Integrand size = 54, antiderivative size = 24 \[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=3-e^{\frac {e^{e^x} \left (-2+\frac {2}{x^2}\right ) x}{\log (2)}} \] Output:
3-exp(exp(exp(x))*(2/x^2-2)/ln(2)*x)
Time = 1.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=-e^{-\frac {2 e^{e^x} \left (-1+x^2\right )}{x \log (2)}} \] Input:
Integrate[(E^(E^x + (E^E^x*(2 - 2*x^2))/(x*Log[2]))*(2 + 2*x^2 + E^x*(-2*x + 2*x^3)))/(x^2*Log[2]),x]
Output:
-E^((-2*E^E^x*(-1 + x^2))/(x*Log[2]))
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.
\(\displaystyle \int \frac {\left (e^x \left (2 x^3-2 x\right )+2 x^2+2\right ) e^{\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}+e^x}}{x^2 \log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 e^{\frac {2 e^{e^x} \left (1-x^2\right )}{x \log (2)}+e^x} \left (x^2-e^x \left (x-x^3\right )+1\right )}{x^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {e^{\frac {2 e^{e^x} \left (1-x^2\right )}{x \log (2)}+e^x} \left (x^2-e^x \left (x-x^3\right )+1\right )}{x^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2 \int \left (\frac {e^{x+e^x+\frac {2 e^{e^x} \left (1-x^2\right )}{\log (2) x}} (x-1) (x+1)}{x}+\frac {e^{\frac {2 e^{e^x} \left (1-x^2\right )}{x \log (2)}+e^x} \left (x^2+1\right )}{x^2}\right )dx}{\log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (\int e^{\frac {2 e^{e^x} \left (1-x^2\right )}{x \log (2)}+e^x}dx+\int \frac {e^{\frac {2 e^{e^x} \left (1-x^2\right )}{x \log (2)}+e^x}}{x^2}dx-\int \frac {e^{x+e^x+\frac {2 e^{e^x} \left (1-x^2\right )}{\log (2) x}}}{x}dx+\int e^{x+e^x+\frac {2 e^{e^x} \left (1-x^2\right )}{\log (2) x}} xdx\right )}{\log (2)}\) |
Input:
Int[(E^(E^x + (E^E^x*(2 - 2*x^2))/(x*Log[2]))*(2 + 2*x^2 + E^x*(-2*x + 2*x ^3)))/(x^2*Log[2]),x]
Output:
$Aborted
Time = 8.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-{\mathrm e}^{-\frac {2 \left (-1+x \right ) \left (1+x \right ) {\mathrm e}^{{\mathrm e}^{x}}}{x \ln \left (2\right )}}\) | \(22\) |
parallelrisch | \(-{\mathrm e}^{\frac {\left (-2 x^{2}+2\right ) {\mathrm e}^{{\mathrm e}^{x}}}{x \ln \left (2\right )}}\) | \(22\) |
Input:
int(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp(x))/x/ ln(2))/x^2/ln(2),x,method=_RETURNVERBOSE)
Output:
-exp(-2*(-1+x)*(1+x)*exp(exp(x))/x/ln(2))
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=-e^{\left (\frac {x e^{x} \log \left (2\right ) - 2 \, {\left (x^{2} - 1\right )} e^{\left (e^{x}\right )}}{x \log \left (2\right )} - e^{x}\right )} \] Input:
integrate(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp( x))/x/log(2))/x^2/log(2),x, algorithm="fricas")
Output:
-e^((x*e^x*log(2) - 2*(x^2 - 1)*e^(e^x))/(x*log(2)) - e^x)
Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=- e^{\frac {\left (2 - 2 x^{2}\right ) e^{e^{x}}}{x \log {\left (2 \right )}}} \] Input:
integrate(((2*x**3-2*x)*exp(x)+2*x**2+2)*exp(exp(x))*exp((-2*x**2+2)*exp(e xp(x))/x/ln(2))/x**2/ln(2),x)
Output:
-exp((2 - 2*x**2)*exp(exp(x))/(x*log(2)))
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=-e^{\left (-\frac {2 \, x e^{\left (e^{x}\right )}}{\log \left (2\right )} + \frac {2 \, e^{\left (e^{x}\right )}}{x \log \left (2\right )}\right )} \] Input:
integrate(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp( x))/x/log(2))/x^2/log(2),x, algorithm="maxima")
Output:
-e^(-2*x*e^(e^x)/log(2) + 2*e^(e^x)/(x*log(2)))
\[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=\int { \frac {2 \, {\left (x^{2} + {\left (x^{3} - x\right )} e^{x} + 1\right )} e^{\left (-\frac {2 \, {\left (x^{2} - 1\right )} e^{\left (e^{x}\right )}}{x \log \left (2\right )} + e^{x}\right )}}{x^{2} \log \left (2\right )} \,d x } \] Input:
integrate(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp( x))/x/log(2))/x^2/log(2),x, algorithm="giac")
Output:
integrate(2*(x^2 + (x^3 - x)*e^x + 1)*e^(-2*(x^2 - 1)*e^(e^x)/(x*log(2)) + e^x)/(x^2*log(2)), x)
Time = 3.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=-{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{{\mathrm {e}}^x}-2\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{x\,\ln \left (2\right )}} \] Input:
int((exp(-(exp(exp(x))*(2*x^2 - 2))/(x*log(2)))*exp(exp(x))*(2*x^2 - exp(x )*(2*x - 2*x^3) + 2))/(x^2*log(2)),x)
Output:
-exp((2*exp(exp(x)) - 2*x^2*exp(exp(x)))/(x*log(2)))
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx=-\frac {e^{\frac {2 e^{e^{x}}}{\mathrm {log}\left (2\right ) x}}}{e^{\frac {2 e^{e^{x}} x}{\mathrm {log}\left (2\right )}}} \] Input:
int(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp(x))/x/ log(2))/x^2/log(2),x)
Output:
( - e**((2*e**(e**x))/(log(2)*x)))/e**((2*e**(e**x)*x)/log(2))