Integrand size = 101, antiderivative size = 29 \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=-x+e^{-x} \left (-\frac {1}{5}+\frac {2}{x}+x \log \left (e^2+4 x\right )\right ) \] Output:
(2/x-1/5+x*ln(exp(2)+4*x))/exp(x)-x
Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=\frac {1}{5} \left (e^{-x} \left (-1+\frac {10}{x}\right )-5 x+5 e^{-x} x \log \left (e^2+4 x\right )\right ) \] Input:
Integrate[(-40*x - 40*x^2 + 24*x^3 + E^2*(-10 - 10*x + x^2) + E^x*(-5*E^2* x^2 - 20*x^3) + (20*x^3 - 20*x^4 + E^2*(5*x^2 - 5*x^3))*Log[E^2 + 4*x])/(E ^x*(5*E^2*x^2 + 20*x^3)),x]
Output:
((-1 + 10/x)/E^x - 5*x + (5*x*Log[E^2 + 4*x])/E^x)/5
Time = 2.85 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2026, 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.
\(\displaystyle \int \frac {e^{-x} \left (24 x^3-40 x^2+e^2 \left (x^2-10 x-10\right )+e^x \left (-20 x^3-5 e^2 x^2\right )+\left (-20 x^4+20 x^3+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (4 x+e^2\right )-40 x\right )}{20 x^3+5 e^2 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (24 x^3-40 x^2+e^2 \left (x^2-10 x-10\right )+e^x \left (-20 x^3-5 e^2 x^2\right )+\left (-20 x^4+20 x^3+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (4 x+e^2\right )-40 x\right )}{x^2 \left (20 x+5 e^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-x} \left (-20 x^4 \log \left (4 x+e^2\right )+24 x^3+20 \left (1-\frac {e^2}{4}\right ) x^3 \log \left (4 x+e^2\right )-40 \left (1-\frac {e^2}{40}\right ) x^2+5 e^2 x^2 \log \left (4 x+e^2\right )-40 \left (1+\frac {e^2}{4}\right ) x-10 e^2\right )}{5 x^2 \left (4 x+e^2\right )}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x-\frac {e^{-x}}{5}+\frac {2 e^{-x}}{x}+e^{-x} \log \left (4 x+e^2\right )-e^{-x} (1-x) \log \left (4 x+e^2\right )\) |
Input:
Int[(-40*x - 40*x^2 + 24*x^3 + E^2*(-10 - 10*x + x^2) + E^x*(-5*E^2*x^2 - 20*x^3) + (20*x^3 - 20*x^4 + E^2*(5*x^2 - 5*x^3))*Log[E^2 + 4*x])/(E^x*(5* E^2*x^2 + 20*x^3)),x]
Output:
-1/5*1/E^x + 2/(E^x*x) - x + Log[E^2 + 4*x]/E^x - ((1 - x)*Log[E^2 + 4*x]) /E^x
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 3.88 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
parts | \(-x +\frac {\left (2+x^{2} \ln \left ({\mathrm e}^{2}+4 x \right )-\frac {x}{5}\right ) {\mathrm e}^{-x}}{x}\) | \(29\) |
default | \(-x +\frac {\left (10-x +5 x^{2} \ln \left ({\mathrm e}^{2}+4 x \right )\right ) {\mathrm e}^{-x}}{5 x}\) | \(31\) |
norman | \(\frac {\left (2+x^{2} \ln \left ({\mathrm e}^{2}+4 x \right )-\frac {x}{5}-{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}}{x}\) | \(32\) |
risch | \(x \,{\mathrm e}^{-x} \ln \left ({\mathrm e}^{2}+4 x \right )-\frac {\left (5 \,{\mathrm e}^{x} x^{2}+x -10\right ) {\mathrm e}^{-x}}{5 x}\) | \(34\) |
parallelrisch | \(\frac {\left (160+40 x \,{\mathrm e}^{2} {\mathrm e}^{x}-80 \,{\mathrm e}^{x} x^{2}+80 x^{2} \ln \left ({\mathrm e}^{2}+4 x \right )-16 x \right ) {\mathrm e}^{-x}}{80 x}\) | \(41\) |
Input:
int((((-5*x^3+5*x^2)*exp(2)-20*x^4+20*x^3)*ln(exp(2)+4*x)+(-5*x^2*exp(2)-2 0*x^3)*exp(x)+(x^2-10*x-10)*exp(2)+24*x^3-40*x^2-40*x)/(5*x^2*exp(2)+20*x^ 3)/exp(x),x,method=_RETURNVERBOSE)
Output:
-x+(2+x^2*ln(exp(2)+4*x)-1/5*x)/x/exp(x)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=-\frac {{\left (5 \, x^{2} e^{x} - 5 \, x^{2} \log \left (4 \, x + e^{2}\right ) + x - 10\right )} e^{\left (-x\right )}}{5 \, x} \] Input:
integrate((((-5*x^3+5*x^2)*exp(2)-20*x^4+20*x^3)*log(exp(2)+4*x)+(-5*x^2*e xp(2)-20*x^3)*exp(x)+(x^2-10*x-10)*exp(2)+24*x^3-40*x^2-40*x)/(5*x^2*exp(2 )+20*x^3)/exp(x),x, algorithm="fricas")
Output:
-1/5*(5*x^2*e^x - 5*x^2*log(4*x + e^2) + x - 10)*e^(-x)/x
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=- x + \frac {\left (5 x^{2} \log {\left (4 x + e^{2} \right )} - x + 10\right ) e^{- x}}{5 x} \] Input:
integrate((((-5*x**3+5*x**2)*exp(2)-20*x**4+20*x**3)*ln(exp(2)+4*x)+(-5*x* *2*exp(2)-20*x**3)*exp(x)+(x**2-10*x-10)*exp(2)+24*x**3-40*x**2-40*x)/(5*x **2*exp(2)+20*x**3)/exp(x),x)
Output:
-x + (5*x**2*log(4*x + exp(2)) - x + 10)*exp(-x)/(5*x)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=\frac {5 \, x^{2} e^{\left (-x\right )} \log \left (4 \, x + e^{2}\right ) - 5 \, x^{2} - {\left (x - 10\right )} e^{\left (-x\right )}}{5 \, x} \] Input:
integrate((((-5*x^3+5*x^2)*exp(2)-20*x^4+20*x^3)*log(exp(2)+4*x)+(-5*x^2*e xp(2)-20*x^3)*exp(x)+(x^2-10*x-10)*exp(2)+24*x^3-40*x^2-40*x)/(5*x^2*exp(2 )+20*x^3)/exp(x),x, algorithm="maxima")
Output:
1/5*(5*x^2*e^(-x)*log(4*x + e^2) - 5*x^2 - (x - 10)*e^(-x))/x
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=-\frac {{\left (5 \, x^{2} e^{x} - 5 \, x^{2} \log \left (4 \, x + e^{2}\right ) + x - 10\right )} e^{\left (-x\right )}}{5 \, x} \] Input:
integrate((((-5*x^3+5*x^2)*exp(2)-20*x^4+20*x^3)*log(exp(2)+4*x)+(-5*x^2*e xp(2)-20*x^3)*exp(x)+(x^2-10*x-10)*exp(2)+24*x^3-40*x^2-40*x)/(5*x^2*exp(2 )+20*x^3)/exp(x),x, algorithm="giac")
Output:
-1/5*(5*x^2*e^x - 5*x^2*log(4*x + e^2) + x - 10)*e^(-x)/x
Timed out. \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=\int -\frac {{\mathrm {e}}^{-x}\,\left (40\,x+{\mathrm {e}}^2\,\left (-x^2+10\,x+10\right )+{\mathrm {e}}^x\,\left (20\,x^3+5\,{\mathrm {e}}^2\,x^2\right )-\ln \left (4\,x+{\mathrm {e}}^2\right )\,\left ({\mathrm {e}}^2\,\left (5\,x^2-5\,x^3\right )+20\,x^3-20\,x^4\right )+40\,x^2-24\,x^3\right )}{20\,x^3+5\,{\mathrm {e}}^2\,x^2} \,d x \] Input:
int(-(exp(-x)*(40*x + exp(2)*(10*x - x^2 + 10) + exp(x)*(5*x^2*exp(2) + 20 *x^3) - log(4*x + exp(2))*(exp(2)*(5*x^2 - 5*x^3) + 20*x^3 - 20*x^4) + 40* x^2 - 24*x^3))/(5*x^2*exp(2) + 20*x^3),x)
Output:
int(-(exp(-x)*(40*x + exp(2)*(10*x - x^2 + 10) + exp(x)*(5*x^2*exp(2) + 20 *x^3) - log(4*x + exp(2))*(exp(2)*(5*x^2 - 5*x^3) + 20*x^3 - 20*x^4) + 40* x^2 - 24*x^3))/(5*x^2*exp(2) + 20*x^3), x)
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (-40 x-40 x^2+24 x^3+e^2 \left (-10-10 x+x^2\right )+e^x \left (-5 e^2 x^2-20 x^3\right )+\left (20 x^3-20 x^4+e^2 \left (5 x^2-5 x^3\right )\right ) \log \left (e^2+4 x\right )\right )}{5 e^2 x^2+20 x^3} \, dx=\frac {-5 e^{x} x^{2}+5 \,\mathrm {log}\left (e^{2}+4 x \right ) x^{2}-x +10}{5 e^{x} x} \] Input:
int((((-5*x^3+5*x^2)*exp(2)-20*x^4+20*x^3)*log(exp(2)+4*x)+(-5*x^2*exp(2)- 20*x^3)*exp(x)+(x^2-10*x-10)*exp(2)+24*x^3-40*x^2-40*x)/(5*x^2*exp(2)+20*x ^3)/exp(x),x)
Output:
( - 5*e**x*x**2 + 5*log(e**2 + 4*x)*x**2 - x + 10)/(5*e**x*x)