Integrand size = 113, antiderivative size = 30 \[ \int \frac {e^{-\frac {5}{x^2}+\frac {e^{-\frac {5}{x^2}} \left (-x^3+e^{\frac {5}{x^2}} \left (-e x^2+x^3+x^2 \log (5)\right )-e^{\frac {5}{x^2}} \log ^2(x)\right )}{x^2}} \left (-10 x-x^3+e^{\frac {5}{x^2}} x^3-2 e^{\frac {5}{x^2}} \log (x)+2 e^{\frac {5}{x^2}} \log ^2(x)\right )}{4 x^3} \, dx=\frac {5}{4} e^{-e+x-e^{-\frac {5}{x^2}} x-\frac {\log ^2(x)}{x^2}} \]
Time = 2.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {5}{x^2}+\frac {e^{-\frac {5}{x^2}} \left (-x^3+e^{\frac {5}{x^2}} \left (-e x^2+x^3+x^2 \log (5)\right )-e^{\frac {5}{x^2}} \log ^2(x)\right )}{x^2}} \left (-10 x-x^3+e^{\frac {5}{x^2}} x^3-2 e^{\frac {5}{x^2}} \log (x)+2 e^{\frac {5}{x^2}} \log ^2(x)\right )}{4 x^3} \, dx=\frac {5}{4} e^{-e+x-e^{-\frac {5}{x^2}} x-\frac {\log ^2(x)}{x^2}} \]
Integrate[(E^(-5/x^2 + (-x^3 + E^(5/x^2)*(-(E*x^2) + x^3 + x^2*Log[5]) - E ^(5/x^2)*Log[x]^2)/(E^(5/x^2)*x^2))*(-10*x - x^3 + E^(5/x^2)*x^3 - 2*E^(5/ x^2)*Log[x] + 2*E^(5/x^2)*Log[x]^2))/(4*x^3),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.
| \(\displaystyle \int \frac {\left (-x^3+2 e^{\frac {5}{x^2}} \log ^2(x)-2 e^{\frac {5}{x^2}} \log (x)+e^{\frac {5}{x^2}} x^3-10 x\right ) \exp \left (\frac {e^{-\frac {5}{x^2}} \left (-x^3-e^{\frac {5}{x^2}} \log ^2(x)+e^{\frac {5}{x^2}} \left (x^3-e x^2+x^2 \log (5)\right )\right )}{x^2}-\frac {5}{x^2}\right )}{4 x^3} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int -\frac {\exp \left (-\frac {e^{-\frac {5}{x^2}} \left (x^3+e^{\frac {5}{x^2}} \log ^2(x)+e^{\frac {5}{x^2}} \left (-x^3-\log (5) x^2+e x^2\right )\right )}{x^2}-\frac {5}{x^2}\right ) \left (-e^{\frac {5}{x^2}} x^3+x^3+10 x-2 e^{\frac {5}{x^2}} \log ^2(x)+2 e^{\frac {5}{x^2}} \log (x)\right )}{x^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{4} \int \frac {\exp \left (-\frac {e^{-\frac {5}{x^2}} \left (x^3+e^{\frac {5}{x^2}} \log ^2(x)+e^{\frac {5}{x^2}} \left (-x^3-\log (5) x^2+e x^2\right )\right )}{x^2}-\frac {5}{x^2}\right ) \left (-e^{\frac {5}{x^2}} x^3+x^3+10 x-2 e^{\frac {5}{x^2}} \log ^2(x)+2 e^{\frac {5}{x^2}} \log (x)\right )}{x^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{4} \int \left (\frac {\exp \left (-\frac {e^{-\frac {5}{x^2}} \left (x^3+e^{\frac {5}{x^2}} \log ^2(x)+e^{\frac {5}{x^2}} \left (-x^3-\log (5) x^2+e x^2\right )\right )}{x^2}-\frac {5}{x^2}\right ) \left (x^2+10\right )}{x^2}-\frac {\exp \left (-\frac {e^{-\frac {5}{x^2}} \left (x^3+e^{\frac {5}{x^2}} \log ^2(x)+e^{\frac {5}{x^2}} \left (-x^3-\log (5) x^2+e x^2\right )\right )}{x^2}\right ) \left (x^3+2 \log ^2(x)-2 \log (x)\right )}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (5 \int e^{-\frac {\log ^2(x)}{x^2}-e^{-\frac {5}{x^2}} x+x-e}dx-5 \int e^{-\frac {\log ^2(x)}{x^2}-e^{-\frac {5}{x^2}} x+x-\frac {5}{x^2}-e}dx-50 \int \frac {e^{-\frac {\log ^2(x)}{x^2}-e^{-\frac {5}{x^2}} x+x-\frac {5}{x^2}-e}}{x^2}dx-10 \int \frac {e^{-\frac {\log ^2(x)}{x^2}-e^{-\frac {5}{x^2}} x+x-e} \log (x)}{x^3}dx+10 \int \frac {e^{-\frac {\log ^2(x)}{x^2}-e^{-\frac {5}{x^2}} x+x-e} \log ^2(x)}{x^3}dx\right )\) |
Int[(E^(-5/x^2 + (-x^3 + E^(5/x^2)*(-(E*x^2) + x^3 + x^2*Log[5]) - E^(5/x^ 2)*Log[x]^2)/(E^(5/x^2)*x^2))*(-10*x - x^3 + E^(5/x^2)*x^3 - 2*E^(5/x^2)*L og[x] + 2*E^(5/x^2)*Log[x]^2))/(4*x^3),x]
3.15.60.3.1 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]]
Time = 0.48 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17
| method | result | size |
| risch | \(\frac {5 \,{\mathrm e}^{-\frac {{\mathrm e}^{-\frac {5}{x^{2}}} x^{3}+x^{2} {\mathrm e}-x^{3}+\ln \left (x \right )^{2}}{x^{2}}}}{4}\) | \(35\) |
| derivativedivides | \(\frac {{\mathrm e}^{\frac {\left (-{\mathrm e}^{\frac {5}{x^{2}}} \ln \left (x \right )^{2}+\left (x^{2} \ln \left (5\right )-x^{2} {\mathrm e}+x^{3}\right ) {\mathrm e}^{\frac {5}{x^{2}}}-x^{3}\right ) {\mathrm e}^{-\frac {5}{x^{2}}}}{x^{2}}}}{4}\) | \(58\) |
| default | \(\frac {{\mathrm e}^{\frac {\left (-{\mathrm e}^{\frac {5}{x^{2}}} \ln \left (x \right )^{2}+\left (x^{2} \ln \left (5\right )-x^{2} {\mathrm e}+x^{3}\right ) {\mathrm e}^{\frac {5}{x^{2}}}-x^{3}\right ) {\mathrm e}^{-\frac {5}{x^{2}}}}{x^{2}}}}{4}\) | \(58\) |
| parallelrisch | \(\frac {{\mathrm e}^{\frac {\left (-{\mathrm e}^{\frac {5}{x^{2}}} \ln \left (x \right )^{2}+\left (x^{2} \ln \left (5\right )-x^{2} {\mathrm e}+x^{3}\right ) {\mathrm e}^{\frac {5}{x^{2}}}-x^{3}\right ) {\mathrm e}^{-\frac {5}{x^{2}}}}{x^{2}}}}{4}\) | \(58\) |
int(1/4*(2*exp(5/x^2)*ln(x)^2-2*exp(5/x^2)*ln(x)+x^3*exp(5/x^2)-x^3-10*x)* exp((-exp(5/x^2)*ln(x)^2+(x^2*ln(5)-x^2*exp(1)+x^3)*exp(5/x^2)-x^3)/x^2/ex p(5/x^2))/x^3/exp(5/x^2),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (29) = 58\).
Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03 \[ \int \frac {e^{-\frac {5}{x^2}+\frac {e^{-\frac {5}{x^2}} \left (-x^3+e^{\frac {5}{x^2}} \left (-e x^2+x^3+x^2 \log (5)\right )-e^{\frac {5}{x^2}} \log ^2(x)\right )}{x^2}} \left (-10 x-x^3+e^{\frac {5}{x^2}} x^3-2 e^{\frac {5}{x^2}} \log (x)+2 e^{\frac {5}{x^2}} \log ^2(x)\right )}{4 x^3} \, dx=\frac {1}{4} \, e^{\left (-\frac {{\left (x^{3} + e^{\left (\frac {5}{x^{2}}\right )} \log \left (x\right )^{2} - {\left (x^{3} - x^{2} e + x^{2} \log \left (5\right ) - 5\right )} e^{\left (\frac {5}{x^{2}}\right )}\right )} e^{\left (-\frac {5}{x^{2}}\right )}}{x^{2}} + \frac {5}{x^{2}}\right )} \]
integrate(1/4*(2*exp(5/x^2)*log(x)^2-2*exp(5/x^2)*log(x)+x^3*exp(5/x^2)-x^ 3-10*x)*exp((-exp(5/x^2)*log(x)^2+(x^2*log(5)-x^2*exp(1)+x^3)*exp(5/x^2)-x ^3)/x^2/exp(5/x^2))/x^3/exp(5/x^2),x, algorithm=\
1/4*e^(-(x^3 + e^(5/x^2)*log(x)^2 - (x^3 - x^2*e + x^2*log(5) - 5)*e^(5/x^ 2))*e^(-5/x^2)/x^2 + 5/x^2)
Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-\frac {5}{x^2}+\frac {e^{-\frac {5}{x^2}} \left (-x^3+e^{\frac {5}{x^2}} \left (-e x^2+x^3+x^2 \log (5)\right )-e^{\frac {5}{x^2}} \log ^2(x)\right )}{x^2}} \left (-10 x-x^3+e^{\frac {5}{x^2}} x^3-2 e^{\frac {5}{x^2}} \log (x)+2 e^{\frac {5}{x^2}} \log ^2(x)\right )}{4 x^3} \, dx=\frac {e^{\frac {\left (- x^{3} + \left (x^{3} - e x^{2} + x^{2} \log {\left (5 \right )}\right ) e^{\frac {5}{x^{2}}} - e^{\frac {5}{x^{2}}} \log {\left (x \right )}^{2}\right ) e^{- \frac {5}{x^{2}}}}{x^{2}}}}{4} \]
integrate(1/4*(2*exp(5/x**2)*ln(x)**2-2*exp(5/x**2)*ln(x)+x**3*exp(5/x**2) -x**3-10*x)*exp((-exp(5/x**2)*ln(x)**2+(x**2*ln(5)-x**2*exp(1)+x**3)*exp(5 /x**2)-x**3)/x**2/exp(5/x**2))/x**3/exp(5/x**2),x)
exp((-x**3 + (x**3 - E*x**2 + x**2*log(5))*exp(5/x**2) - exp(5/x**2)*log(x )**2)*exp(-5/x**2)/x**2)/4
Time = 0.63 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {5}{x^2}+\frac {e^{-\frac {5}{x^2}} \left (-x^3+e^{\frac {5}{x^2}} \left (-e x^2+x^3+x^2 \log (5)\right )-e^{\frac {5}{x^2}} \log ^2(x)\right )}{x^2}} \left (-10 x-x^3+e^{\frac {5}{x^2}} x^3-2 e^{\frac {5}{x^2}} \log (x)+2 e^{\frac {5}{x^2}} \log ^2(x)\right )}{4 x^3} \, dx=\frac {5}{4} \, e^{\left (-x e^{\left (-\frac {5}{x^{2}}\right )} + x - \frac {\log \left (x\right )^{2}}{x^{2}} - e\right )} \]
integrate(1/4*(2*exp(5/x^2)*log(x)^2-2*exp(5/x^2)*log(x)+x^3*exp(5/x^2)-x^ 3-10*x)*exp((-exp(5/x^2)*log(x)^2+(x^2*log(5)-x^2*exp(1)+x^3)*exp(5/x^2)-x ^3)/x^2/exp(5/x^2))/x^3/exp(5/x^2),x, algorithm=\
\[ \int \frac {e^{-\frac {5}{x^2}+\frac {e^{-\frac {5}{x^2}} \left (-x^3+e^{\frac {5}{x^2}} \left (-e x^2+x^3+x^2 \log (5)\right )-e^{\frac {5}{x^2}} \log ^2(x)\right )}{x^2}} \left (-10 x-x^3+e^{\frac {5}{x^2}} x^3-2 e^{\frac {5}{x^2}} \log (x)+2 e^{\frac {5}{x^2}} \log ^2(x)\right )}{4 x^3} \, dx=\int { \frac {{\left (x^{3} e^{\left (\frac {5}{x^{2}}\right )} - x^{3} + 2 \, e^{\left (\frac {5}{x^{2}}\right )} \log \left (x\right )^{2} - 2 \, e^{\left (\frac {5}{x^{2}}\right )} \log \left (x\right ) - 10 \, x\right )} e^{\left (-\frac {{\left (x^{3} + e^{\left (\frac {5}{x^{2}}\right )} \log \left (x\right )^{2} - {\left (x^{3} - x^{2} e + x^{2} \log \left (5\right )\right )} e^{\left (\frac {5}{x^{2}}\right )}\right )} e^{\left (-\frac {5}{x^{2}}\right )}}{x^{2}} - \frac {5}{x^{2}}\right )}}{4 \, x^{3}} \,d x } \]
integrate(1/4*(2*exp(5/x^2)*log(x)^2-2*exp(5/x^2)*log(x)+x^3*exp(5/x^2)-x^ 3-10*x)*exp((-exp(5/x^2)*log(x)^2+(x^2*log(5)-x^2*exp(1)+x^3)*exp(5/x^2)-x ^3)/x^2/exp(5/x^2))/x^3/exp(5/x^2),x, algorithm=\
integrate(1/4*(x^3*e^(5/x^2) - x^3 + 2*e^(5/x^2)*log(x)^2 - 2*e^(5/x^2)*lo g(x) - 10*x)*e^(-(x^3 + e^(5/x^2)*log(x)^2 - (x^3 - x^2*e + x^2*log(5))*e^ (5/x^2))*e^(-5/x^2)/x^2 - 5/x^2)/x^3, x)
Time = 8.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {5}{x^2}+\frac {e^{-\frac {5}{x^2}} \left (-x^3+e^{\frac {5}{x^2}} \left (-e x^2+x^3+x^2 \log (5)\right )-e^{\frac {5}{x^2}} \log ^2(x)\right )}{x^2}} \left (-10 x-x^3+e^{\frac {5}{x^2}} x^3-2 e^{\frac {5}{x^2}} \log (x)+2 e^{\frac {5}{x^2}} \log ^2(x)\right )}{4 x^3} \, dx=\frac {5\,{\mathrm {e}}^{-\mathrm {e}}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{-\frac {5}{x^2}}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{-\frac {{\ln \left (x\right )}^2}{x^2}}}{4} \]