Integrand size = 54, antiderivative size = 26 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=x^2-\frac {1}{10} e^{2-2 e^x} x \left (-4+\log ^2(6 x)\right ) \] Output:
x^2-1/2*x/exp(-1+exp(x))^2*(1/5*ln(6*x)^2-4/5)
Time = 1.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\frac {1}{10} \left (4 e^{2-2 e^x} x+10 x^2-e^{2-2 e^x} x \log ^2(6 x)\right ) \] Input:
Integrate[(E^(2 - 2*E^x)*(4 + 20*E^(-2 + 2*E^x)*x - 8*E^x*x - 2*Log[6*x] + (-1 + 2*E^x*x)*Log[6*x]^2))/10,x]
Output:
(4*E^(2 - 2*E^x)*x + 10*x^2 - E^(2 - 2*E^x)*x*Log[6*x]^2)/10
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.
\(\displaystyle \int \frac {1}{10} e^{2-2 e^x} \left (20 e^{2 e^x-2} x-8 e^x x+\left (2 e^x x-1\right ) \log ^2(6 x)-2 \log (6 x)+4\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int e^{2-2 e^x} \left (-\left (\left (1-2 e^x x\right ) \log ^2(6 x)\right )-2 \log (6 x)+20 e^{-2+2 e^x} x-8 e^x x+4\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{10} \int e^{-2 \left (-1+e^x\right )} \left (-\left (\left (1-2 e^x x\right ) \log ^2(6 x)\right )-2 \log (6 x)+20 e^{-2+2 e^x} x-8 e^x x+4\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{10} \int \left (e^{-2 \left (-1+e^x\right )} \left (2 e^x x-1\right ) \log ^2(6 x)-2 e^{-2 \left (-1+e^x\right )} \log (6 x)+4 e^{-2 \left (-1+e^x\right )}-8 e^{x-2 \left (-1+e^x\right )} x+20 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} \left (2 e^2 \int \frac {\operatorname {ExpIntegralEi}\left (-2 e^x\right )}{x}dx-8 \int e^{x-2 \left (-1+e^x\right )} xdx-\int e^{-2 \left (-1+e^x\right )} \log ^2(6 x)dx+2 \int e^{x-2 \left (-1+e^x\right )} x \log ^2(6 x)dx+4 e^2 \operatorname {ExpIntegralEi}\left (-2 e^x\right )-2 e^2 \operatorname {ExpIntegralEi}\left (-2 e^x\right ) \log (6 x)+10 x^2\right )\) |
Input:
Int[(E^(2 - 2*E^x)*(4 + 20*E^(-2 + 2*E^x)*x - 8*E^x*x - 2*Log[6*x] + (-1 + 2*E^x*x)*Log[6*x]^2))/10,x]
Output:
$Aborted
Time = 0.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
risch | \(x^{2}+\frac {\left (-x \ln \left (6 x \right )^{2}+4 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}+2}}{10}\) | \(27\) |
parallelrisch | \(\frac {\left (10 \,{\mathrm e}^{2 \,{\mathrm e}^{x}-2} x^{2}-x \ln \left (6 x \right )^{2}+4 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}+2}}{10}\) | \(35\) |
Input:
int(1/10*(20*x*exp(exp(x)-1)^2+(2*exp(x)*x-1)*ln(6*x)^2-2*ln(6*x)-8*exp(x) *x+4)/exp(exp(x)-1)^2,x,method=_RETURNVERBOSE)
Output:
x^2+1/10*(-x*ln(6*x)^2+4*x)*exp(-2*exp(x)+2)
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\frac {1}{10} \, {\left (10 \, x^{2} e^{\left (2 \, e^{x} - 2\right )} - x \log \left (6 \, x\right )^{2} + 4 \, x\right )} e^{\left (-2 \, e^{x} + 2\right )} \] Input:
integrate(1/10*(20*x*exp(-1+exp(x))^2+(2*exp(x)*x-1)*log(6*x)^2-2*log(6*x) -8*exp(x)*x+4)/exp(-1+exp(x))^2,x, algorithm="fricas")
Output:
1/10*(10*x^2*e^(2*e^x - 2) - x*log(6*x)^2 + 4*x)*e^(-2*e^x + 2)
Time = 5.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=x^{2} + \frac {\left (- x \log {\left (6 x \right )}^{2} + 4 x\right ) e^{2 - 2 e^{x}}}{10} \] Input:
integrate(1/10*(20*x*exp(-1+exp(x))**2+(2*exp(x)*x-1)*ln(6*x)**2-2*ln(6*x) -8*exp(x)*x+4)/exp(-1+exp(x))**2,x)
Output:
x**2 + (-x*log(6*x)**2 + 4*x)*exp(2 - 2*exp(x))/10
\[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\int { \frac {1}{10} \, {\left ({\left (2 \, x e^{x} - 1\right )} \log \left (6 \, x\right )^{2} - 8 \, x e^{x} + 20 \, x e^{\left (2 \, e^{x} - 2\right )} - 2 \, \log \left (6 \, x\right ) + 4\right )} e^{\left (-2 \, e^{x} + 2\right )} \,d x } \] Input:
integrate(1/10*(20*x*exp(-1+exp(x))^2+(2*exp(x)*x-1)*log(6*x)^2-2*log(6*x) -8*exp(x)*x+4)/exp(-1+exp(x))^2,x, algorithm="maxima")
Output:
x^2 + 2/5*Ei(-2*e^x)*e^2 - 1/10*(2*x*(log(3) + log(2))*e^2*log(x) + x*e^2* log(x)^2 + (log(3)^2 + 2*log(3)*log(2) + log(2)^2 - 4)*x*e^2)*e^(-2*e^x) - 2/5*integrate(e^(-2*e^x + 2), x)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.62 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=-\frac {1}{10} \, {\left (x e^{\left (x - 2 \, e^{x} + 2\right )} \log \left (6\right )^{2} + 2 \, x e^{\left (x - 2 \, e^{x} + 2\right )} \log \left (6\right ) \log \left (x\right ) + x e^{\left (x - 2 \, e^{x} + 2\right )} \log \left (x\right )^{2} - 10 \, x^{2} e^{x} - 4 \, x e^{\left (x - 2 \, e^{x} + 2\right )}\right )} e^{\left (-x\right )} \] Input:
integrate(1/10*(20*x*exp(-1+exp(x))^2+(2*exp(x)*x-1)*log(6*x)^2-2*log(6*x) -8*exp(x)*x+4)/exp(-1+exp(x))^2,x, algorithm="giac")
Output:
-1/10*(x*e^(x - 2*e^x + 2)*log(6)^2 + 2*x*e^(x - 2*e^x + 2)*log(6)*log(x) + x*e^(x - 2*e^x + 2)*log(x)^2 - 10*x^2*e^x - 4*x*e^(x - 2*e^x + 2))*e^(-x )
Time = 4.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx={\mathrm {e}}^{2-2\,{\mathrm {e}}^x}\,\left (\frac {2\,x}{5}-\frac {x\,{\ln \left (6\,x\right )}^2}{10}\right )+x^2 \] Input:
int(exp(2 - 2*exp(x))*((log(6*x)^2*(2*x*exp(x) - 1))/10 - log(6*x)/5 - (4* x*exp(x))/5 + 2*x*exp(2*exp(x) - 2) + 2/5),x)
Output:
exp(2 - 2*exp(x))*((2*x)/5 - (x*log(6*x)^2)/10) + x^2
Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\frac {x \left (10 e^{2 e^{x}} x -\mathrm {log}\left (6 x \right )^{2} e^{2}+4 e^{2}\right )}{10 e^{2 e^{x}}} \] Input:
int(1/10*(20*x*exp(-1+exp(x))^2+(2*exp(x)*x-1)*log(6*x)^2-2*log(6*x)-8*exp (x)*x+4)/exp(-1+exp(x))^2,x)
Output:
(x*(10*e**(2*e**x)*x - log(6*x)**2*e**2 + 4*e**2))/(10*e**(2*e**x))