Integrand size = 71, antiderivative size = 26 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log \left (e^{x (1+x)} \left (\frac {1}{5}+e^x+x+\frac {x^2}{4}\right )\right )} \]
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=-\frac {1}{\log (20)-\log \left (e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \]
Integrate[(-24 + E^x*(-40 - 40*x) - 38*x - 45*x^2 - 10*x^3)/((4 + 20*E^x + 20*x + 5*x^2)*Log[(E^(x + x^2)*(4 + 20*E^x + 20*x + 5*x^2))/20]^2),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.
\(\displaystyle \int \frac {-10 x^3-45 x^2-38 x+e^x (-40 x-40)-24}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 x^2+10 x-16}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}-\frac {2 (x+1)}{\log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{\log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx-2 \int \frac {x}{\log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx-16 \int \frac {1}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx+10 \int \frac {x}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx+5 \int \frac {x^2}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx\) |
Int[(-24 + E^x*(-40 - 40*x) - 38*x - 45*x^2 - 10*x^3)/((4 + 20*E^x + 20*x + 5*x^2)*Log[(E^(x + x^2)*(4 + 20*E^x + 20*x + 5*x^2))/20]^2),x]
3.18.23.3.1 Defintions of rubi rules used
Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {1}{\ln \left (\frac {\left (20 \,{\mathrm e}^{x}+5 x^{2}+20 x +4\right ) {\mathrm e}^{x^{2}+x}}{20}\right )}\) | \(26\) |
risch | \(-\frac {2}{4 \ln \left (2\right )-2 \ln \left ({\mathrm e}^{\left (1+x \right ) x}\right )-2 \ln \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )+i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )-i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{3}}\) | \(196\) |
int(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+4)/ln( 1/20*(20*exp(x)+5*x^2+20*x+4)*exp(x^2+x))^2,x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log \left (\frac {1}{20} \, {\left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} e^{\left (x^{2} + x\right )}\right )} \]
integrate(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+ 4)/log(1/20*(20*exp(x)+5*x^2+20*x+4)*exp(x^2+x))^2,x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log {\left (\left (\frac {x^{2}}{4} + x + e^{x} + \frac {1}{5}\right ) e^{x^{2} + x} \right )}} \]
integrate(((-40*x-40)*exp(x)-10*x**3-45*x**2-38*x-24)/(20*exp(x)+5*x**2+20 *x+4)/ln(1/20*(20*exp(x)+5*x**2+20*x+4)*exp(x**2+x))**2,x)
Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x^{2} + x - \log \left (5\right ) - 2 \, \log \left (2\right ) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \]
integrate(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+ 4)/log(1/20*(20*exp(x)+5*x^2+20*x+4)*exp(x^2+x))^2,x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x^{2} + x - \log \left (20\right ) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \]
integrate(((-40*x-40)*exp(x)-10*x^3-45*x^2-38*x-24)/(20*exp(x)+5*x^2+20*x+ 4)/log(1/20*(20*exp(x)+5*x^2+20*x+4)*exp(x^2+x))^2,x, algorithm=\
Time = 0.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x+\ln \left (x+{\mathrm {e}}^x+\frac {x^2}{4}+\frac {1}{5}\right )+x^2} \]