Integrand size = 146, antiderivative size = 25 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=(-225+x)^{\frac {e^4}{e^{2/x}-e^{5+x}}} \]
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \]
Integrate[(E^(4 + 2/x)*x^2 - E^(9 + x)*x^2 + (E^(4 + 2/x)*(-450 + 2*x) + E ^(9 + x)*(-225*x^2 + x^3))*Log[-225 + x])/((-225 + x)^(E^4/(-E^(2/x) + E^( 5 + x)))*(E^(5 + 2/x + x)*(450*x^2 - 2*x^3) + E^(4/x)*(-225*x^2 + x^3) + E ^(10 + 2*x)*(-225*x^2 + x^3))),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.
\(\displaystyle \int \frac {(x-225)^{-\frac {e^4}{e^{x+5}-e^{2/x}}} \left (e^{\frac {2}{x}+4} x^2-e^{x+9} x^2+\left (e^{x+9} \left (x^3-225 x^2\right )+e^{\frac {2}{x}+4} (2 x-450)\right ) \log (x-225)\right )}{e^{x+\frac {2}{x}+5} \left (450 x^2-2 x^3\right )+e^{4/x} \left (x^3-225 x^2\right )+e^{2 x+10} \left (x^3-225 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(x-225)^{-\frac {e^4}{e^{x+5}-e^{2/x}}-1} \left (e^{\frac {2}{x}+4} x^2-e^{x+9} x^2+\left (e^{x+9} \left (x^3-225 x^2\right )+e^{\frac {2}{x}+4} (2 x-450)\right ) \log (x-225)\right )}{\left (e^{2/x}-e^{x+5}\right )^2 x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {2}{x}+4} \left (x^3-225 x^2+2 x-450\right ) (x-225)^{-\frac {e^4}{e^{x+5}-e^{2/x}}-1} \log (x-225)}{\left (e^{2/x}-e^{x+5}\right )^2 x^2}+\frac {e^4 (x-225)^{-\frac {e^4}{e^{x+5}-e^{2/x}}-1} (x \log (x-225)-225 \log (x-225)-1)}{e^{x+5}-e^{2/x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -225 \log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{\left (e^{2/x}-e^{x+5}\right )^2}dx+225 e^4 \log (x-225) \int \frac {(x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{e^{2/x}-e^{x+5}}dx+e^4 \int \frac {(x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{e^{2/x}-e^{x+5}}dx-450 \log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{\left (e^{2/x}-e^{x+5}\right )^2 x^2}dx+2 \log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{\left (e^{2/x}-e^{x+5}\right )^2 x}dx+\log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}} x}{\left (e^{2/x}-e^{x+5}\right )^2}dx-e^4 \log (x-225) \int \frac {(x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}} x}{e^{2/x}-e^{x+5}}dx+225 \int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{\left (e^{2/x}-e^{x+5}\right )^2}dx}{x-225}dx-225 e^4 \int \frac {\int \frac {(x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{e^{2/x}-e^{x+5}}dx}{x-225}dx+450 \int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{\left (e^{2/x}-e^{x+5}\right )^2 x^2}dx}{x-225}dx-2 \int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{\left (e^{2/x}-e^{x+5}\right )^2 x}dx}{x-225}dx-\int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1} x}{\left (e^{2/x}-e^{x+5}\right )^2}dx}{x-225}dx+e^4 \int \frac {\int \frac {(x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1} x}{e^{2/x}-e^{x+5}}dx}{x-225}dx\) |
Int[(E^(4 + 2/x)*x^2 - E^(9 + x)*x^2 + (E^(4 + 2/x)*(-450 + 2*x) + E^(9 + x)*(-225*x^2 + x^3))*Log[-225 + x])/((-225 + x)^(E^4/(-E^(2/x) + E^(5 + x) ))*(E^(5 + 2/x + x)*(450*x^2 - 2*x^3) + E^(4/x)*(-225*x^2 + x^3) + E^(10 + 2*x)*(-225*x^2 + x^3))),x]
3.9.89.3.1 Defintions of rubi rules used
Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
\[\left (x -225\right )^{-\frac {{\mathrm e}^{4}}{{\mathrm e}^{5+x}-{\mathrm e}^{\frac {2}{x}}}}\]
int((((x^3-225*x^2)*exp(4)*exp(5+x)+(2*x-450)*exp(4)*exp(2/x))*ln(x-225)-x ^2*exp(4)*exp(5+x)+x^2*exp(4)*exp(2/x))*exp(-exp(4)*ln(x-225)/(exp(5+x)-ex p(2/x)))/((x^3-225*x^2)*exp(5+x)^2+(-2*x^3+450*x^2)*exp(2/x)*exp(5+x)+(x^3 -225*x^2)*exp(2/x)^2),x)
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x - 225\right )}^{\frac {e^{\left (x + 17\right )}}{e^{\left (2 \, x + 18\right )} - e^{\left (\frac {x^{2} + 5 \, x + 2}{x} + 8\right )}}}} \]
integrate((((x^3-225*x^2)*exp(4)*exp(5+x)+(2*x-450)*exp(4)*exp(2/x))*log(x -225)-x^2*exp(4)*exp(5+x)+x^2*exp(4)*exp(2/x))*exp(-exp(4)*log(x-225)/(exp (5+x)-exp(2/x)))/((x^3-225*x^2)*exp(5+x)^2+(-2*x^3+450*x^2)*exp(2/x)*exp(5 +x)+(x^3-225*x^2)*exp(2/x)^2),x, algorithm=\
Time = 5.60 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=e^{- \frac {e^{4} \log {\left (x - 225 \right )}}{- e^{\frac {2}{x}} + e^{x + 5}}} \]
integrate((((x**3-225*x**2)*exp(4)*exp(5+x)+(2*x-450)*exp(4)*exp(2/x))*ln( x-225)-x**2*exp(4)*exp(5+x)+x**2*exp(4)*exp(2/x))*exp(-exp(4)*ln(x-225)/(e xp(5+x)-exp(2/x)))/((x**3-225*x**2)*exp(5+x)**2+(-2*x**3+450*x**2)*exp(2/x )*exp(5+x)+(x**3-225*x**2)*exp(2/x)**2),x)
Time = 0.65 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x - 225\right )}^{\frac {e^{4}}{e^{\left (x + 5\right )} - e^{\frac {2}{x}}}}} \]
integrate((((x^3-225*x^2)*exp(4)*exp(5+x)+(2*x-450)*exp(4)*exp(2/x))*log(x -225)-x^2*exp(4)*exp(5+x)+x^2*exp(4)*exp(2/x))*exp(-exp(4)*log(x-225)/(exp (5+x)-exp(2/x)))/((x^3-225*x^2)*exp(5+x)^2+(-2*x^3+450*x^2)*exp(2/x)*exp(5 +x)+(x^3-225*x^2)*exp(2/x)^2),x, algorithm=\
\[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\int { -\frac {x^{2} e^{\left (x + 9\right )} - x^{2} e^{\left (\frac {2}{x} + 4\right )} - {\left ({\left (x^{3} - 225 \, x^{2}\right )} e^{\left (x + 9\right )} + 2 \, {\left (x - 225\right )} e^{\left (\frac {2}{x} + 4\right )}\right )} \log \left (x - 225\right )}{{\left ({\left (x^{3} - 225 \, x^{2}\right )} e^{\left (2 \, x + 10\right )} - 2 \, {\left (x^{3} - 225 \, x^{2}\right )} e^{\left (x + \frac {2}{x} + 5\right )} + {\left (x^{3} - 225 \, x^{2}\right )} e^{\frac {4}{x}}\right )} {\left (x - 225\right )}^{\frac {e^{4}}{e^{\left (x + 5\right )} - e^{\frac {2}{x}}}}} \,d x } \]
integrate((((x^3-225*x^2)*exp(4)*exp(5+x)+(2*x-450)*exp(4)*exp(2/x))*log(x -225)-x^2*exp(4)*exp(5+x)+x^2*exp(4)*exp(2/x))*exp(-exp(4)*log(x-225)/(exp (5+x)-exp(2/x)))/((x^3-225*x^2)*exp(5+x)^2+(-2*x^3+450*x^2)*exp(2/x)*exp(5 +x)+(x^3-225*x^2)*exp(2/x)^2),x, algorithm=\
integrate(-(x^2*e^(x + 9) - x^2*e^(2/x + 4) - ((x^3 - 225*x^2)*e^(x + 9) + 2*(x - 225)*e^(2/x + 4))*log(x - 225))/(((x^3 - 225*x^2)*e^(2*x + 10) - 2 *(x^3 - 225*x^2)*e^(x + 2/x + 5) + (x^3 - 225*x^2)*e^(4/x))*(x - 225)^(e^4 /(e^(x + 5) - e^(2/x)))), x)
Time = 10.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x-225\right )}^{\frac {{\mathrm {e}}^4}{{\mathrm {e}}^{x+5}-{\mathrm {e}}^{2/x}}}} \]