Integrand size = 73, antiderivative size = 33 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {4 e^{-x} x \left (e^2+x+\frac {3}{\log (x)}\right )}{3 \left (-2+x+\frac {3+x}{4}\right )} \] Output:
4/3*x/exp(x)/(5/4*x-5/4)*(x+exp(2)+3/ln(x))
Time = 3.67 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {16 e^{-x} x \left (3+\left (e^2+x\right ) \log (x)\right )}{15 (-1+x) \log (x)} \] Input:
Integrate[(48 - 48*x + (-48 + 48*x - 48*x^2)*Log[x] + (-32*x + 32*x^2 - 16 *x^3 + E^2*(-16 + 16*x - 16*x^2))*Log[x]^2)/(E^x*(15 - 30*x + 15*x^2)*Log[ x]^2),x]
Output:
(16*x*(3 + (E^2 + x)*Log[x]))/(15*E^x*(-1 + x)*Log[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.
\(\displaystyle \int \frac {e^{-x} \left (\left (-48 x^2+48 x-48\right ) \log (x)+\left (-16 x^3+32 x^2+e^2 \left (-16 x^2+16 x-16\right )-32 x\right ) \log ^2(x)-48 x+48\right )}{\left (15 x^2-30 x+15\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 60 \int \frac {4 e^{-x} \left (-\left (\left (x^3-2 x^2+2 x+e^2 \left (x^2-x+1\right )\right ) \log ^2(x)\right )-3 \left (x^2-x+1\right ) \log (x)-3 x+3\right )}{225 (1-x)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {16}{15} \int \frac {e^{-x} \left (-\left (\left (x^3-2 x^2+2 x+e^2 \left (x^2-x+1\right )\right ) \log ^2(x)\right )-3 \left (x^2-x+1\right ) \log (x)-3 x+3\right )}{(1-x)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {16}{15} \int \left (-\frac {3 e^{-x} \left (x^2-x+1\right )}{(x-1)^2 \log (x)}+\frac {e^{-x} \left (-x^3+\left (2-e^2\right ) x^2-\left (2-e^2\right ) x-e^2\right )}{(1-x)^2}-\frac {3 e^{-x}}{(x-1) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {16}{15} \left (-3 \int \frac {e^{-x}}{(x-1) \log ^2(x)}dx-3 \int \frac {e^{-x}}{\log (x)}dx-3 \int \frac {e^{-x}}{(x-1)^2 \log (x)}dx-3 \int \frac {e^{-x}}{(x-1) \log (x)}dx+e^{-x} x+e^{2-x}+e^{-x}-\frac {\left (1+e^2\right ) e^{-x}}{1-x}\right )\) |
Input:
Int[(48 - 48*x + (-48 + 48*x - 48*x^2)*Log[x] + (-32*x + 32*x^2 - 16*x^3 + E^2*(-16 + 16*x - 16*x^2))*Log[x]^2)/(E^x*(15 - 30*x + 15*x^2)*Log[x]^2), x]
Output:
$Aborted
Time = 1.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {16 x \left (x +{\mathrm e}^{2}\right ) {\mathrm e}^{-x}}{15 \left (-1+x \right )}+\frac {16 x \,{\mathrm e}^{-x}}{5 \left (-1+x \right ) \ln \left (x \right )}\) | \(34\) |
parallelrisch | \(\frac {\left (16 x \,{\mathrm e}^{2} \ln \left (x \right )+16 x^{2} \ln \left (x \right )+48 x \right ) {\mathrm e}^{-x}}{15 \ln \left (x \right ) \left (-1+x \right )}\) | \(34\) |
Input:
int((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*ln(x)^2+(-48*x^2+48*x-4 8)*ln(x)-48*x+48)/(15*x^2-30*x+15)/exp(x)/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
16/15*x*(x+exp(2))/(-1+x)*exp(-x)+16/5*x*exp(-x)/(-1+x)/ln(x)
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {16 \, {\left ({\left (x^{2} + x e^{2}\right )} e^{\left (-x\right )} \log \left (x\right ) + 3 \, x e^{\left (-x\right )}\right )}}{15 \, {\left (x - 1\right )} \log \left (x\right )} \] Input:
integrate((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*log(x)^2+(-48*x^2 +48*x-48)*log(x)-48*x+48)/(15*x^2-30*x+15)/exp(x)/log(x)^2,x, algorithm="f ricas")
Output:
16/15*((x^2 + x*e^2)*e^(-x)*log(x) + 3*x*e^(-x))/((x - 1)*log(x))
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {\left (16 x^{2} \log {\left (x \right )} + 16 x e^{2} \log {\left (x \right )} + 48 x\right ) e^{- x}}{15 x \log {\left (x \right )} - 15 \log {\left (x \right )}} \] Input:
integrate((((-16*x**2+16*x-16)*exp(2)-16*x**3+32*x**2-32*x)*ln(x)**2+(-48* x**2+48*x-48)*ln(x)-48*x+48)/(15*x**2-30*x+15)/exp(x)/ln(x)**2,x)
Output:
(16*x**2*log(x) + 16*x*exp(2)*log(x) + 48*x)*exp(-x)/(15*x*log(x) - 15*log (x))
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {16 \, {\left ({\left (x^{2} + x e^{2}\right )} \log \left (x\right ) + 3 \, x\right )} e^{\left (-x\right )}}{15 \, {\left (x - 1\right )} \log \left (x\right )} \] Input:
integrate((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*log(x)^2+(-48*x^2 +48*x-48)*log(x)-48*x+48)/(15*x^2-30*x+15)/exp(x)/log(x)^2,x, algorithm="m axima")
Output:
16/15*((x^2 + x*e^2)*log(x) + 3*x)*e^(-x)/((x - 1)*log(x))
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {16 \, {\left (x^{2} e^{\left (-x\right )} \log \left (x\right ) + x e^{\left (-x + 2\right )} \log \left (x\right ) + 3 \, x e^{\left (-x\right )}\right )}}{15 \, {\left (x \log \left (x\right ) - \log \left (x\right )\right )}} \] Input:
integrate((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*log(x)^2+(-48*x^2 +48*x-48)*log(x)-48*x+48)/(15*x^2-30*x+15)/exp(x)/log(x)^2,x, algorithm="g iac")
Output:
16/15*(x^2*e^(-x)*log(x) + x*e^(-x + 2)*log(x) + 3*x*e^(-x))/(x*log(x) - l og(x))
Time = 2.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {16\,x\,{\mathrm {e}}^{-x}\,\left (x+{\mathrm {e}}^2\right )}{15\,\left (x-1\right )}+\frac {16\,x\,{\mathrm {e}}^{-x}}{5\,\ln \left (x\right )\,\left (x-1\right )} \] Input:
int(-(exp(-x)*(48*x + log(x)^2*(32*x + exp(2)*(16*x^2 - 16*x + 16) - 32*x^ 2 + 16*x^3) + log(x)*(48*x^2 - 48*x + 48) - 48))/(log(x)^2*(15*x^2 - 30*x + 15)),x)
Output:
(16*x*exp(-x)*(x + exp(2)))/(15*(x - 1)) + (16*x*exp(-x))/(5*log(x)*(x - 1 ))
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-x} \left (48-48 x+\left (-48+48 x-48 x^2\right ) \log (x)+\left (-32 x+32 x^2-16 x^3+e^2 \left (-16+16 x-16 x^2\right )\right ) \log ^2(x)\right )}{\left (15-30 x+15 x^2\right ) \log ^2(x)} \, dx=\frac {16 x \left (\mathrm {log}\left (x \right ) e^{2}+\mathrm {log}\left (x \right ) x +3\right )}{15 e^{x} \mathrm {log}\left (x \right ) \left (x -1\right )} \] Input:
int((((-16*x^2+16*x-16)*exp(2)-16*x^3+32*x^2-32*x)*log(x)^2+(-48*x^2+48*x- 48)*log(x)-48*x+48)/(15*x^2-30*x+15)/exp(x)/log(x)^2,x)
Output:
(16*x*(log(x)*e**2 + log(x)*x + 3))/(15*e**x*log(x)*(x - 1))