Integrand size = 104, antiderivative size = 28 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=x-\frac {5 e^{-\frac {x^2}{(1-x)^2}} x}{(4+x) \log (5)} \] Output:
x-5/(4+x)/exp(x^2/(1-x)^2)/ln(5)*x
Time = 10.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\frac {-\frac {5 e^{-\frac {x^2}{(-1+x)^2}} x}{4+x}+(-1+x) \log (5)}{\log (5)} \] Input:
Integrate[(20 - 60*x + 20*x^2 - 30*x^3 + E^(x^2/(1 - 2*x + x^2))*(-16 + 40 *x - 25*x^2 - 5*x^3 + 5*x^4 + x^5)*Log[5])/(E^(x^2/(1 - 2*x + x^2))*(-16 + 40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5)*Log[5]),x]
Output:
((-5*x)/(E^(x^2/(-1 + x)^2)*(4 + x)) + (-1 + x)*Log[5])/Log[5]
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.
| \(\displaystyle \int \frac {e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2+e^{\frac {x^2}{x^2-2 x+1}} \left (x^5+5 x^4-5 x^3-25 x^2+40 x-16\right ) \log (5)-60 x+20\right )}{\left (x^5+5 x^4-5 x^3-25 x^2+40 x-16\right ) \log (5)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\frac {e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2-60 x-e^{\frac {x^2}{x^2-2 x+1}} \left (-x^5-5 x^4+5 x^3+25 x^2-40 x+16\right ) \log (5)+20\right )}{-x^5-5 x^4+5 x^3+25 x^2-40 x+16}dx}{\log (5)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2-60 x-e^{\frac {x^2}{x^2-2 x+1}} \left (-x^5-5 x^4+5 x^3+25 x^2-40 x+16\right ) \log (5)+20\right )}{-x^5-5 x^4+5 x^3+25 x^2-40 x+16}dx}{\log (5)}\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle -\frac {\int \left (-\frac {3 e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2-60 x-e^{\frac {x^2}{x^2-2 x+1}} \left (-x^5-5 x^4+5 x^3+25 x^2-40 x+16\right ) \log (5)+20\right )}{625 (x-1)}+\frac {3 e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2-60 x-e^{\frac {x^2}{x^2-2 x+1}} \left (-x^5-5 x^4+5 x^3+25 x^2-40 x+16\right ) \log (5)+20\right )}{625 (x+4)}+\frac {2 e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2-60 x-e^{\frac {x^2}{x^2-2 x+1}} \left (-x^5-5 x^4+5 x^3+25 x^2-40 x+16\right ) \log (5)+20\right )}{125 (x-1)^2}+\frac {e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2-60 x-e^{\frac {x^2}{x^2-2 x+1}} \left (-x^5-5 x^4+5 x^3+25 x^2-40 x+16\right ) \log (5)+20\right )}{125 (x+4)^2}-\frac {e^{-\frac {x^2}{x^2-2 x+1}} \left (-30 x^3+20 x^2-60 x-e^{\frac {x^2}{x^2-2 x+1}} \left (-x^5-5 x^4+5 x^3+25 x^2-40 x+16\right ) \log (5)+20\right )}{25 (x-1)^3}\right )dx}{\log (5)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {2}{5} \int \frac {e^{-\frac {x^2}{x^2-2 x+1}}}{(x-1)^3}dx+\frac {6}{5} \int \frac {e^{-\frac {x^2}{x^2-2 x+1}}}{(x-1)^2}dx+\frac {32}{25} \int \frac {e^{-\frac {x^2}{x^2-2 x+1}}}{x-1}dx+20 \int \frac {e^{-\frac {x^2}{x^2-2 x+1}}}{(x+4)^2}dx-\frac {32}{25} \int \frac {e^{-\frac {x^2}{x^2-2 x+1}}}{x+4}dx+\frac {6}{5} e^{-\frac {x^2}{x^2-2 x+1}}-\frac {1}{250} (1-x)^4 \log (5)+\frac {1}{25} (1-x)^3 \log (5)+\frac {1}{250} (x+4)^4 \log (5)-\frac {1}{25} (x+4)^3 \log (5)}{\log (5)}\) |
Input:
Int[(20 - 60*x + 20*x^2 - 30*x^3 + E^(x^2/(1 - 2*x + x^2))*(-16 + 40*x - 2 5*x^2 - 5*x^3 + 5*x^4 + x^5)*Log[5])/(E^(x^2/(1 - 2*x + x^2))*(-16 + 40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5)*Log[5]),x]
Output:
$Aborted
Time = 0.81 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
| method | result | size |
| risch | \(x -\frac {5 x \,{\mathrm e}^{-\frac {x^{2}}{\left (-1+x \right )^{2}}}}{\ln \left (5\right ) \left (4+x \right )}\) | \(26\) |
| parts | \(x +\frac {\left (-\frac {5 x}{\ln \left (5\right )}+\frac {10 x^{2}}{\ln \left (5\right )}-\frac {5 x^{3}}{\ln \left (5\right )}\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\left (4+x \right ) \left (-1+x \right )^{2}}\) | \(57\) |
| norman | \(\frac {\left (x^{4} {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-8 \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}+18 x \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-11 x^{2} {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-\frac {5 x}{\ln \left (5\right )}+\frac {10 x^{2}}{\ln \left (5\right )}-\frac {5 x^{3}}{\ln \left (5\right )}\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\left (4+x \right ) \left (-1+x \right )^{2}}\) | \(129\) |
| parallelrisch | \(\frac {\left (\ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}} x^{4}+4 \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}} x^{3}-3 \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}} x^{2}-10 \ln \left (5\right ) x \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-5 x^{3}+8 \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}+10 x^{2}-5 x \right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\ln \left (5\right ) \left (4+x \right ) \left (-1+x \right )^{2}}\) | \(151\) |
| orering | \(\frac {\left (x -\frac {5}{6}\right ) \left (\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right ) \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-30 x^{3}+20 x^{2}-60 x +20\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right ) \ln \left (5\right )}-\frac {\left (3 x^{5}-15 x^{4}-13 x^{2}+30 x -10\right ) \left (4+x \right ) \left (-1+x \right )^{3} \left (\frac {\left (\left (5 x^{4}+20 x^{3}-15 x^{2}-50 x +40\right ) \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}+\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right ) \ln \left (5\right ) \left (\frac {2 x}{x^{2}-2 x +1}-\frac {x^{2} \left (-2+2 x \right )}{\left (x^{2}-2 x +1\right )^{2}}\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-90 x^{2}+40 x -60\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right ) \ln \left (5\right )}-\frac {\left (\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right ) \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-30 x^{3}+20 x^{2}-60 x +20\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}} \left (5 x^{4}+20 x^{3}-15 x^{2}-50 x +40\right )}{\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right )^{2} \ln \left (5\right )}-\frac {\left (\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right ) \ln \left (5\right ) {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-30 x^{3}+20 x^{2}-60 x +20\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}} \left (\frac {2 x}{x^{2}-2 x +1}-\frac {x^{2} \left (-2+2 x \right )}{\left (x^{2}-2 x +1\right )^{2}}\right )}{\left (x^{5}+5 x^{4}-5 x^{3}-25 x^{2}+40 x -16\right ) \ln \left (5\right )}\right )}{6 \left (6 x^{6}-21 x^{5}+44 x^{4}-87 x^{3}-20 x^{2}+24 x +4\right )}\) | \(603\) |
Input:
int(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*ln(5)*exp(x^2/(x^2-2*x+1))-30*x^3+20 *x^2-60*x+20)/(x^5+5*x^4-5*x^3-25*x^2+40*x-16)/ln(5)/exp(x^2/(x^2-2*x+1)), x,method=_RETURNVERBOSE)
Output:
x-5/ln(5)*x/(4+x)*exp(-x^2/(-1+x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (25) = 50\).
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\frac {{\left ({\left (x^{2} + 4 \, x\right )} e^{\left (\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )} \log \left (5\right ) - 5 \, x\right )} e^{\left (-\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )}}{{\left (x + 4\right )} \log \left (5\right )} \] Input:
integrate(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*log(5)*exp(x^2/(x^2-2*x+1))-30 *x^3+20*x^2-60*x+20)/(x^5+5*x^4-5*x^3-25*x^2+40*x-16)/log(5)/exp(x^2/(x^2- 2*x+1)),x, algorithm="fricas")
Output:
((x^2 + 4*x)*e^(x^2/(x^2 - 2*x + 1))*log(5) - 5*x)*e^(-x^2/(x^2 - 2*x + 1) )/((x + 4)*log(5))
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=x - \frac {5 x e^{- \frac {x^{2}}{x^{2} - 2 x + 1}}}{x \log {\left (5 \right )} + 4 \log {\left (5 \right )}} \] Input:
integrate(((x**5+5*x**4-5*x**3-25*x**2+40*x-16)*ln(5)*exp(x**2/(x**2-2*x+1 ))-30*x**3+20*x**2-60*x+20)/(x**5+5*x**4-5*x**3-25*x**2+40*x-16)/ln(5)/exp (x**2/(x**2-2*x+1)),x)
Output:
x - 5*x*exp(-x**2/(x**2 - 2*x + 1))/(x*log(5) + 4*log(5))
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (25) = 50\).
Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.43 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=-\frac {{\left (5 \, x e^{\left (-\frac {1}{x^{2} - 2 \, x + 1}\right )} - {\left (x^{2} e \log \left (5\right ) + 4 \, x e \log \left (5\right )\right )} e^{\left (\frac {2}{x - 1}\right )}\right )} e^{\left (-\frac {2}{x - 1}\right )}}{{\left (x e + 4 \, e\right )} \log \left (5\right )} \] Input:
integrate(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*log(5)*exp(x^2/(x^2-2*x+1))-30 *x^3+20*x^2-60*x+20)/(x^5+5*x^4-5*x^3-25*x^2+40*x-16)/log(5)/exp(x^2/(x^2- 2*x+1)),x, algorithm="maxima")
Output:
-(5*x*e^(-1/(x^2 - 2*x + 1)) - (x^2*e*log(5) + 4*x*e*log(5))*e^(2/(x - 1)) )*e^(-2/(x - 1))/((x*e + 4*e)*log(5))
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\frac {x^{2} \log \left (5\right ) - 5 \, x e^{\left (-\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )} + 4 \, x \log \left (5\right )}{{\left (x + 4\right )} \log \left (5\right )} \] Input:
integrate(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*log(5)*exp(x^2/(x^2-2*x+1))-30 *x^3+20*x^2-60*x+20)/(x^5+5*x^4-5*x^3-25*x^2+40*x-16)/log(5)/exp(x^2/(x^2- 2*x+1)),x, algorithm="giac")
Output:
(x^2*log(5) - 5*x*e^(-x^2/(x^2 - 2*x + 1)) + 4*x*log(5))/((x + 4)*log(5))
Time = 4.81 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=x-\frac {5\,x\,{\mathrm {e}}^{-\frac {x^2}{x^2-2\,x+1}}}{\ln \left (5\right )\,\left (x+4\right )} \] Input:
int((exp(-x^2/(x^2 - 2*x + 1))*(20*x^2 - 60*x - 30*x^3 + exp(x^2/(x^2 - 2* x + 1))*log(5)*(40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5 - 16) + 20))/(log(5)*( 40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5 - 16)),x)
Output:
x - (5*x*exp(-x^2/(x^2 - 2*x + 1)))/(log(5)*(x + 4))
Time = 0.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.07 \[ \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx=\frac {x \left (e^{\frac {2 x}{x^{2}-2 x +1}} \mathrm {log}\left (5\right ) e x +4 e^{\frac {2 x}{x^{2}-2 x +1}} \mathrm {log}\left (5\right ) e -5 e^{\frac {1}{x^{2}-2 x +1}}\right )}{e^{\frac {2 x}{x^{2}-2 x +1}} \mathrm {log}\left (5\right ) e \left (x +4\right )} \] Input:
int(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*log(5)*exp(x^2/(x^2-2*x+1))-30*x^3+2 0*x^2-60*x+20)/(x^5+5*x^4-5*x^3-25*x^2+40*x-16)/log(5)/exp(x^2/(x^2-2*x+1) ),x)
Output:
(x*(e**((2*x)/(x**2 - 2*x + 1))*log(5)*e*x + 4*e**((2*x)/(x**2 - 2*x + 1)) *log(5)*e - 5*e**(1/(x**2 - 2*x + 1))))/(e**((2*x)/(x**2 - 2*x + 1))*log(5 )*e*(x + 4))