\(\int \frac {e^{\frac {9}{-8-x^2+\log (x)}} (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+(-80+80 x-10 x^2+10 x^3) \log (x)+(5-5 x) \log ^2(x))}{e^x (64+16 x^2+x^4)+e^x (-16-2 x^2) \log (x)+e^x \log ^2(x)} \, dx\) [5698]

Optimal result

Integrand size = 107, antiderivative size = 22 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 e^{-x-\frac {9}{8+x^2-\log (x)}} x \]

[Out]

5*x/exp(9/(8+x^2-ln(x)))/exp(x)

Rubi [F]

\[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=\int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx \]

[In]

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]

[Out]

5*Defer[Int][E^(-x + 9/(-8 - x^2 + Log[x])), x] - 5*Defer[Int][E^(-x + 9/(-8 - x^2 + Log[x]))*x, x] - 45*Defer
[Int][E^(-x + 9/(-8 - x^2 + Log[x]))/(8 + x^2 - Log[x])^2, x] + 90*Defer[Int][(E^(-x + 9/(-8 - x^2 + Log[x]))*
x^2)/(8 + x^2 - Log[x])^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x+\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 )}{\left (8+x^2-\log (x)\right )^2} \, dx \\ & = \int \left (-5 e^{-x+\frac {9}{-8-x^2+\log (x)}} (-1+x)+\frac {45 e^{-x+\frac {9}{-8-x^2+\log (x)}} \left (-1+2 x^2\right )}{\left (8+x^2-\log (x)\right )^2}\right ) \, dx \\ & = -\left (5 \int e^{-x+\frac {9}{-8-x^2+\log (x)}} (-1+x) \, dx\right )+45 \int \frac {e^{-x+\frac {9}{-8-x^2+\log (x)}} \left (-1+2 x^2\right )}{\left (8+x^2-\log (x)\right )^2} \, dx \\ & = -\left (5 \int \left (-e^{-x+\frac {9}{-8-x^2+\log (x)}}+e^{-x+\frac {9}{-8-x^2+\log (x)}} x\right ) \, dx\right )+45 \int \left (-\frac {e^{-x+\frac {9}{-8-x^2+\log (x)}}}{\left (8+x^2-\log (x)\right )^2}+\frac {2 e^{-x+\frac {9}{-8-x^2+\log (x)}} x^2}{\left (8+x^2-\log (x)\right )^2}\right ) \, dx \\ & = 5 \int e^{-x+\frac {9}{-8-x^2+\log (x)}} \, dx-5 \int e^{-x+\frac {9}{-8-x^2+\log (x)}} x \, dx-45 \int \frac {e^{-x+\frac {9}{-8-x^2+\log (x)}}}{\left (8+x^2-\log (x)\right )^2} \, dx+90 \int \frac {e^{-x+\frac {9}{-8-x^2+\log (x)}} x^2}{\left (8+x^2-\log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 e^{-x+\frac {9}{-8-x^2+\log (x)}} x \]

[In]

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]

[Out]

5*E^(-x + 9/(-8 - x^2 + Log[x]))*x

Maple [A] (verified)

Time = 17.50 (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	\(-\frac {\left (-61440 x +7680 x \ln \left (x \right )-7680 x^{3}\right ) {\mathrm e}^{\frac {9}{\ln \left (x \right )-x^{2}-8}} {\mathrm e}^{-x}}{1536 \left (8+x^{2}-\ln \left (x \right )\right )}\)	\(48\)

[In]

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)*exp(x))/exp(-9/(ln(x)-x^2-8)),x,method=_RETURNVERBOSE)

[Out]

5*x*exp(-(-x^3+x*ln(x)-8*x-9)/(ln(x)-x^2-8))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 \, x e^{\left (-x - \frac {9}{x^{2} - \log \left (x\right ) + 8}\right )} \]

[In]

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")

[Out]

5*x*e^(-x - 9/(x^2 - log(x) + 8))

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 x e^{- x} e^{\frac {9}{- x^{2} + \log {\left (x \right )} - 8}} \]

[In]

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)

[Out]

5*x*exp(-x)*exp(9/(-x**2 + log(x) - 8))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 \, x e^{\left (-x - \frac {9}{x^{2} - \log \left (x\right ) + 8}\right )} \]

[In]

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")

[Out]

5*x*e^(-x - 9/(x^2 - log(x) + 8))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=5 \, x e^{\left (-\frac {8 \, x^{3} - 9 \, x^{2} - 8 \, x \log \left (x\right ) + 64 \, x + 9 \, \log \left (x\right )}{8 \, {\left (x^{2} - \log \left (x\right ) + 8\right )}} - \frac {9}{8}\right )} \]

[In]

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")

[Out]

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)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^{-\frac {9}{x^2-\ln \left (x\right )+8}}\,\left (320\,x-170\,x^2+80\,x^3-5\,x^4+5\,x^5+{\ln \left (x\right )}^2\,\left (5\,x-5\right )-\ln \left (x\right )\,\left (10\,x^3-10\,x^2+80\,x-80\right )-275\right )}{{\mathrm {e}}^x\,{\ln \left (x\right )}^2-{\mathrm {e}}^x\,\left (2\,x^2+16\right )\,\ln \left (x\right )+{\mathrm {e}}^x\,\left (x^4+16\,x^2+64\right )} \,d x \]

[In]

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
)

[Out]

-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)