Integrand size = 72, antiderivative size = 29 \[ \int e^{-7+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} \left (e^2 \left (1+2 x^2\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-8 x-4 x \log \left (25 x^2\right )\right )\right ) \, dx=e^{-5+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} x \]
Time = 1.48 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int e^{-7+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} \left (e^2 \left (1+2 x^2\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-8 x-4 x \log \left (25 x^2\right )\right )\right ) \, dx=e^{-5+x^2-390625^{\frac {x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} x \]
Integrate[E^(-7 + x^2 - 25^((4*x)/E^2)*(x^2)^((4*x)/E^2))*(E^2*(1 + 2*x^2) + 25^((4*x)/E^2)*(x^2)^((4*x)/E^2)*(-8*x - 4*x*Log[25*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.
\(\displaystyle \int e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-7} \left (e^2 \left (2 x^2+1\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-4 x \log \left (25 x^2\right )-8 x\right )\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-5} \left (2 x^2+1\right )-4\ 25^{\frac {4 x}{e^2}} e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-7} x \left (x^2\right )^{\frac {4 x}{e^2}} \left (\log \left (25 x^2\right )+2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-5}dx+2 \int e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-5} x^2dx-8 \int 25^{\frac {4 x}{e^2}} e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-7} x \left (x^2\right )^{\frac {4 x}{e^2}}dx+8 \int \frac {\int 25^{\frac {4 x}{e^2}} e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-7} x \left (x^2\right )^{\frac {4 x}{e^2}}dx}{x}dx-4 \log \left (25 x^2\right ) \int 25^{\frac {4 x}{e^2}} e^{-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}+x^2-7} x \left (x^2\right )^{\frac {4 x}{e^2}}dx\) |
Int[E^(-7 + x^2 - 25^((4*x)/E^2)*(x^2)^((4*x)/E^2))*(E^2*(1 + 2*x^2) + 25^ ((4*x)/E^2)*(x^2)^((4*x)/E^2)*(-8*x - 4*x*Log[25*x^2])),x]
3.7.80.3.1 Defintions of rubi rules used
Time = 4.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(x \,{\mathrm e}^{-\left (25 x^{2}\right )^{4 \,{\mathrm e}^{-2} x}+x^{2}-5}\) | \(24\) |
parallelrisch | \(x \,{\mathrm e}^{-{\mathrm e}^{4 x \ln \left (25 x^{2}\right ) {\mathrm e}^{-2}}+x^{2}-5}\) | \(27\) |
int(((-4*x*ln(25*x^2)-8*x)*exp(2*x*ln(25*x^2)/exp(2))^2+(2*x^2+1)*exp(2))* exp(-exp(2*x*ln(25*x^2)/exp(2))^2+x^2-5)/exp(2),x,method=_RETURNVERBOSE)
Time = 0.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int e^{-7+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} \left (e^2 \left (1+2 x^2\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-8 x-4 x \log \left (25 x^2\right )\right )\right ) \, dx=x e^{\left (x^{2} - \left (25 \, x^{2}\right )^{4 \, x e^{\left (-2\right )}} - 5\right )} \]
integrate(((-4*x*log(25*x^2)-8*x)*exp(2*x*log(25*x^2)/exp(2))^2+(2*x^2+1)* exp(2))*exp(-exp(2*x*log(25*x^2)/exp(2))^2+x^2-5)/exp(2),x, algorithm=\
Time = 23.83 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int e^{-7+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} \left (e^2 \left (1+2 x^2\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-8 x-4 x \log \left (25 x^2\right )\right )\right ) \, dx=x e^{x^{2} - e^{\frac {4 x \log {\left (25 x^{2} \right )}}{e^{2}}} - 5} \]
integrate(((-4*x*ln(25*x**2)-8*x)*exp(2*x*ln(25*x**2)/exp(2))**2+(2*x**2+1 )*exp(2))*exp(-exp(2*x*ln(25*x**2)/exp(2))**2+x**2-5)/exp(2),x)
Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int e^{-7+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} \left (e^2 \left (1+2 x^2\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-8 x-4 x \log \left (25 x^2\right )\right )\right ) \, dx=x e^{\left (x^{2} - e^{\left (8 \, x e^{\left (-2\right )} \log \left (5\right ) + 8 \, x e^{\left (-2\right )} \log \left (x\right )\right )} - 5\right )} \]
integrate(((-4*x*log(25*x^2)-8*x)*exp(2*x*log(25*x^2)/exp(2))^2+(2*x^2+1)* exp(2))*exp(-exp(2*x*log(25*x^2)/exp(2))^2+x^2-5)/exp(2),x, algorithm=\
\[ \int e^{-7+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} \left (e^2 \left (1+2 x^2\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-8 x-4 x \log \left (25 x^2\right )\right )\right ) \, dx=\int { -{\left (4 \, {\left (x \log \left (25 \, x^{2}\right ) + 2 \, x\right )} \left (25 \, x^{2}\right )^{4 \, x e^{\left (-2\right )}} - {\left (2 \, x^{2} + 1\right )} e^{2}\right )} e^{\left (x^{2} - \left (25 \, x^{2}\right )^{4 \, x e^{\left (-2\right )}} - 7\right )} \,d x } \]
integrate(((-4*x*log(25*x^2)-8*x)*exp(2*x*log(25*x^2)/exp(2))^2+(2*x^2+1)* exp(2))*exp(-exp(2*x*log(25*x^2)/exp(2))^2+x^2-5)/exp(2),x, algorithm=\
integrate(-(4*(x*log(25*x^2) + 2*x)*(25*x^2)^(4*x*e^(-2)) - (2*x^2 + 1)*e^ 2)*e^(x^2 - (25*x^2)^(4*x*e^(-2)) - 7), x)
Time = 12.89 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int e^{-7+x^2-25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}}} \left (e^2 \left (1+2 x^2\right )+25^{\frac {4 x}{e^2}} \left (x^2\right )^{\frac {4 x}{e^2}} \left (-8 x-4 x \log \left (25 x^2\right )\right )\right ) \, dx=x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-{\left (390625\,x^8\right )}^{x\,{\mathrm {e}}^{-2}}} \]