Integrand size = 85, antiderivative size = 28 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\frac {5}{2}-\frac {1}{25} e^x \left (5+\frac {x}{-e^{x^2}+x}\right )+\log (x) \] Output:
ln(x)-exp(x)*(1/25*x/(-exp(x^2)+x)+1/5)+5/2
Time = 8.96 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\frac {1}{25} \left (\frac {e^x \left (-5 e^{x^2}+6 x\right )}{e^{x^2}-x}+25 \log (x)\right ) \] Input:
Integrate[(25*x^2 - 6*E^x*x^3 + E^(2*x^2)*(25 - 5*E^x*x) + E^x^2*(-50*x + E^x*(x + 11*x^2 - 2*x^3)))/(25*E^(2*x^2)*x - 50*E^x^2*x^2 + 25*x^3),x]
Output:
((E^x*(-5*E^x^2 + 6*x))/(E^x^2 - x) + 25*Log[x])/25
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.
\(\displaystyle \int \frac {-6 e^x x^3+25 x^2+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (e^x \left (-2 x^3+11 x^2+x\right )-50 x\right )}{25 x^3-50 e^{x^2} x^2+25 e^{2 x^2} x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-6 e^x x^3+25 x^2+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (e^x \left (-2 x^3+11 x^2+x\right )-50 x\right )}{25 \left (e^{x^2}-x\right )^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \frac {-6 e^x x^3+25 x^2+5 e^{2 x^2} \left (5-e^x x\right )-e^{x^2} \left (50 x-e^x \left (-2 x^3+11 x^2+x\right )\right )}{\left (e^{x^2}-x\right )^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{25} \int \left (-\frac {5 \left (e^x x-5\right )}{x}-\frac {e^x x \left (2 x^2-1\right )}{\left (e^{x^2}-x\right )^2}-\frac {e^x \left (2 x^2-x-1\right )}{e^{x^2}-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (\int \frac {e^x}{e^{x^2}-x}dx+\int \frac {e^x x}{\left (e^{x^2}-x\right )^2}dx+\int \frac {e^x x}{e^{x^2}-x}dx-2 \int \frac {e^x x^2}{e^{x^2}-x}dx-2 \int \frac {e^x x^3}{\left (e^{x^2}-x\right )^2}dx-5 e^x+25 \log (x)\right )\) |
Input:
Int[(25*x^2 - 6*E^x*x^3 + E^(2*x^2)*(25 - 5*E^x*x) + E^x^2*(-50*x + E^x*(x + 11*x^2 - 2*x^3)))/(25*E^(2*x^2)*x - 50*E^x^2*x^2 + 25*x^3),x]
Output:
$Aborted
Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
risch | \(\ln \left (x \right )-\frac {{\mathrm e}^{x}}{5}-\frac {x \,{\mathrm e}^{x}}{25 \left (-{\mathrm e}^{x^{2}}+x \right )}\) | \(23\) |
norman | \(\frac {-\frac {6 \,{\mathrm e}^{x} x}{25}+\frac {{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{5}}{-{\mathrm e}^{x^{2}}+x}+\ln \left (x \right )\) | \(29\) |
parts | \(\frac {-\frac {6 \,{\mathrm e}^{x} x}{25}+\frac {{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{5}}{-{\mathrm e}^{x^{2}}+x}+\ln \left (x \right )\) | \(29\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}-6 \,{\mathrm e}^{x} x -25 \,{\mathrm e}^{x^{2}} \ln \left (x \right )+25 x \ln \left (x \right )}{-25 \,{\mathrm e}^{x^{2}}+25 x}\) | \(40\) |
Input:
int(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp(x^2)- 6*exp(x)*x^3+25*x^2)/(25*x*exp(x^2)^2-50*x^2*exp(x^2)+25*x^3),x,method=_RE TURNVERBOSE)
Output:
ln(x)-1/5*exp(x)-1/25*x*exp(x)/(-exp(x^2)+x)
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=-\frac {6 \, x e^{x} - 25 \, {\left (x - e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - 5 \, e^{\left (x^{2} + x\right )}}{25 \, {\left (x - e^{\left (x^{2}\right )}\right )}} \] Input:
integrate(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp (x^2)-6*exp(x)*x^3+25*x^2)/(25*x*exp(x^2)^2-50*exp(x^2)*x^2+25*x^3),x, alg orithm="fricas")
Output:
-1/25*(6*x*e^x - 25*(x - e^(x^2))*log(x) - 5*e^(x^2 + x))/(x - e^(x^2))
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\frac {x e^{x}}{- 25 x + 25 e^{x^{2}}} - \frac {e^{x}}{5} + \log {\left (x \right )} \] Input:
integrate(((-5*exp(x)*x+25)*exp(x**2)**2+((-2*x**3+11*x**2+x)*exp(x)-50*x) *exp(x**2)-6*exp(x)*x**3+25*x**2)/(25*x*exp(x**2)**2-50*exp(x**2)*x**2+25* x**3),x)
Output:
x*exp(x)/(-25*x + 25*exp(x**2)) - exp(x)/5 + log(x)
Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=-\frac {6 \, x e^{x} - 5 \, e^{\left (x^{2} + x\right )}}{25 \, {\left (x - e^{\left (x^{2}\right )}\right )}} + \log \left (x\right ) \] Input:
integrate(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp (x^2)-6*exp(x)*x^3+25*x^2)/(25*x*exp(x^2)^2-50*exp(x^2)*x^2+25*x^3),x, alg orithm="maxima")
Output:
-1/25*(6*x*e^x - 5*e^(x^2 + x))/(x - e^(x^2)) + log(x)
Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=-\frac {6 \, x e^{x} - 25 \, x \log \left (x\right ) + 25 \, e^{\left (x^{2}\right )} \log \left (x\right ) - 5 \, e^{\left (x^{2} + x\right )}}{25 \, {\left (x - e^{\left (x^{2}\right )}\right )}} \] Input:
integrate(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp (x^2)-6*exp(x)*x^3+25*x^2)/(25*x*exp(x^2)^2-50*exp(x^2)*x^2+25*x^3),x, alg orithm="giac")
Output:
-1/25*(6*x*e^x - 25*x*log(x) + 25*e^(x^2)*log(x) - 5*e^(x^2 + x))/(x - e^( x^2))
Time = 2.91 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\ln \left (x\right )+\frac {5\,{\mathrm {e}}^{x^2+x}-6\,x\,{\mathrm {e}}^x}{25\,x-25\,{\mathrm {e}}^{x^2}} \] Input:
int(-(exp(2*x^2)*(5*x*exp(x) - 25) + 6*x^3*exp(x) + exp(x^2)*(50*x - exp(x )*(x + 11*x^2 - 2*x^3)) - 25*x^2)/(25*x*exp(2*x^2) - 50*x^2*exp(x^2) + 25* x^3),x)
Output:
log(x) + (5*exp(x + x^2) - 6*x*exp(x))/(25*x - 25*exp(x^2))
Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {25 x^2-6 e^x x^3+e^{2 x^2} \left (25-5 e^x x\right )+e^{x^2} \left (-50 x+e^x \left (x+11 x^2-2 x^3\right )\right )}{25 e^{2 x^2} x-50 e^{x^2} x^2+25 x^3} \, dx=\frac {-5 e^{x^{2}+x}+25 e^{x^{2}} \mathrm {log}\left (x \right )+6 e^{x} x -25 \,\mathrm {log}\left (x \right ) x}{25 e^{x^{2}}-25 x} \] Input:
int(((-5*exp(x)*x+25)*exp(x^2)^2+((-2*x^3+11*x^2+x)*exp(x)-50*x)*exp(x^2)- 6*exp(x)*x^3+25*x^2)/(25*x*exp(x^2)^2-50*exp(x^2)*x^2+25*x^3),x)
Output:
( - 5*e**(x**2 + x) + 25*e**(x**2)*log(x) + 6*e**x*x - 25*log(x)*x)/(25*(e **(x**2) - x))