Integrand size = 81, antiderivative size = 29 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=3+x-\frac {x^2-\log \left (5+e^{-6 (4-2 x) x} x\right )}{x} \] Output:
x+3-(x^2-ln(5+x/exp(6*(4-2*x)*x)))/x
Time = 0.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=\frac {\log \left (5+e^{-24 x+12 x^2} x\right )}{x} \] Input:
Integrate[(x - 24*x^2 + 24*x^3 + (-5*E^(24*x - 12*x^2) - x)*Log[E^(-24*x + 12*x^2)*(5*E^(24*x - 12*x^2) + x)])/(5*E^(24*x - 12*x^2)*x^2 + x^3),x]
Output:
Log[5 + E^(-24*x + 12*x^2)*x]/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.
\(\displaystyle \int \frac {24 x^3-24 x^2+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{12 x^2-24 x} \left (5 e^{24 x-12 x^2}+x\right )\right )+x}{x^3+5 e^{24 x-12 x^2} x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {24 x^2-24 x-\log \left (e^{12 (x-2) x} x+5\right )+1}{x^2}-\frac {5 e^{24 x} \left (24 x^2-24 x+1\right )}{x^2 \left (e^{12 x^2} x+5 e^{24 x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -120 \int \frac {e^{24 x}}{e^{12 x^2} x+5 e^{24 x}}dx-5 \int \frac {e^{24 x}}{x^2 \left (e^{12 x^2} x+5 e^{24 x}\right )}dx+120 \int \frac {e^{24 x}}{x \left (e^{12 x^2} x+5 e^{24 x}\right )}dx+120 \int \frac {1}{e^{12 (x-2) x} x+5}dx+24 \int \frac {e^{12 (x-2) x}}{e^{12 (x-2) x} x+5}dx-\int \frac {e^{12 (x-2) x}}{x \left (e^{12 (x-2) x} x+5\right )}dx-\frac {1}{x}-24 \log (x)+\frac {\log \left (e^{-12 (2-x) x} x+5\right )}{x}\) |
Input:
Int[(x - 24*x^2 + 24*x^3 + (-5*E^(24*x - 12*x^2) - x)*Log[E^(-24*x + 12*x^ 2)*(5*E^(24*x - 12*x^2) + x)])/(5*E^(24*x - 12*x^2)*x^2 + x^3),x]
Output:
$Aborted
Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
method | result | size |
norman | \(\frac {\ln \left (\left (5 \,{\mathrm e}^{-12 x^{2}+24 x}+x \right ) {\mathrm e}^{12 x^{2}-24 x}\right )}{x}\) | \(33\) |
parallelrisch | \(\frac {\ln \left (\left (5 \,{\mathrm e}^{-12 x^{2}+24 x}+x \right ) {\mathrm e}^{12 x^{2}-24 x}\right )}{x}\) | \(33\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{-12 \left (-2+x \right ) x}\right )}{x}+\frac {-i \pi \,\operatorname {csgn}\left (i \left (5 \,{\mathrm e}^{-12 \left (-2+x \right ) x}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{12 \left (-2+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{12 \left (-2+x \right ) x} \left (5 \,{\mathrm e}^{-12 \left (-2+x \right ) x}+x \right )\right )+i \pi \,\operatorname {csgn}\left (i \left (5 \,{\mathrm e}^{-12 \left (-2+x \right ) x}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{12 \left (-2+x \right ) x} \left (5 \,{\mathrm e}^{-12 \left (-2+x \right ) x}+x \right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{12 \left (-2+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{12 \left (-2+x \right ) x} \left (5 \,{\mathrm e}^{-12 \left (-2+x \right ) x}+x \right )\right )^{2}-i \pi \operatorname {csgn}\left (i {\mathrm e}^{12 \left (-2+x \right ) x} \left (5 \,{\mathrm e}^{-12 \left (-2+x \right ) x}+x \right )\right )^{3}+2 \ln \left (5 \,{\mathrm e}^{-12 \left (-2+x \right ) x}+x \right )}{2 x}\) | \(197\) |
Input:
int(((-5*exp(-12*x^2+24*x)-x)*ln((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x) )+24*x^3-24*x^2+x)/(5*x^2*exp(-12*x^2+24*x)+x^3),x,method=_RETURNVERBOSE)
Output:
ln((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x))/x
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=\frac {\log \left (x e^{\left (12 \, x^{2} - 24 \, x\right )} + 5\right )}{x} \] Input:
integrate(((-5*exp(-12*x^2+24*x)-x)*log((5*exp(-12*x^2+24*x)+x)/exp(-12*x^ 2+24*x))+24*x^3-24*x^2+x)/(5*x^2*exp(-12*x^2+24*x)+x^3),x, algorithm="fric as")
Output:
log(x*e^(12*x^2 - 24*x) + 5)/x
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=\frac {\log {\left (\left (x + 5 e^{- 12 x^{2} + 24 x}\right ) e^{12 x^{2} - 24 x} \right )}}{x} \] Input:
integrate(((-5*exp(-12*x**2+24*x)-x)*ln((5*exp(-12*x**2+24*x)+x)/exp(-12*x **2+24*x))+24*x**3-24*x**2+x)/(5*x**2*exp(-12*x**2+24*x)+x**3),x)
Output:
log((x + 5*exp(-12*x**2 + 24*x))*exp(12*x**2 - 24*x))/x
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=\frac {\log \left (x e^{\left (12 \, x^{2}\right )} + 5 \, e^{\left (24 \, x\right )}\right )}{x} \] Input:
integrate(((-5*exp(-12*x^2+24*x)-x)*log((5*exp(-12*x^2+24*x)+x)/exp(-12*x^ 2+24*x))+24*x^3-24*x^2+x)/(5*x^2*exp(-12*x^2+24*x)+x^3),x, algorithm="maxi ma")
Output:
log(x*e^(12*x^2) + 5*e^(24*x))/x
Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=\frac {\log \left (x e^{\left (12 \, x^{2} - 24 \, x\right )} + 5\right )}{x} \] Input:
integrate(((-5*exp(-12*x^2+24*x)-x)*log((5*exp(-12*x^2+24*x)+x)/exp(-12*x^ 2+24*x))+24*x^3-24*x^2+x)/(5*x^2*exp(-12*x^2+24*x)+x^3),x, algorithm="giac ")
Output:
log(x*e^(12*x^2 - 24*x) + 5)/x
Time = 2.83 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=\frac {\ln \left (x\,{\mathrm {e}}^{-24\,x}\,{\mathrm {e}}^{12\,x^2}+5\right )}{x} \] Input:
int((x - log(exp(12*x^2 - 24*x)*(x + 5*exp(24*x - 12*x^2)))*(x + 5*exp(24* x - 12*x^2)) - 24*x^2 + 24*x^3)/(5*x^2*exp(24*x - 12*x^2) + x^3),x)
Output:
log(x*exp(-24*x)*exp(12*x^2) + 5)/x
Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx=\frac {\mathrm {log}\left (\frac {e^{12 x^{2}} x +5 e^{24 x}}{e^{24 x}}\right )}{x} \] Input:
int(((-5*exp(-12*x^2+24*x)-x)*log((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x ))+24*x^3-24*x^2+x)/(5*x^2*exp(-12*x^2+24*x)+x^3),x)
Output:
log((e**(12*x**2)*x + 5*e**(24*x))/e**(24*x))/x