Integrand size = 73, antiderivative size = 33 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{25} e^{\frac {2 e^{3-x}}{x}-2 x} x+x^{\left .\frac {1}{2}\right /x} \] Output:
1/25*x*exp(exp(3-x)/x-x)^2+exp(1/2*ln(x)/x)
Time = 1.71 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{50} \left (2 e^{\frac {2 e^{3-x}}{x}-2 x} x+50 x^{\left .\frac {1}{2}\right /x}\right ) \] Input:
Integrate[(E^((2*(E^(3 - x) - x^2))/x)*(2*x^2 - 4*x^3 + E^(3 - x)*(-4*x - 4*x^2)) + x^(1/(2*x))*(25 - 25*Log[x]))/(50*x^2),x]
Output:
(2*E^((2*E^(3 - x))/x - 2*x)*x + 50*x^(1/(2*x)))/50
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.
\(\displaystyle \int \frac {x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))+e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (-4 x^3+2 x^2+e^{3-x} \left (-4 x^2-4 x\right )\right )}{50 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{50} \int \frac {25 (1-\log (x)) x^{\left .\frac {1}{2}\right /x}+2 e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (-2 x^3+x^2-2 e^{3-x} \left (x^2+x\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{50} \int \left (-25 (\log (x)-1) x^{\frac {1}{2 x}-2}-\frac {2 e^{\frac {2 e^{3-x}}{x}-3 x} \left (2 e^x x^2-e^x x+2 e^3 x+2 e^3\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{50} \left (25 \int x^{\frac {1}{2 x}-2}dx+25 \int \frac {\int x^{\frac {1}{2 x}-2}dx}{x}dx-25 \log (x) \int x^{\frac {1}{2 x}-2}dx-4 \int e^{-3 x+3+\frac {2 e^{3-x}}{x}}dx+2 \int e^{\frac {2 e^{3-x}}{x}-2 x}dx-4 \int \frac {e^{-3 x+3+\frac {2 e^{3-x}}{x}}}{x}dx-4 \int e^{\frac {2 e^{3-x}}{x}-2 x} xdx\right )\) |
Input:
Int[(E^((2*(E^(3 - x) - x^2))/x)*(2*x^2 - 4*x^3 + E^(3 - x)*(-4*x - 4*x^2) ) + x^(1/(2*x))*(25 - 25*Log[x]))/(50*x^2),x]
Output:
$Aborted
Time = 1.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91
method | result | size |
risch | \(x^{\frac {1}{2 x}}+\frac {x \,{\mathrm e}^{-\frac {2 \left (-{\mathrm e}^{-x +3}+x^{2}\right )}{x}}}{25}\) | \(30\) |
default | \({\mathrm e}^{\frac {\ln \left (x \right )}{2 x}}+\frac {x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{-x +3}-2 x^{2}}{x}}}{25}\) | \(32\) |
parallelrisch | \({\mathrm e}^{\frac {\ln \left (x \right )}{2 x}}+\frac {x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{-x +3}-2 x^{2}}{x}}}{25}\) | \(32\) |
parts | \({\mathrm e}^{\frac {\ln \left (x \right )}{2 x}}+\frac {x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{-x +3}-2 x^{2}}{x}}}{25}\) | \(32\) |
Input:
int(1/50*((-25*ln(x)+25)*exp(1/2*ln(x)/x)+((-4*x^2-4*x)*exp(-x+3)-4*x^3+2* x^2)*exp((exp(-x+3)-x^2)/x)^2)/x^2,x,method=_RETURNVERBOSE)
Output:
x^(1/2/x)+1/25*x*exp(-2*(-exp(-x+3)+x^2)/x)
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{25} \, x e^{\left (-\frac {2 \, {\left (x^{2} - e^{\left (-x + 3\right )}\right )}}{x}\right )} + x^{\frac {1}{2 \, x}} \] Input:
integrate(1/50*((-25*log(x)+25)*exp(1/2*log(x)/x)+((-4*x^2-4*x)*exp(3-x)-4 *x^3+2*x^2)*exp((exp(3-x)-x^2)/x)^2)/x^2,x, algorithm="fricas")
Output:
1/25*x*e^(-2*(x^2 - e^(-x + 3))/x) + x^(1/2/x)
Timed out. \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\text {Timed out} \] Input:
integrate(1/50*((-25*ln(x)+25)*exp(1/2*ln(x)/x)+((-4*x**2-4*x)*exp(3-x)-4* x**3+2*x**2)*exp((exp(3-x)-x**2)/x)**2)/x**2,x)
Output:
Timed out
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{25} \, {\left (x e^{\left (\frac {2 \, e^{\left (-x + 3\right )}}{x}\right )} + 25 \, e^{\left (2 \, x + \frac {\log \left (x\right )}{2 \, x}\right )}\right )} e^{\left (-2 \, x\right )} \] Input:
integrate(1/50*((-25*log(x)+25)*exp(1/2*log(x)/x)+((-4*x^2-4*x)*exp(3-x)-4 *x^3+2*x^2)*exp((exp(3-x)-x^2)/x)^2)/x^2,x, algorithm="maxima")
Output:
1/25*(x*e^(2*e^(-x + 3)/x) + 25*e^(2*x + 1/2*log(x)/x))*e^(-2*x)
\[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\int { -\frac {25 \, x^{\frac {1}{2 \, x}} {\left (\log \left (x\right ) - 1\right )} + 2 \, {\left (2 \, x^{3} - x^{2} + 2 \, {\left (x^{2} + x\right )} e^{\left (-x + 3\right )}\right )} e^{\left (-\frac {2 \, {\left (x^{2} - e^{\left (-x + 3\right )}\right )}}{x}\right )}}{50 \, x^{2}} \,d x } \] Input:
integrate(1/50*((-25*log(x)+25)*exp(1/2*log(x)/x)+((-4*x^2-4*x)*exp(3-x)-4 *x^3+2*x^2)*exp((exp(3-x)-x^2)/x)^2)/x^2,x, algorithm="giac")
Output:
integrate(-1/50*(25*x^(1/2/x)*(log(x) - 1) + 2*(2*x^3 - x^2 + 2*(x^2 + x)* e^(-x + 3))*e^(-2*(x^2 - e^(-x + 3))/x))/x^2, x)
Time = 4.91 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=x^{\frac {1}{2\,x}}+\frac {x\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}}}{25} \] Input:
int(-((exp((2*(exp(3 - x) - x^2))/x)*(exp(3 - x)*(4*x + 4*x^2) - 2*x^2 + 4 *x^3))/50 + (exp(log(x)/(2*x))*(25*log(x) - 25))/50)/x^2,x)
Output:
x^(1/(2*x)) + (x*exp(-2*x)*exp((2*exp(-x)*exp(3))/x))/25
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {e^{\frac {2 e^{3}}{e^{x} x}} x +25 e^{\frac {\mathrm {log}\left (x \right )+4 x^{2}}{2 x}}}{25 e^{2 x}} \] Input:
int(1/50*((-25*log(x)+25)*exp(1/2*log(x)/x)+((-4*x^2-4*x)*exp(3-x)-4*x^3+2 *x^2)*exp((exp(3-x)-x^2)/x)^2)/x^2,x)
Output:
(e**((2*e**3)/(e**x*x))*x + 25*e**((log(x) + 4*x**2)/(2*x)))/(25*e**(2*x))