Integrand size = 72, antiderivative size = 24 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=-1+e^{e^{15} x \left (x+e^{-x} x\right )}+\frac {\log (x)}{x} \] Output:
exp(exp(15)*x*(x/exp(x)+x))+ln(x)/x-1
Time = 1.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=e^{e^{15-x} \left (1+e^x\right ) x^2}+\frac {\log (x)}{x} \] Input:
Integrate[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E ^15*(2*x^3 - x^4)) - E^x*Log[x])/(E^x*x^2),x]
Output:
E^(E^(15 - x)*(1 + E^x)*x^2) + Log[x]/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.
\(\displaystyle \int \frac {e^{-x} \left (e^{e^{-x} \left (e^{x+15} x^2+e^{15} x^2\right )} \left (2 e^{x+15} x^3+e^{15} \left (2 x^3-x^4\right )\right )+e^x-e^x \log (x)\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{e^{15-x} x^2+e^{15} x^2-x+15} \left (-x+2 e^x+2\right ) x-\frac {\log (x)-1}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int e^{e^{15-x} x^2+e^{15} x^2+15} xdx+2 \int e^{e^{15-x} x^2+e^{15} x^2-x+15} xdx-\int e^{e^{15-x} x^2+e^{15} x^2-x+15} x^2dx+\frac {\log (x)}{x}\) |
Input:
Int[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E^15*(2 *x^3 - x^4)) - E^x*Log[x])/(E^x*x^2),x]
Output:
$Aborted
Time = 3.48 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\ln \left (x \right )}{x}+{\mathrm e}^{x^{2} \left ({\mathrm e}^{15}+{\mathrm e}^{x +15}\right ) {\mathrm e}^{-x}}\) | \(24\) |
default | \({\mathrm e}^{x^{2} \left ({\mathrm e}^{15}+{\mathrm e}^{15} {\mathrm e}^{x}\right ) {\mathrm e}^{-x}}+\frac {\ln \left (x \right )}{x}\) | \(25\) |
parallelrisch | \(\frac {{\mathrm e}^{x^{2} {\mathrm e}^{15} \left ({\mathrm e}^{x}+1\right ) {\mathrm e}^{-x}} x +\ln \left (x \right )}{x}\) | \(25\) |
Input:
int(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x ^2*exp(15))/exp(x))-exp(x)*ln(x)+exp(x))/exp(x)/x^2,x,method=_RETURNVERBOS E)
Output:
ln(x)/x+exp(x^2*(exp(15)+exp(x+15))*exp(-x))
Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {x e^{\left ({\left (x^{2} e^{30} + x^{2} e^{\left (x + 30\right )}\right )} e^{\left (-x - 15\right )}\right )} + \log \left (x\right )}{x} \] Input:
integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*ex p(x)+x^2*exp(15))/exp(x))-exp(x)*log(x)+exp(x))/exp(x)/x^2,x, algorithm="f ricas")
Output:
(x*e^((x^2*e^30 + x^2*e^(x + 30))*e^(-x - 15)) + log(x))/x
Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=e^{\left (x^{2} e^{15} e^{x} + x^{2} e^{15}\right ) e^{- x}} + \frac {\log {\left (x \right )}}{x} \] Input:
integrate(((2*x**3*exp(15)*exp(x)+(-x**4+2*x**3)*exp(15))*exp((x**2*exp(15 )*exp(x)+x**2*exp(15))/exp(x))-exp(x)*ln(x)+exp(x))/exp(x)/x**2,x)
Output:
exp((x**2*exp(15)*exp(x) + x**2*exp(15))*exp(-x)) + log(x)/x
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {x e^{\left (x^{2} e^{15} + x^{2} e^{\left (-x + 15\right )}\right )} + \log \left (x\right ) + 1}{x} - \frac {1}{x} \] Input:
integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*ex p(x)+x^2*exp(15))/exp(x))-exp(x)*log(x)+exp(x))/exp(x)/x^2,x, algorithm="m axima")
Output:
(x*e^(x^2*e^15 + x^2*e^(-x + 15)) + log(x) + 1)/x - 1/x
\[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\int { \frac {{\left ({\left (2 \, x^{3} e^{\left (x + 15\right )} - {\left (x^{4} - 2 \, x^{3}\right )} e^{15}\right )} e^{\left ({\left (x^{2} e^{15} + x^{2} e^{\left (x + 15\right )}\right )} e^{\left (-x\right )}\right )} - e^{x} \log \left (x\right ) + e^{x}\right )} e^{\left (-x\right )}}{x^{2}} \,d x } \] Input:
integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*ex p(x)+x^2*exp(15))/exp(x))-exp(x)*log(x)+exp(x))/exp(x)/x^2,x, algorithm="g iac")
Output:
integrate(((2*x^3*e^(x + 15) - (x^4 - 2*x^3)*e^15)*e^((x^2*e^15 + x^2*e^(x + 15))*e^(-x)) - e^x*log(x) + e^x)*e^(-x)/x^2, x)
Time = 3.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {\ln \left (x\right )}{x}+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{15}}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{15}} \] Input:
int((exp(-x)*(exp(x) - exp(x)*log(x) + exp(exp(-x)*(x^2*exp(15) + x^2*exp( 15)*exp(x)))*(exp(15)*(2*x^3 - x^4) + 2*x^3*exp(15)*exp(x))))/x^2,x)
Output:
log(x)/x + exp(x^2*exp(15))*exp(x^2*exp(-x)*exp(15))
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {e^{\frac {e^{x} e^{15} x^{2}+e^{15} x^{2}}{e^{x}}} x +\mathrm {log}\left (x \right )}{x} \] Input:
int(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x ^2*exp(15))/exp(x))-exp(x)*log(x)+exp(x))/exp(x)/x^2,x)
Output:
(e**((e**x*e**15*x**2 + e**15*x**2)/e**x)*x + log(x))/x