Integrand size = 76, antiderivative size = 21 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)} \] Output:
exp((-2-2*x)*exp(-exp(16)^2/(ln(x)-5)))
Time = 2.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{-2 e^{-\frac {e^{32}}{-5+\log (x)}} (1+x)} \] Input:
Integrate[(E^((-2 - 2*x)/E^(E^32/(-5 + Log[x])) - E^32/(-5 + Log[x]))*(E^3 2*(-2 - 2*x) - 50*x + 20*x*Log[x] - 2*x*Log[x]^2))/(25*x - 10*x*Log[x] + x *Log[x]^2),x]
Output:
E^((-2*(1 + x))/E^(E^32/(-5 + 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.
\(\displaystyle \int \frac {\left (e^{32} (-2 x-2)-50 x-2 x \log ^2(x)+20 x \log (x)\right ) \exp \left ((-2 x-2) e^{-\frac {e^{32}}{\log (x)-5}}-\frac {e^{32}}{\log (x)-5}\right )}{25 x+x \log ^2(x)-10 x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (-25 \left (1+\frac {e^{32}}{25}\right ) x-x \log ^2(x)+10 x \log (x)-e^{32}\right ) \exp \left ((-2 x-2) e^{-\frac {e^{32}}{\log (x)-5}}-\frac {e^{32}}{\log (x)-5}\right )}{x (5-\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {e^{\frac {e^{32}}{5-\log (x)}-2 e^{\frac {e^{32}}{5-\log (x)}} (x+1)} \left (x \log ^2(x)-10 x \log (x)+\left (25+e^{32}\right ) x+e^{32}\right )}{x (5-\log (x))^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {e^{\frac {e^{32}}{5-\log (x)}-2 e^{\frac {e^{32}}{5-\log (x)}} (x+1)} \left (x \log ^2(x)-10 x \log (x)+\left (25+e^{32}\right ) x+e^{32}\right )}{x (5-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {\exp \left (-2 e^{\frac {e^{32}}{5-\log (x)}} (x+1)+\frac {e^{32}}{5-\log (x)}+32\right ) (x+1)}{x (\log (x)-5)^2}+e^{\frac {e^{32}}{5-\log (x)}-2 e^{\frac {e^{32}}{5-\log (x)}} (x+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\int \frac {\exp \left (-2 e^{\frac {e^{32}}{5-\log (x)}} (x+1)+\frac {e^{32}}{5-\log (x)}+32\right )}{(\log (x)-5)^2}dx+\int \frac {\exp \left (-2 e^{\frac {e^{32}}{5-\log (x)}} (x+1)+\frac {e^{32}}{5-\log (x)}+32\right )}{x (\log (x)-5)^2}dx+\int e^{\frac {e^{32}}{5-\log (x)}-2 e^{\frac {e^{32}}{5-\log (x)}} (x+1)}dx\right )\) |
Input:
Int[(E^((-2 - 2*x)/E^(E^32/(-5 + Log[x])) - E^32/(-5 + Log[x]))*(E^32*(-2 - 2*x) - 50*x + 20*x*Log[x] - 2*x*Log[x]^2))/(25*x - 10*x*Log[x] + x*Log[x ]^2),x]
Output:
$Aborted
Time = 10.41 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
risch | \({\mathrm e}^{-2 \left (1+x \right ) {\mathrm e}^{-\frac {{\mathrm e}^{32}}{\ln \left (x \right )-5}}}\) | \(18\) |
parallelrisch | \({\mathrm e}^{\left (-2-2 x \right ) {\mathrm e}^{-\frac {{\mathrm e}^{32}}{\ln \left (x \right )-5}}}\) | \(21\) |
Input:
int((-2*x*ln(x)^2+20*x*ln(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16)^2/(ln(x )-5))*exp((-2-2*x)*exp(-exp(16)^2/(ln(x)-5)))/(x*ln(x)^2-10*x*ln(x)+25*x), x,method=_RETURNVERBOSE)
Output:
exp(-2*(1+x)*exp(-exp(32)/(ln(x)-5)))
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (17) = 34\).
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.19 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{\left (-\frac {2 \, {\left ({\left (x + 1\right )} \log \left (x\right ) - 5 \, x - 5\right )} e^{\left (-\frac {e^{32}}{\log \left (x\right ) - 5}\right )} + e^{32}}{\log \left (x\right ) - 5} + \frac {e^{32}}{\log \left (x\right ) - 5}\right )} \] Input:
integrate((-2*x*log(x)^2+20*x*log(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16) ^2/(log(x)-5))*exp((-2-2*x)*exp(-exp(16)^2/(log(x)-5)))/(x*log(x)^2-10*x*l og(x)+25*x),x, algorithm="fricas")
Output:
e^(-(2*((x + 1)*log(x) - 5*x - 5)*e^(-e^32/(log(x) - 5)) + e^32)/(log(x) - 5) + e^32/(log(x) - 5))
Time = 1.59 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{\left (- 2 x - 2\right ) e^{- \frac {e^{32}}{\log {\left (x \right )} - 5}}} \] Input:
integrate((-2*x*ln(x)**2+20*x*ln(x)+(-2-2*x)*exp(16)**2-50*x)*exp(-exp(16) **2/(ln(x)-5))*exp((-2-2*x)*exp(-exp(16)**2/(ln(x)-5)))/(x*ln(x)**2-10*x*l n(x)+25*x),x)
Output:
exp((-2*x - 2)*exp(-exp(32)/(log(x) - 5)))
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{\left (-2 \, x e^{\left (-\frac {e^{32}}{\log \left (x\right ) - 5}\right )} - 2 \, e^{\left (-\frac {e^{32}}{\log \left (x\right ) - 5}\right )}\right )} \] Input:
integrate((-2*x*log(x)^2+20*x*log(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16) ^2/(log(x)-5))*exp((-2-2*x)*exp(-exp(16)^2/(log(x)-5)))/(x*log(x)^2-10*x*l og(x)+25*x),x, algorithm="maxima")
Output:
e^(-2*x*e^(-e^32/(log(x) - 5)) - 2*e^(-e^32/(log(x) - 5)))
\[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=\int { -\frac {2 \, {\left (x \log \left (x\right )^{2} + {\left (x + 1\right )} e^{32} - 10 \, x \log \left (x\right ) + 25 \, x\right )} e^{\left (-2 \, {\left (x + 1\right )} e^{\left (-\frac {e^{32}}{\log \left (x\right ) - 5}\right )} - \frac {e^{32}}{\log \left (x\right ) - 5}\right )}}{x \log \left (x\right )^{2} - 10 \, x \log \left (x\right ) + 25 \, x} \,d x } \] Input:
integrate((-2*x*log(x)^2+20*x*log(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16) ^2/(log(x)-5))*exp((-2-2*x)*exp(-exp(16)^2/(log(x)-5)))/(x*log(x)^2-10*x*l og(x)+25*x),x, algorithm="giac")
Output:
integrate(-2*(x*log(x)^2 + (x + 1)*e^32 - 10*x*log(x) + 25*x)*e^(-2*(x + 1 )*e^(-e^32/(log(x) - 5)) - e^32/(log(x) - 5))/(x*log(x)^2 - 10*x*log(x) + 25*x), x)
Time = 3.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx={\mathrm {e}}^{-2\,x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{32}}{\ln \left (x\right )-5}}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{32}}{\ln \left (x\right )-5}}} \] Input:
int(-(exp(-exp(-exp(32)/(log(x) - 5))*(2*x + 2))*exp(-exp(32)/(log(x) - 5) )*(50*x + 2*x*log(x)^2 - 20*x*log(x) + exp(32)*(2*x + 2)))/(25*x + x*log(x )^2 - 10*x*log(x)),x)
Output:
exp(-2*x*exp(-exp(32)/(log(x) - 5)))*exp(-2*exp(-exp(32)/(log(x) - 5)))
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=\frac {1}{e^{\frac {2 x +2}{e^{\frac {e^{32}}{\mathrm {log}\left (x \right )-5}}}}} \] Input:
int((-2*x*log(x)^2+20*x*log(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16)^2/(lo g(x)-5))*exp((-2-2*x)*exp(-exp(16)^2/(log(x)-5)))/(x*log(x)^2-10*x*log(x)+ 25*x),x)
Output:
1/e**((2*x + 2)/e**(e**32/(log(x) - 5)))