\(\int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} (230400 e^x-103680 x)}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx\) [36]

Optimal result

Integrand size = 61, antiderivative size = 23 \[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx=e^{-1+\frac {40}{-\frac {5 e^x}{18}+\frac {x^2}{16}}} \] Output:

exp(40/(1/16*x^2-5/18*exp(x))-1)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx=e^{-1-\frac {5760}{40 e^x-9 x^2}} \] Input:

Integrate[(E^((-5760 - 40*E^x + 9*x^2)/(40*E^x - 9*x^2))*(230400*E^x - 103 
680*x))/(1600*E^(2*x) - 720*E^x*x^2 + 81*x^4),x]
 

Output:

E^(-1 - 5760/(40*E^x - 9*x^2))
 

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.

\(\displaystyle \int \frac {e^{\frac {9 x^2-40 e^x-5760}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{81 x^4-720 e^x x^2+1600 e^{2 x}} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {11520 e^{\frac {9 x^2-40 e^x-5760}{40 e^x-9 x^2}} \left (20 e^x-9 x\right )}{\left (40 e^x-9 x^2\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 11520 \int \frac {e^{-\frac {-9 x^2+40 e^x+5760}{40 e^x-9 x^2}} \left (20 e^x-9 x\right )}{\left (40 e^x-9 x^2\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 11520 \int \left (\frac {9 e^{-\frac {-9 x^2+40 e^x+5760}{40 e^x-9 x^2}} (x-2) x}{2 \left (9 x^2-40 e^x\right )^2}+\frac {e^{-\frac {-9 x^2+40 e^x+5760}{40 e^x-9 x^2}}}{2 \left (40 e^x-9 x^2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 11520 \left (\frac {1}{2} \int \frac {e^{-\frac {-9 x^2+40 e^x+5760}{40 e^x-9 x^2}}}{40 e^x-9 x^2}dx-9 \int \frac {e^{-\frac {-9 x^2+40 e^x+5760}{40 e^x-9 x^2}} x}{\left (9 x^2-40 e^x\right )^2}dx+\frac {9}{2} \int \frac {e^{-\frac {-9 x^2+40 e^x+5760}{40 e^x-9 x^2}} x^2}{\left (9 x^2-40 e^x\right )^2}dx\right )\)

Input:

Int[(E^((-5760 - 40*E^x + 9*x^2)/(40*E^x - 9*x^2))*(230400*E^x - 103680*x) 
)/(1600*E^(2*x) - 720*E^x*x^2 + 81*x^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17

method	result	size
risch	\({\mathrm e}^{-\frac {-9 x^{2}+40 \,{\mathrm e}^{x}+5760}{40 \,{\mathrm e}^{x}-9 x^{2}}}\)	\(27\)
parallelrisch	\({\mathrm e}^{-\frac {-9 x^{2}+40 \,{\mathrm e}^{x}+5760}{40 \,{\mathrm e}^{x}-9 x^{2}}}\)	\(27\)
norman	\(\frac {9 x^{2} {\mathrm e}^{\frac {-40 \,{\mathrm e}^{x}+9 x^{2}-5760}{40 \,{\mathrm e}^{x}-9 x^{2}}}-40 \,{\mathrm e}^{x} {\mathrm e}^{\frac {-40 \,{\mathrm e}^{x}+9 x^{2}-5760}{40 \,{\mathrm e}^{x}-9 x^{2}}}}{9 x^{2}-40 \,{\mathrm e}^{x}}\)	\(74\)

Input:

int((230400*exp(x)-103680*x)*exp((-40*exp(x)+9*x^2-5760)/(40*exp(x)-9*x^2) 
)/(1600*exp(x)^2-720*exp(x)*x^2+81*x^4),x,method=_RETURNVERBOSE)
 

Output:

exp(-(-9*x^2+40*exp(x)+5760)/(40*exp(x)-9*x^2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx=e^{\left (-\frac {9 \, x^{2} - 40 \, e^{x} - 5760}{9 \, x^{2} - 40 \, e^{x}}\right )} \] Input:

integrate((230400*exp(x)-103680*x)*exp((-40*exp(x)+9*x^2-5760)/(40*exp(x)- 
9*x^2))/(1600*exp(x)^2-720*exp(x)*x^2+81*x^4),x, algorithm="fricas")
 

Output:

e^(-(9*x^2 - 40*e^x - 5760)/(9*x^2 - 40*e^x))
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx=e^{\frac {9 x^{2} - 40 e^{x} - 5760}{- 9 x^{2} + 40 e^{x}}} \] Input:

integrate((230400*exp(x)-103680*x)*exp((-40*exp(x)+9*x**2-5760)/(40*exp(x) 
-9*x**2))/(1600*exp(x)**2-720*exp(x)*x**2+81*x**4),x)
 

Output:

exp((9*x**2 - 40*exp(x) - 5760)/(-9*x**2 + 40*exp(x)))
 

Maxima [A] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx=e^{\left (\frac {5760}{9 \, x^{2} - 40 \, e^{x}} - 1\right )} \] Input:

integrate((230400*exp(x)-103680*x)*exp((-40*exp(x)+9*x^2-5760)/(40*exp(x)- 
9*x^2))/(1600*exp(x)^2-720*exp(x)*x^2+81*x^4),x, algorithm="maxima")
 

Output:

e^(5760/(9*x^2 - 40*e^x) - 1)
 

Giac [F]

\[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx=\int { -\frac {11520 \, {\left (9 \, x - 20 \, e^{x}\right )} e^{\left (-\frac {9 \, x^{2} - 40 \, e^{x} - 5760}{9 \, x^{2} - 40 \, e^{x}}\right )}}{81 \, x^{4} - 720 \, x^{2} e^{x} + 1600 \, e^{\left (2 \, x\right )}} \,d x } \] Input:

integrate((230400*exp(x)-103680*x)*exp((-40*exp(x)+9*x^2-5760)/(40*exp(x)- 
9*x^2))/(1600*exp(x)^2-720*exp(x)*x^2+81*x^4),x, algorithm="giac")
 

Output:

integrate(-11520*(9*x - 20*e^x)*e^(-(9*x^2 - 40*e^x - 5760)/(9*x^2 - 40*e^ 
x))/(81*x^4 - 720*x^2*e^x + 1600*e^(2*x)), x)
 

Mupad [B] (verification not implemented)

Time = 2.76 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx={\mathrm {e}}^{\frac {9\,x^2}{40\,{\mathrm {e}}^x-9\,x^2}}\,{\mathrm {e}}^{-\frac {40\,{\mathrm {e}}^x}{40\,{\mathrm {e}}^x-9\,x^2}}\,{\mathrm {e}}^{-\frac {5760}{40\,{\mathrm {e}}^x-9\,x^2}} \] Input:

int(-(exp(-(40*exp(x) - 9*x^2 + 5760)/(40*exp(x) - 9*x^2))*(103680*x - 230 
400*exp(x)))/(1600*exp(2*x) - 720*x^2*exp(x) + 81*x^4),x)
 

Output:

exp((9*x^2)/(40*exp(x) - 9*x^2))*exp(-(40*exp(x))/(40*exp(x) - 9*x^2))*exp 
(-5760/(40*exp(x) - 9*x^2))
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-5760-40 e^x+9 x^2}{40 e^x-9 x^2}} \left (230400 e^x-103680 x\right )}{1600 e^{2 x}-720 e^x x^2+81 x^4} \, dx=\frac {1}{e^{\frac {5760}{40 e^{x}-9 x^{2}}} e} \] Input:

int((230400*exp(x)-103680*x)*exp((-40*exp(x)+9*x^2-5760)/(40*exp(x)-9*x^2) 
)/(1600*exp(x)^2-720*exp(x)*x^2+81*x^4),x)
 

Output:

1/(e**(5760/(40*e**x - 9*x**2))*e)