Integrand size = 71, antiderivative size = 26 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log \left (e^{x (1+x)} \left (\frac {1}{5}+e^x+x+\frac {x^2}{4}\right )\right )} \] Output:
1/ln(exp(x*(1+x))*(1/5+1/4*x^2+x+exp(x)))
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=-\frac {1}{\log (20)-\log \left (e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \] Input:
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]
Output:
-(Log[20] - Log[E^(x + x^2)*(4 + 20*E^x + 20*x + 5*x^2)])^(-1)
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 |
\(\displaystyle -2 \int \frac {1}{\log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx-2 \int \frac {x}{\log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx-16 \int \frac {1}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx+10 \int \frac {x}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx+5 \int \frac {x^2}{\left (5 x^2+20 x+20 e^x+4\right ) \log ^2\left (\frac {1}{20} e^{x^2+x} \left (5 x^2+20 x+20 e^x+4\right )\right )}dx\) |
Input:
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]
Output:
$Aborted
Time = 1.00 (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 | \(-\frac {2}{4 \ln \left (2\right )-2 \ln \left ({\mathrm e}^{\left (1+x \right ) x}\right )-2 \ln \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )+i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )-i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{\left (1+x \right ) x} \left (x^{2}+4 x +4 \,{\mathrm e}^{x}+\frac {4}{5}\right )\right )^{3}}\) | \(196\) |
Input:
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)
Output:
1/ln(1/20*(20*exp(x)+5*x^2+20*x+4)*exp(x^2+x))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log \left (\frac {1}{20} \, {\left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} e^{\left (x^{2} + x\right )}\right )} \] Input:
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="fricas")
Output:
1/log(1/20*(5*x^2 + 20*x + 20*e^x + 4)*e^(x^2 + x))
Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{\log {\left (\left (\frac {x^{2}}{4} + x + e^{x} + \frac {1}{5}\right ) e^{x^{2} + x} \right )}} \] Input:
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)
Output:
1/log((x**2/4 + x + exp(x) + 1/5)*exp(x**2 + x))
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x^{2} + x - \log \left (5\right ) - 2 \, \log \left (2\right ) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \] Input:
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="maxima")
Output:
1/(x^2 + x - log(5) - 2*log(2) + log(5*x^2 + 20*x + 20*e^x + 4))
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x^{2} + x - \log \left (20\right ) + \log \left (5 \, x^{2} + 20 \, x + 20 \, e^{x} + 4\right )} \] Input:
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="giac")
Output:
1/(x^2 + x - log(20) + log(5*x^2 + 20*x + 20*e^x + 4))
Time = 4.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx=\frac {1}{x+\ln \left (x+{\mathrm {e}}^x+\frac {x^2}{4}+\frac {1}{5}\right )+x^2} \] Input:
int(-(38*x + exp(x)*(40*x + 40) + 45*x^2 + 10*x^3 + 24)/(log((exp(x + x^2) *(20*x + 20*exp(x) + 5*x^2 + 4))/20)^2*(20*x + 20*exp(x) + 5*x^2 + 4)),x)
Output:
1/(x + log(x + exp(x) + x^2/4 + 1/5) + x^2)
\[ \int \frac {-24+e^x (-40-40 x)-38 x-45 x^2-10 x^3}{\left (4+20 e^x+20 x+5 x^2\right ) \log ^2\left (\frac {1}{20} e^{x+x^2} \left (4+20 e^x+20 x+5 x^2\right )\right )} \, dx =\text {Too large to display} \] Input:
int(((-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)
Output:
- 40*int(e**x/(20*e**x*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 2 0*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2 + 5*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2*x**2 + 20*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2*x + 4*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x* *2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2),x) - 10*int(x**3/(20*e* *x*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4 *e**(x**2 + x))/20)**2 + 5*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2*x**2 + 20*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)* *2*x + 4*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x) *x + 4*e**(x**2 + x))/20)**2),x) - 45*int(x**2/(20*e**x*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)** 2 + 5*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2*x**2 + 20*log((20*e**(x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2*x + 4*log((20*e**( x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x)) /20)**2),x) - 40*int((e**x*x)/(20*e**x*log((20*e**(x**2 + 2*x) + 5*e**(x** 2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 + x))/20)**2 + 5*log((20*e** (x**2 + 2*x) + 5*e**(x**2 + x)*x**2 + 20*e**(x**2 + x)*x + 4*e**(x**2 +...