Integrand size = 107, antiderivative size = 22 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 e^{-x-\frac {9}{8+x^2-\log (x)}} x \] Output:
5*x/exp(9/(8+x^2-ln(x)))/exp(x)
Time = 5.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 e^{-x+\frac {9}{-8-x^2+\log (x)}} x \] Input:
Integrate[(E^(9/(-8 - x^2 + Log[x]))*(275 - 320*x + 170*x^2 - 80*x^3 + 5*x ^4 - 5*x^5 + (-80 + 80*x - 10*x^2 + 10*x^3)*Log[x] + (5 - 5*x)*Log[x]^2))/ (E^x*(64 + 16*x^2 + x^4) + E^x*(-16 - 2*x^2)*Log[x] + E^x*Log[x]^2),x]
Output:
5*E^(-x + 9/(-8 - 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.
\(\displaystyle \int \frac {e^{\frac {9}{-x^2+\log (x)-8}} \left (-5 x^5+5 x^4-80 x^3+170 x^2+\left (10 x^3-10 x^2+80 x-80\right ) \log (x)-320 x+(5-5 x) \log ^2(x)+275\right )}{e^x \left (-2 x^2-16\right ) \log (x)+e^x \left (x^4+16 x^2+64\right )+e^x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {9}{-x^2+\log (x)-8}-x} \left (-5 x^5+5 x^4-80 x^3+170 x^2+\left (10 x^3-10 x^2+80 x-80\right ) \log (x)-320 x+(5-5 x) \log ^2(x)+275\right )}{\left (x^2-\log (x)+8\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {45 \left (2 x^2-1\right ) e^{\frac {9}{-x^2+\log (x)-8}-x}}{\left (x^2-\log (x)+8\right )^2}-5 (x-1) e^{\frac {9}{-x^2+\log (x)-8}-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \int e^{\frac {9}{-x^2+\log (x)-8}-x}dx-5 \int e^{\frac {9}{-x^2+\log (x)-8}-x} xdx-45 \int \frac {e^{\frac {9}{-x^2+\log (x)-8}-x}}{\left (x^2-\log (x)+8\right )^2}dx+90 \int \frac {e^{\frac {9}{-x^2+\log (x)-8}-x} x^2}{\left (x^2-\log (x)+8\right )^2}dx\) |
Input:
Int[(E^(9/(-8 - x^2 + Log[x]))*(275 - 320*x + 170*x^2 - 80*x^3 + 5*x^4 - 5 *x^5 + (-80 + 80*x - 10*x^2 + 10*x^3)*Log[x] + (5 - 5*x)*Log[x]^2))/(E^x*( 64 + 16*x^2 + x^4) + E^x*(-16 - 2*x^2)*Log[x] + E^x*Log[x]^2),x]
Output:
$Aborted
Time = 17.51 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45
method | result | size |
risch | \(5 x \,{\mathrm e}^{-\frac {-x^{3}+x \ln \left (x \right )-8 x -9}{\ln \left (x \right )-x^{2}-8}}\) | \(32\) |
parallelrisch | \(\frac {\left (4320 x -540 x \ln \left (x \right )+540 x^{3}\right ) {\mathrm e}^{\frac {9}{\ln \left (x \right )-x^{2}-8}} {\mathrm e}^{-x}}{864+108 x^{2}-108 \ln \left (x \right )}\) | \(48\) |
Input:
int(((-5*x+5)*ln(x)^2+(10*x^3-10*x^2+80*x-80)*ln(x)-5*x^5+5*x^4-80*x^3+170 *x^2-320*x+275)/(exp(x)*ln(x)^2+(-2*x^2-16)*exp(x)*ln(x)+(x^4+16*x^2+64)*e xp(x))/exp(-9/(ln(x)-x^2-8)),x,method=_RETURNVERBOSE)
Output:
5*x*exp(-(-x^3+x*ln(x)-8*x-9)/(ln(x)-x^2-8))
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 \, x e^{\left (-x - \frac {9}{x^{2} - \log \left (x\right ) + 8}\right )} \] Input:
integrate(((-5*x+5)*log(x)^2+(10*x^3-10*x^2+80*x-80)*log(x)-5*x^5+5*x^4-80 *x^3+170*x^2-320*x+275)/(exp(x)*log(x)^2+(-2*x^2-16)*exp(x)*log(x)+(x^4+16 *x^2+64)*exp(x))/exp(-9/(log(x)-x^2-8)),x, algorithm="fricas")
Output:
5*x*e^(-x - 9/(x^2 - log(x) + 8))
Time = 0.46 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 x e^{- x} e^{\frac {9}{- x^{2} + \log {\left (x \right )} - 8}} \] Input:
integrate(((-5*x+5)*ln(x)**2+(10*x**3-10*x**2+80*x-80)*ln(x)-5*x**5+5*x**4 -80*x**3+170*x**2-320*x+275)/(exp(x)*ln(x)**2+(-2*x**2-16)*exp(x)*ln(x)+(x **4+16*x**2+64)*exp(x))/exp(-9/(ln(x)-x**2-8)),x)
Output:
5*x*exp(-x)*exp(9/(-x**2 + log(x) - 8))
Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 \, x e^{\left (-x - \frac {9}{x^{2} - \log \left (x\right ) + 8}\right )} \] Input:
integrate(((-5*x+5)*log(x)^2+(10*x^3-10*x^2+80*x-80)*log(x)-5*x^5+5*x^4-80 *x^3+170*x^2-320*x+275)/(exp(x)*log(x)^2+(-2*x^2-16)*exp(x)*log(x)+(x^4+16 *x^2+64)*exp(x))/exp(-9/(log(x)-x^2-8)),x, algorithm="maxima")
Output:
5*x*e^(-x - 9/(x^2 - log(x) + 8))
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 \, x e^{\left (-\frac {8 \, x^{3} - 9 \, x^{2} - 8 \, x \log \left (x\right ) + 64 \, x + 9 \, \log \left (x\right )}{8 \, {\left (x^{2} - \log \left (x\right ) + 8\right )}} - \frac {9}{8}\right )} \] Input:
integrate(((-5*x+5)*log(x)^2+(10*x^3-10*x^2+80*x-80)*log(x)-5*x^5+5*x^4-80 *x^3+170*x^2-320*x+275)/(exp(x)*log(x)^2+(-2*x^2-16)*exp(x)*log(x)+(x^4+16 *x^2+64)*exp(x))/exp(-9/(log(x)-x^2-8)),x, algorithm="giac")
Output:
5*x*e^(-1/8*(8*x^3 - 9*x^2 - 8*x*log(x) + 64*x + 9*log(x))/(x^2 - log(x) + 8) - 9/8)
Timed out. \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^{-\frac {9}{x^2-\ln \left (x\right )+8}}\,\left (320\,x-170\,x^2+80\,x^3-5\,x^4+5\,x^5+{\ln \left (x\right )}^2\,\left (5\,x-5\right )-\ln \left (x\right )\,\left (10\,x^3-10\,x^2+80\,x-80\right )-275\right )}{{\mathrm {e}}^x\,{\ln \left (x\right )}^2-{\mathrm {e}}^x\,\left (2\,x^2+16\right )\,\ln \left (x\right )+{\mathrm {e}}^x\,\left (x^4+16\,x^2+64\right )} \,d x \] Input:
int(-(exp(-9/(x^2 - log(x) + 8))*(320*x - 170*x^2 + 80*x^3 - 5*x^4 + 5*x^5 + log(x)^2*(5*x - 5) - log(x)*(80*x - 10*x^2 + 10*x^3 - 80) - 275))/(exp( x)*log(x)^2 + exp(x)*(16*x^2 + x^4 + 64) - exp(x)*log(x)*(2*x^2 + 16)),x)
Output:
-int((exp(-9/(x^2 - log(x) + 8))*(320*x - 170*x^2 + 80*x^3 - 5*x^4 + 5*x^5 + log(x)^2*(5*x - 5) - log(x)*(80*x - 10*x^2 + 10*x^3 - 80) - 275))/(exp( x)*log(x)^2 + exp(x)*(16*x^2 + x^4 + 64) - exp(x)*log(x)*(2*x^2 + 16)), x)
Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=\frac {5 e^{\frac {9}{\mathrm {log}\left (x \right )-x^{2}-8}} x}{e^{x}} \] Input:
int(((-5*x+5)*log(x)^2+(10*x^3-10*x^2+80*x-80)*log(x)-5*x^5+5*x^4-80*x^3+1 70*x^2-320*x+275)/(exp(x)*log(x)^2+(-2*x^2-16)*exp(x)*log(x)+(x^4+16*x^2+6 4)*exp(x))/exp(-9/(log(x)-x^2-8)),x)
Output:
(5*e**(9/(log(x) - x**2 - 8))*x)/e**x