Integrand size = 73, antiderivative size = 33 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{25} e^{\frac {2 e^{3-x}}{x}-2 x} x+x^{\left .\frac {1}{2}\right /x} \]
Time = 2.72 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{50} \left (2 e^{\frac {2 e^{3-x}}{x}-2 x} x+50 x^{\left .\frac {1}{2}\right /x}\right ) \]
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]
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 |
\(\displaystyle \frac {1}{50} \left (25 \int x^{\frac {1}{2 x}-2}dx+25 \int \frac {\int x^{\frac {1}{2 x}-2}dx}{x}dx-25 \log (x) \int x^{\frac {1}{2 x}-2}dx-4 \int e^{-3 x+3+\frac {2 e^{3-x}}{x}}dx+2 \int e^{\frac {2 e^{3-x}}{x}-2 x}dx-4 \int \frac {e^{-3 x+3+\frac {2 e^{3-x}}{x}}}{x}dx-4 \int e^{\frac {2 e^{3-x}}{x}-2 x} xdx\right )\) |
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]
3.22.88.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.72 (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 | \({\mathrm e}^{\frac {\ln \left (x \right )}{2 x}}+\frac {x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{-x +3}-2 x^{2}}{x}}}{25}\) | \(32\) |
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)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{25} \, x e^{\left (-\frac {2 \, {\left (x^{2} - e^{\left (-x + 3\right )}\right )}}{x}\right )} + x^{\frac {1}{2 \, x}} \]
integrate(1/50*((-25*log(x)+25)*exp(1/2*log(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, algorithm=\
Timed out. \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\text {Timed out} \]
integrate(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)
Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\frac {1}{25} \, {\left (x e^{\left (\frac {2 \, e^{\left (-x + 3\right )}}{x}\right )} + 25 \, e^{\left (2 \, x + \frac {\log \left (x\right )}{2 \, x}\right )}\right )} e^{\left (-2 \, x\right )} \]
integrate(1/50*((-25*log(x)+25)*exp(1/2*log(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, algorithm=\
\[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=\int { -\frac {25 \, x^{\frac {1}{2 \, x}} {\left (\log \left (x\right ) - 1\right )} + 2 \, {\left (2 \, x^{3} - x^{2} + 2 \, {\left (x^{2} + x\right )} e^{\left (-x + 3\right )}\right )} e^{\left (-\frac {2 \, {\left (x^{2} - e^{\left (-x + 3\right )}\right )}}{x}\right )}}{50 \, x^{2}} \,d x } \]
integrate(1/50*((-25*log(x)+25)*exp(1/2*log(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, algorithm=\
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 = 11.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx=x^{\frac {1}{2\,x}}+\frac {x\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}}}{25} \]