Integrand size = 83, antiderivative size = 29 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\frac {e^{5 \left (2+e^3+\left (-e^5+x^2\right )^2\right )}}{2 x}} \]
Time = 3.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\frac {e^{5 \left (2+e^3+e^{10}-2 e^5 x^2+x^4\right )}}{2 x}} \]
Integrate[(E^(10 + 5*E^3 + 5*E^10 + E^(10 + 5*E^3 + 5*E^10 - 10*E^5*x^2 + 5*x^4)/(2*x) - 10*E^5*x^2 + 5*x^4)*(-1 - 20*E^5*x^2 + 20*x^4))/(2*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.
\(\displaystyle \int \frac {\left (20 x^4-20 e^5 x^2-1\right ) \exp \left (5 x^4-10 e^5 x^2+\frac {e^{5 x^4-10 e^5 x^2+5 e^{10}+5 e^3+10}}{2 x}+5 e^{10}+5 e^3+10\right )}{2 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) \left (-20 x^4+20 e^5 x^2+1\right )}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) \left (-20 x^4+20 e^5 x^2+1\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (-20 \exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) x^2+20 \exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+5+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )+\frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-20 \int \exp \left (5 x^4-10 e^5 x^2+5 \left (3+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )dx-\int \frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )}{x^2}dx+20 \int \exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) x^2dx\right )\) |
Int[(E^(10 + 5*E^3 + 5*E^10 + E^(10 + 5*E^3 + 5*E^10 - 10*E^5*x^2 + 5*x^4) /(2*x) - 10*E^5*x^2 + 5*x^4)*(-1 - 20*E^5*x^2 + 20*x^4))/(2*x^2),x]
3.27.64.3.1 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 = 0.94 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03
method | result | size |
risch | \({\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{10}-10 x^{2} {\mathrm e}^{5}+5 \,{\mathrm e}^{3}+5 x^{4}+10}}{2 x}}\) | \(30\) |
norman | \({\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{10}-10 x^{2} {\mathrm e}^{5}+5 \,{\mathrm e}^{3}+5 x^{4}+10}}{2 x}}\) | \(32\) |
parallelrisch | \({\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{10}-10 x^{2} {\mathrm e}^{5}+5 \,{\mathrm e}^{3}+5 x^{4}+10}}{2 x}}\) | \(32\) |
int(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5* x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x^2,x,m ethod=_RETURNVERBOSE)
Timed out. \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=\text {Timed out} \]
integrate(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp (3)+5*x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x ^2,x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\frac {e^{5 x^{4} - 10 x^{2} e^{5} + 10 + 5 e^{3} + 5 e^{10}}}{2 x}} \]
integrate(1/2*(-20*x**2*exp(5)+20*x**4-1)*exp(5*exp(5)**2-10*x**2*exp(5)+5 *exp(3)+5*x**4+10)*exp(1/2*exp(5*exp(5)**2-10*x**2*exp(5)+5*exp(3)+5*x**4+ 10)/x)/x**2,x)
Time = 0.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\left (\frac {e^{\left (5 \, x^{4} - 10 \, x^{2} e^{5} + 5 \, e^{10} + 5 \, e^{3} + 10\right )}}{2 \, x}\right )} \]
integrate(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp (3)+5*x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x ^2,x, algorithm=\
\[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=\int { \frac {{\left (20 \, x^{4} - 20 \, x^{2} e^{5} - 1\right )} e^{\left (5 \, x^{4} - 10 \, x^{2} e^{5} + \frac {e^{\left (5 \, x^{4} - 10 \, x^{2} e^{5} + 5 \, e^{10} + 5 \, e^{3} + 10\right )}}{2 \, x} + 5 \, e^{10} + 5 \, e^{3} + 10\right )}}{2 \, x^{2}} \,d x } \]
integrate(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp (3)+5*x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x ^2,x, algorithm=\
integrate(1/2*(20*x^4 - 20*x^2*e^5 - 1)*e^(5*x^4 - 10*x^2*e^5 + 1/2*e^(5*x ^4 - 10*x^2*e^5 + 5*e^10 + 5*e^3 + 10)/x + 5*e^10 + 5*e^3 + 10)/x^2, x)
Time = 9.38 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{-10\,x^2\,{\mathrm {e}}^5}\,{\mathrm {e}}^{5\,{\mathrm {e}}^3}\,{\mathrm {e}}^{5\,{\mathrm {e}}^{10}}\,{\mathrm {e}}^{10}\,{\mathrm {e}}^{5\,x^4}}{2\,x}} \]