\(\int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+(e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} (-225 x^2+x^3)) \log (-225+x))}{e^{5+\frac {2}{x}+x} (450 x^2-2 x^3)+e^{4/x} (-225 x^2+x^3)+e^{10+2 x} (-225 x^2+x^3)} \, dx\) [889]

Optimal result

Integrand size = 146, antiderivative size = 25 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=(-225+x)^{\frac {e^4}{e^{2/x}-e^{5+x}}} \] Output:

exp(ln(x-225)*exp(4)/(exp(2/x)-exp(5+x)))
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \] Input:

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]
 

Output:

(-225 + x)^(-(E^4/(-E^(2/x) + E^(5 + x))))
 

Rubi [F]

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

\(\displaystyle -225 \log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{\left (e^{2/x}-e^{x+5}\right )^2}dx+225 e^4 \log (x-225) \int \frac {(x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{e^{2/x}-e^{x+5}}dx+e^4 \int \frac {(x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{e^{2/x}-e^{x+5}}dx-450 \log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{\left (e^{2/x}-e^{x+5}\right )^2 x^2}dx+2 \log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}}}{\left (e^{2/x}-e^{x+5}\right )^2 x}dx+\log (x-225) \int \frac {e^{4+\frac {2}{x}} (x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}} x}{\left (e^{2/x}-e^{x+5}\right )^2}dx-e^4 \log (x-225) \int \frac {(x-225)^{-1-\frac {e^4}{-e^{2/x}+e^{x+5}}} x}{e^{2/x}-e^{x+5}}dx+225 \int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{\left (e^{2/x}-e^{x+5}\right )^2}dx}{x-225}dx-225 e^4 \int \frac {\int \frac {(x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{e^{2/x}-e^{x+5}}dx}{x-225}dx+450 \int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{\left (e^{2/x}-e^{x+5}\right )^2 x^2}dx}{x-225}dx-2 \int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1}}{\left (e^{2/x}-e^{x+5}\right )^2 x}dx}{x-225}dx-\int \frac {\int \frac {e^{4+\frac {2}{x}} (x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1} x}{\left (e^{2/x}-e^{x+5}\right )^2}dx}{x-225}dx+e^4 \int \frac {\int \frac {(x-225)^{\frac {e^4}{e^{2/x}-e^{x+5}}-1} x}{e^{2/x}-e^{x+5}}dx}{x-225}dx\)

Input:

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]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

\[\left (x -225\right )^{-\frac {{\mathrm e}^{4}}{{\mathrm e}^{5+x}-{\mathrm e}^{\frac {2}{x}}}}\]

Input:

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)
 

Output:

(x-225)^(-exp(4)/(exp(5+x)-exp(2/x)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x - 225\right )}^{\frac {e^{\left (x + 17\right )}}{e^{\left (2 \, x + 18\right )} - e^{\left (\frac {x^{2} + 5 \, x + 2}{x} + 8\right )}}}} \] Input:

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="fricas")
 

Output:

1/((x - 225)^(e^(x + 17)/(e^(2*x + 18) - e^((x^2 + 5*x + 2)/x + 8))))
 

Sympy [A] (verification not implemented)

Time = 5.91 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=e^{- \frac {e^{4} \log {\left (x - 225 \right )}}{- e^{\frac {2}{x}} + e^{x + 5}}} \] Input:

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)
 

Output:

exp(-exp(4)*log(x - 225)/(-exp(2/x) + exp(x + 5)))
 

Maxima [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x - 225\right )}^{\frac {e^{4}}{e^{\left (x + 5\right )} - e^{\frac {2}{x}}}}} \] Input:

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="maxima")
 

Output:

1/((x - 225)^(e^4/(e^(x + 5) - e^(2/x))))
 

Giac [F]

\[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\int { -\frac {x^{2} e^{\left (x + 9\right )} - x^{2} e^{\left (\frac {2}{x} + 4\right )} - {\left ({\left (x^{3} - 225 \, x^{2}\right )} e^{\left (x + 9\right )} + 2 \, {\left (x - 225\right )} e^{\left (\frac {2}{x} + 4\right )}\right )} \log \left (x - 225\right )}{{\left ({\left (x^{3} - 225 \, x^{2}\right )} e^{\left (2 \, x + 10\right )} - 2 \, {\left (x^{3} - 225 \, x^{2}\right )} e^{\left (x + \frac {2}{x} + 5\right )} + {\left (x^{3} - 225 \, x^{2}\right )} e^{\frac {4}{x}}\right )} {\left (x - 225\right )}^{\frac {e^{4}}{e^{\left (x + 5\right )} - e^{\frac {2}{x}}}}} \,d x } \] Input:

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="giac")
 

Output:

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)
 

Mupad [B] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x-225\right )}^{\frac {{\mathrm {e}}^4}{{\mathrm {e}}^{x+5}-{\mathrm {e}}^{2/x}}}} \] Input:

int(-(exp(-(log(x - 225)*exp(4))/(exp(x + 5) - exp(2/x)))*(log(x - 225)*(e 
xp(4)*exp(2/x)*(2*x - 450) - exp(x + 5)*exp(4)*(225*x^2 - x^3)) + x^2*exp( 
4)*exp(2/x) - x^2*exp(x + 5)*exp(4)))/(exp(2*x + 10)*(225*x^2 - x^3) + exp 
(4/x)*(225*x^2 - x^3) - exp(x + 5)*exp(2/x)*(450*x^2 - 2*x^3)),x)
 

Output:

1/(x - 225)^(exp(4)/(exp(x + 5) - exp(2/x)))
 

Reduce [B] (verification not implemented)

Time = 8.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=e^{\frac {\mathrm {log}\left (x -225\right ) e^{4}}{e^{\frac {2}{x}}-e^{x} e^{5}}} \] Input:

int((((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)
 

Output:

e**((log(x - 225)*e**4)/(e**(2/x) - e**x*e**5))