - log(x)))/x
Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\frac {\log \left (x\right ) + \log \left (-2 \, \log \left (2\right ) + \log \left (9 \, x e^{2} - 4 \, e^{2} \log \left (x\right ) + 4 \, e^{x}\right ) - 2\right )}{x} \] Input:
integrate(((-4*log(x)+4*exp(-2+x)+9*x)*log(-log(x)+exp(-2+x)+9/4*x)*log(x* log(-log(x)+exp(-2+x)+9/4*x))+(4*log(x)-4*exp(-2+x)-9*x)*log(-log(x)+exp(- 2+x)+9/4*x)-4*x*exp(-2+x)-9*x+4)/(4*x^2*log(x)-4*x^2*exp(-2+x)-9*x^3)/log( -log(x)+exp(-2+x)+9/4*x),x, algorithm="maxima")
Output:
(log(x) + log(-2*log(2) + log(9*x*e^2 - 4*e^2*log(x) + 4*e^x) - 2))/x
\[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\int { -\frac {{\left (9 \, x + 4 \, e^{\left (x - 2\right )} - 4 \, \log \left (x\right )\right )} \log \left (x \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \left (x\right )\right )\right ) \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \left (x\right )\right ) - 4 \, x e^{\left (x - 2\right )} - {\left (9 \, x + 4 \, e^{\left (x - 2\right )} - 4 \, \log \left (x\right )\right )} \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \left (x\right )\right ) - 9 \, x + 4}{{\left (9 \, x^{3} + 4 \, x^{2} e^{\left (x - 2\right )} - 4 \, x^{2} \log \left (x\right )\right )} \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \left (x\right )\right )} \,d x } \] Input:
integrate(((-4*log(x)+4*exp(-2+x)+9*x)*log(-log(x)+exp(-2+x)+9/4*x)*log(x* log(-log(x)+exp(-2+x)+9/4*x))+(4*log(x)-4*exp(-2+x)-9*x)*log(-log(x)+exp(- 2+x)+9/4*x)-4*x*exp(-2+x)-9*x+4)/(4*x^2*log(x)-4*x^2*exp(-2+x)-9*x^3)/log( -log(x)+exp(-2+x)+9/4*x),x, algorithm="giac")
Output:
integrate(-((9*x + 4*e^(x - 2) - 4*log(x))*log(x*log(9/4*x + e^(x - 2) - l og(x)))*log(9/4*x + e^(x - 2) - log(x)) - 4*x*e^(x - 2) - (9*x + 4*e^(x - 2) - 4*log(x))*log(9/4*x + e^(x - 2) - log(x)) - 9*x + 4)/((9*x^3 + 4*x^2* e^(x - 2) - 4*x^2*log(x))*log(9/4*x + e^(x - 2) - log(x))), x)
Time = 3.84 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\frac {\ln \left (x\,\ln \left (\frac {9\,x}{4}-\ln \left (x\right )+{\mathrm {e}}^{-2}\,{\mathrm {e}}^x\right )\right )}{x} \] Input:
int((9*x + 4*x*exp(x - 2) + log((9*x)/4 + exp(x - 2) - log(x))*(9*x + 4*ex p(x - 2) - 4*log(x)) - log((9*x)/4 + exp(x - 2) - log(x))*log(x*log((9*x)/ 4 + exp(x - 2) - log(x)))*(9*x + 4*exp(x - 2) - 4*log(x)) - 4)/(log((9*x)/ 4 + exp(x - 2) - log(x))*(4*x^2*exp(x - 2) - 4*x^2*log(x) + 9*x^3)),x)
Output:
log(x*log((9*x)/4 - log(x) + exp(-2)*exp(x)))/x
Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {4-9 x-4 e^{-2+x} x+\left (-4 e^{-2+x}-9 x+4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )+\left (4 e^{-2+x}+9 x-4 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right ) \log \left (x \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )\right )}{\left (-4 e^{-2+x} x^2-9 x^3+4 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^{-2+x}+9 x-4 \log (x)\right )\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {4 e^{x}-4 \,\mathrm {log}\left (x \right ) e^{2}+9 e^{2} x}{4 e^{2}}\right ) x \right )}{x} \] Input:
int(((-4*log(x)+4*exp(-2+x)+9*x)*log(-log(x)+exp(-2+x)+9/4*x)*log(x*log(-l og(x)+exp(-2+x)+9/4*x))+(4*log(x)-4*exp(-2+x)-9*x)*log(-log(x)+exp(-2+x)+9 /4*x)-4*x*exp(-2+x)-9*x+4)/(4*x^2*log(x)-4*x^2*exp(-2+x)-9*x^3)/log(-log(x )+exp(-2+x)+9/4*x),x)
Output:
log(log((4*e**x - 4*log(x)*e**2 + 9*e**2*x)/(4*e**2))*x)/x