Integrand size = 98, antiderivative size = 23 \[ \int \frac {e^{\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}} \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(x)} \, dx=e^{\frac {e^{-x} x}{-1+\frac {x}{5-9 \log (x)}}} \]
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}} \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(x)} \, dx=e^{\frac {5 e^{-x} x}{-5+x+9 \log (x)}} x^{-\frac {9 e^{-x} x}{-5+x+9 \log (x)}} \]
Integrate[(E^((5*x - 9*x*Log[x])/(E^x*(-5 + x) + 9*E^x*Log[x]))*(-25 + 16* x - 5*x^2 + (90 - 90*x + 9*x^2)*Log[x] + (-81 + 81*x)*Log[x]^2))/(E^x*(25 - 10*x + x^2) + E^x*(-90 + 18*x)*Log[x] + 81*E^x*Log[x]^2),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.
\(\displaystyle \int \frac {\left (-5 x^2+\left (9 x^2-90 x+90\right ) \log (x)+16 x+(81 x-81) \log ^2(x)-25\right ) \exp \left (\frac {5 x-9 x \log (x)}{e^x (x-5)+9 e^x \log (x)}\right )}{e^x \left (x^2-10 x+25\right )+81 e^x \log ^2(x)+e^x (18 x-90) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-5 x^2+\left (9 x^2-90 x+90\right ) \log (x)+16 x+(81 x-81) \log ^2(x)-25\right ) \exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right )}{(-x-9 \log (x)+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (x \exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right )-\frac {(x-2) x \exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right )}{x+9 \log (x)-5}-\frac {(x+9) x \exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right )}{(x+9 \log (x)-5)^2}-\exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right ) x^2}{(x+9 \log (x)-5)^2}dx-\int \frac {\exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right ) x^2}{x+9 \log (x)-5}dx-\int \exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right )dx+\int \exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right ) xdx-9 \int \frac {\exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right ) x}{(x+9 \log (x)-5)^2}dx+2 \int \frac {\exp \left (\frac {e^{-x} (5 x-9 x \log (x))}{x+9 \log (x)-5}-x\right ) x}{x+9 \log (x)-5}dx\) |
Int[(E^((5*x - 9*x*Log[x])/(E^x*(-5 + x) + 9*E^x*Log[x]))*(-25 + 16*x - 5* x^2 + (90 - 90*x + 9*x^2)*Log[x] + (-81 + 81*x)*Log[x]^2))/(E^x*(25 - 10*x + x^2) + E^x*(-90 + 18*x)*Log[x] + 81*E^x*Log[x]^2),x]
3.24.91.3.1 Defintions of rubi rules used
Time = 4.40 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \({\mathrm e}^{-\frac {x \left (9 \ln \left (x \right )-5\right ) {\mathrm e}^{-x}}{9 \ln \left (x \right )-5+x}}\) | \(24\) |
parallelrisch | \({\mathrm e}^{\frac {\left (-9 x \ln \left (x \right )+5 x \right ) {\mathrm e}^{-x}}{9 \ln \left (x \right )-5+x}}\) | \(25\) |
int(((81*x-81)*ln(x)^2+(9*x^2-90*x+90)*ln(x)-5*x^2+16*x-25)*exp((-9*x*ln(x )+5*x)/(9*exp(x)*ln(x)+(-5+x)*exp(x)))/(81*exp(x)*ln(x)^2+(18*x-90)*exp(x) *ln(x)+(x^2-10*x+25)*exp(x)),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}} \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(x)} \, dx=e^{\left (-\frac {9 \, x \log \left (x\right ) - 5 \, x}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \left (x\right )}\right )} \]
integrate(((81*x-81)*log(x)^2+(9*x^2-90*x+90)*log(x)-5*x^2+16*x-25)*exp((- 9*x*log(x)+5*x)/(9*exp(x)*log(x)+(-5+x)*exp(x)))/(81*exp(x)*log(x)^2+(18*x -90)*exp(x)*log(x)+(x^2-10*x+25)*exp(x)),x, algorithm=\
Time = 0.61 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}} \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(x)} \, dx=e^{\frac {- 9 x \log {\left (x \right )} + 5 x}{\left (x - 5\right ) e^{x} + 9 e^{x} \log {\left (x \right )}}} \]
integrate(((81*x-81)*ln(x)**2+(9*x**2-90*x+90)*ln(x)-5*x**2+16*x-25)*exp(( -9*x*ln(x)+5*x)/(9*exp(x)*ln(x)+(-5+x)*exp(x)))/(81*exp(x)*ln(x)**2+(18*x- 90)*exp(x)*ln(x)+(x**2-10*x+25)*exp(x)),x)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (22) = 44\).
Time = 0.41 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.17 \[ \int \frac {e^{\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}} \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(x)} \, dx=e^{\left (-9 \, e^{\left (-x\right )} \log \left (x\right ) + \frac {81 \, \log \left (x\right )^{2}}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \left (x\right )} - \frac {90 \, \log \left (x\right )}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \left (x\right )} + \frac {25}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \left (x\right )} + 5 \, e^{\left (-x\right )}\right )} \]
integrate(((81*x-81)*log(x)^2+(9*x^2-90*x+90)*log(x)-5*x^2+16*x-25)*exp((- 9*x*log(x)+5*x)/(9*exp(x)*log(x)+(-5+x)*exp(x)))/(81*exp(x)*log(x)^2+(18*x -90)*exp(x)*log(x)+(x^2-10*x+25)*exp(x)),x, algorithm=\
e^(-9*e^(-x)*log(x) + 81*log(x)^2/((x - 5)*e^x + 9*e^x*log(x)) - 90*log(x) /((x - 5)*e^x + 9*e^x*log(x)) + 25/((x - 5)*e^x + 9*e^x*log(x)) + 5*e^(-x) )
Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {e^{\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}} \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(x)} \, dx=e^{\left (-\frac {9 \, x \log \left (x\right )}{x e^{x} + 9 \, e^{x} \log \left (x\right ) - 5 \, e^{x}} + \frac {5 \, x}{x e^{x} + 9 \, e^{x} \log \left (x\right ) - 5 \, e^{x}}\right )} \]
integrate(((81*x-81)*log(x)^2+(9*x^2-90*x+90)*log(x)-5*x^2+16*x-25)*exp((- 9*x*log(x)+5*x)/(9*exp(x)*log(x)+(-5+x)*exp(x)))/(81*exp(x)*log(x)^2+(18*x -90)*exp(x)*log(x)+(x^2-10*x+25)*exp(x)),x, algorithm=\
Time = 14.34 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {e^{\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}} \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(x)} \, dx={\mathrm {e}}^{\frac {5\,x}{9\,{\mathrm {e}}^x\,\ln \left (x\right )-5\,{\mathrm {e}}^x+x\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {9\,x\,\ln \left (x\right )}{9\,{\mathrm {e}}^x\,\ln \left (x\right )-5\,{\mathrm {e}}^x+x\,{\mathrm {e}}^x}} \]
int((exp((5*x - 9*x*log(x))/(exp(x)*(x - 5) + 9*exp(x)*log(x)))*(16*x + lo g(x)*(9*x^2 - 90*x + 90) - 5*x^2 + log(x)^2*(81*x - 81) - 25))/(exp(x)*(x^ 2 - 10*x + 25) + 81*exp(x)*log(x)^2 + exp(x)*log(x)*(18*x - 90)),x)