Integrand size = 64, antiderivative size = 28 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx=e^x+\frac {1}{4} x^2 \log (x) \left (10-x^2+e^{-x} \log (x)\right ) \] Output:
exp(x)+1/4*x^2*ln(x)*(ln(x)/exp(x)+10-x^2)
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx=\frac {1}{4} \left (4 e^x+10 x^2 \log (x)-x^4 \log (x)+e^{-x} x^2 \log ^2(x)\right ) \] Input:
Integrate[(4*E^(2*x) + E^x*(10*x - x^3) + (2*x + E^x*(20*x - 4*x^3))*Log[x ] + (2*x - x^2)*Log[x]^2)/(4*E^x),x]
Output:
(4*E^x + 10*x^2*Log[x] - x^4*Log[x] + (x^2*Log[x]^2)/E^x)/4
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.
\(\displaystyle \int \frac {1}{4} e^{-x} \left (e^x \left (10 x-x^3\right )+\left (e^x \left (20 x-4 x^3\right )+2 x\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)+4 e^{2 x}\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int e^{-x} \left (\left (2 x-x^2\right ) \log ^2(x)+2 \left (x+2 e^x \left (5 x-x^3\right )\right ) \log (x)+4 e^{2 x}+e^x \left (10 x-x^3\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (-e^{-x} (x-2) x \log ^2(x)-2 e^{-x} x \left (2 e^x x^2-10 e^x-1\right ) \log (x)+4 e^x-x \left (x^2-10\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\int e^{-x} x^2 \log ^2(x)dx+2 \int e^{-x} x \log ^2(x)dx+2 \operatorname {ExpIntegralEi}(-x)+x^4 (-\log (x))+10 x^2 \log (x)-2 e^{-x}+4 e^x-2 e^{-x} x \log (x)-2 e^{-x} \log (x)\right )\) |
Input:
Int[(4*E^(2*x) + E^x*(10*x - x^3) + (2*x + E^x*(20*x - 4*x^3))*Log[x] + (2 *x - x^2)*Log[x]^2)/(4*E^x),x]
Output:
$Aborted
Time = 14.58 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
method | result | size |
default | \(\frac {\ln \left (x \right )^{2} x^{2} {\mathrm e}^{-x}}{4}-\frac {x^{4} \ln \left (x \right )}{4}+\frac {5 x^{2} \ln \left (x \right )}{2}+{\mathrm e}^{x}\) | \(31\) |
parts | \(\frac {\ln \left (x \right )^{2} x^{2} {\mathrm e}^{-x}}{4}-\frac {x^{4} \ln \left (x \right )}{4}+\frac {5 x^{2} \ln \left (x \right )}{2}+{\mathrm e}^{x}\) | \(31\) |
risch | \(\frac {\ln \left (x \right )^{2} x^{2} {\mathrm e}^{-x}}{4}-\frac {\left (x^{2}-5\right )^{2} \ln \left (x \right )}{4}+\frac {25 \ln \left (x \right )}{4}+{\mathrm e}^{x}\) | \(32\) |
parallelrisch | \(-\frac {\left (\ln \left (x \right ) {\mathrm e}^{x} x^{4}-x^{2} \ln \left (x \right )^{2}-10 x^{2} {\mathrm e}^{x} \ln \left (x \right )-4 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}}{4}\) | \(40\) |
orering | \(\text {Expression too large to display}\) | \(4086\) |
Input:
int(1/4*((-x^2+2*x)*ln(x)^2+((-4*x^3+20*x)*exp(x)+2*x)*ln(x)+4*exp(x)^2+(- x^3+10*x)*exp(x))/exp(x),x,method=_RETURNVERBOSE)
Output:
1/4*ln(x)^2*x^2/exp(x)-1/4*x^4*ln(x)+5/2*x^2*ln(x)+exp(x)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx=\frac {1}{4} \, {\left (x^{2} \log \left (x\right )^{2} - {\left (x^{4} - 10 \, x^{2}\right )} e^{x} \log \left (x\right ) + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} \] Input:
integrate(1/4*((-x^2+2*x)*log(x)^2+((-4*x^3+20*x)*exp(x)+2*x)*log(x)+4*exp (x)^2+(-x^3+10*x)*exp(x))/exp(x),x, algorithm="fricas")
Output:
1/4*(x^2*log(x)^2 - (x^4 - 10*x^2)*e^x*log(x) + 4*e^(2*x))*e^(-x)
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx=\frac {x^{2} e^{- x} \log {\left (x \right )}^{2}}{4} + \left (- \frac {x^{4}}{4} + \frac {5 x^{2}}{2}\right ) \log {\left (x \right )} + e^{x} \] Input:
integrate(1/4*((-x**2+2*x)*ln(x)**2+((-4*x**3+20*x)*exp(x)+2*x)*ln(x)+4*ex p(x)**2+(-x**3+10*x)*exp(x))/exp(x),x)
Output:
x**2*exp(-x)*log(x)**2/4 + (-x**4/4 + 5*x**2/2)*log(x) + exp(x)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx=-\frac {1}{4} \, x^{4} \log \left (x\right ) + \frac {1}{4} \, x^{2} e^{\left (-x\right )} \log \left (x\right )^{2} + \frac {5}{2} \, x^{2} \log \left (x\right ) + e^{x} \] Input:
integrate(1/4*((-x^2+2*x)*log(x)^2+((-4*x^3+20*x)*exp(x)+2*x)*log(x)+4*exp (x)^2+(-x^3+10*x)*exp(x))/exp(x),x, algorithm="maxima")
Output:
-1/4*x^4*log(x) + 1/4*x^2*e^(-x)*log(x)^2 + 5/2*x^2*log(x) + e^x
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx=-\frac {1}{4} \, x^{4} \log \left (x\right ) + \frac {1}{4} \, x^{2} e^{\left (-x\right )} \log \left (x\right )^{2} + \frac {5}{2} \, x^{2} \log \left (x\right ) + e^{x} \] Input:
integrate(1/4*((-x^2+2*x)*log(x)^2+((-4*x^3+20*x)*exp(x)+2*x)*log(x)+4*exp (x)^2+(-x^3+10*x)*exp(x))/exp(x),x, algorithm="giac")
Output:
-1/4*x^4*log(x) + 1/4*x^2*e^(-x)*log(x)^2 + 5/2*x^2*log(x) + e^x
Time = 4.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.89 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx={\mathrm {e}}^x+\frac {2\,x\,{\mathrm {e}}^{-x}+2\,x^2\,{\mathrm {e}}^{-x}\,\ln \left (x\right )+x^3\,{\mathrm {e}}^{-x}\,{\ln \left (x\right )}^2+2\,x\,{\mathrm {e}}^{-x}\,\ln \left (x\right )}{4\,x}+\frac {x^2\,\left (40\,\ln \left (x\right )-4\,x^2\,\ln \left (x\right )+x^2-20\right )}{16}+\frac {5\,x^2}{4}-\frac {x^4}{16}-\frac {x\,{\mathrm {e}}^{-x}+x^2\,{\mathrm {e}}^{-x}\,\ln \left (x\right )+x\,{\mathrm {e}}^{-x}\,\ln \left (x\right )}{2\,x} \] Input:
int(exp(-x)*(exp(2*x) + (log(x)^2*(2*x - x^2))/4 + (log(x)*(2*x + exp(x)*( 20*x - 4*x^3)))/4 + (exp(x)*(10*x - x^3))/4),x)
Output:
exp(x) + (2*x*exp(-x) + 2*x^2*exp(-x)*log(x) + x^3*exp(-x)*log(x)^2 + 2*x* exp(-x)*log(x))/(4*x) + (x^2*(40*log(x) - 4*x^2*log(x) + x^2 - 20))/16 + ( 5*x^2)/4 - x^4/16 - (x*exp(-x) + x^2*exp(-x)*log(x) + x*exp(-x)*log(x))/(2 *x)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {1}{4} e^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx=\frac {4 e^{2 x}-e^{x} \mathrm {log}\left (x \right ) x^{4}+10 e^{x} \mathrm {log}\left (x \right ) x^{2}+\mathrm {log}\left (x \right )^{2} x^{2}}{4 e^{x}} \] Input:
int(1/4*((-x^2+2*x)*log(x)^2+((-4*x^3+20*x)*exp(x)+2*x)*log(x)+4*exp(x)^2+ (-x^3+10*x)*exp(x))/exp(x),x)
Output:
(4*e**(2*x) - e**x*log(x)*x**4 + 10*e**x*log(x)*x**2 + log(x)**2*x**2)/(4* e**x)