Integrand size = 93, antiderivative size = 26 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {6}{2-\frac {8 (2-x)^2}{e^x-x}}} \] Output:
exp(3/(2-4*(2-x)*(4-2*x)/(exp(x)-x)))^2
Time = 1.59 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} \] Input:
Integrate[(E^((2*(3*E^x - 3*x))/(-32 + 2*E^x + 30*x - 8*x^2))*(48 - 12*x^2 + E^x*(-96 + 72*x - 12*x^2)))/(256 + E^(2*x) - 480*x + 353*x^2 - 120*x^3 + 16*x^4 + E^x*(-32 + 30*x - 8*x^2)),x]
Output:
E^((3*(E^x - x))/(-16 + E^x + 15*x - 4*x^2))
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.
\(\displaystyle \int \frac {\left (-12 x^2+e^x \left (-12 x^2+72 x-96\right )+48\right ) \exp \left (\frac {2 \left (3 e^x-3 x\right )}{-8 x^2+30 x+2 e^x-32}\right )}{16 x^4-120 x^3+353 x^2+e^x \left (-8 x^2+30 x-32\right )-480 x+e^{2 x}+256} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {12 e^{\frac {3 \left (e^x-x\right )}{-4 x^2+15 x+e^x-16}} (2-x) \left (e^x x+x-4 e^x+2\right )}{\left (4 x^2-15 x-e^x+16\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 12 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} (2-x) \left (e^x x+x-4 e^x+2\right )}{\left (4 x^2-15 x-e^x+16\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 12 \int \left (\frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} \left (x^2-6 x+8\right )}{4 x^2-15 x-e^x+16}-\frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} (x-2)^2 \left (4 x^2-23 x+31\right )}{\left (4 x^2-15 x-e^x+16\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 12 \left (-124 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}}}{\left (-4 x^2+15 x+e^x-16\right )^2}dx-8 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}}}{-4 x^2+15 x+e^x-16}dx+216 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} x}{\left (4 x^2-15 x-e^x+16\right )^2}dx-139 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} x^2}{\left (4 x^2-15 x-e^x+16\right )^2}dx-6 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} x}{4 x^2-15 x-e^x+16}dx+\int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} x^2}{4 x^2-15 x-e^x+16}dx-4 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} x^4}{\left (4 x^2-15 x-e^x+16\right )^2}dx+39 \int \frac {e^{-\frac {3 \left (e^x-x\right )}{4 x^2-15 x-e^x+16}} x^3}{\left (4 x^2-15 x-e^x+16\right )^2}dx\right )\) |
Input:
Int[(E^((2*(3*E^x - 3*x))/(-32 + 2*E^x + 30*x - 8*x^2))*(48 - 12*x^2 + E^x *(-96 + 72*x - 12*x^2)))/(256 + E^(2*x) - 480*x + 353*x^2 - 120*x^3 + 16*x ^4 + E^x*(-32 + 30*x - 8*x^2)),x]
Output:
$Aborted
Time = 0.64 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \({\mathrm e}^{\frac {3 \,{\mathrm e}^{x}-3 x}{-4 x^{2}+{\mathrm e}^{x}+15 x -16}}\) | \(24\) |
parallelrisch | \({\mathrm e}^{\frac {3 \,{\mathrm e}^{x}-3 x}{-4 x^{2}+{\mathrm e}^{x}+15 x -16}}\) | \(26\) |
Input:
int(((-12*x^2+72*x-96)*exp(x)-12*x^2+48)*exp((3*exp(x)-3*x)/(2*exp(x)-8*x^ 2+30*x-32))^2/(exp(x)^2+(-8*x^2+30*x-32)*exp(x)+16*x^4-120*x^3+353*x^2-480 *x+256),x,method=_RETURNVERBOSE)
Output:
exp(3*(exp(x)-x)/(-4*x^2+exp(x)+15*x-16))
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, {\left (x - e^{x}\right )}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \] Input:
integrate(((-12*x^2+72*x-96)*exp(x)-12*x^2+48)*exp((3*exp(x)-3*x)/(2*exp(x )-8*x^2+30*x-32))^2/(exp(x)^2+(-8*x^2+30*x-32)*exp(x)+16*x^4-120*x^3+353*x ^2-480*x+256),x, algorithm="fricas")
Output:
e^(3*(x - e^x)/(4*x^2 - 15*x - e^x + 16))
Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {2 \left (- 3 x + 3 e^{x}\right )}{- 8 x^{2} + 30 x + 2 e^{x} - 32}} \] Input:
integrate(((-12*x**2+72*x-96)*exp(x)-12*x**2+48)*exp((3*exp(x)-3*x)/(2*exp (x)-8*x**2+30*x-32))**2/(exp(x)**2+(-8*x**2+30*x-32)*exp(x)+16*x**4-120*x* *3+353*x**2-480*x+256),x)
Output:
exp(2*(-3*x + 3*exp(x))/(-8*x**2 + 30*x + 2*exp(x) - 32))
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, x}{4 \, x^{2} - 15 \, x - e^{x} + 16} - \frac {3 \, e^{x}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \] Input:
integrate(((-12*x^2+72*x-96)*exp(x)-12*x^2+48)*exp((3*exp(x)-3*x)/(2*exp(x )-8*x^2+30*x-32))^2/(exp(x)^2+(-8*x^2+30*x-32)*exp(x)+16*x^4-120*x^3+353*x ^2-480*x+256),x, algorithm="maxima")
Output:
e^(3*x/(4*x^2 - 15*x - e^x + 16) - 3*e^x/(4*x^2 - 15*x - e^x + 16))
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, x}{4 \, x^{2} - 15 \, x - e^{x} + 16} - \frac {3 \, e^{x}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \] Input:
integrate(((-12*x^2+72*x-96)*exp(x)-12*x^2+48)*exp((3*exp(x)-3*x)/(2*exp(x )-8*x^2+30*x-32))^2/(exp(x)^2+(-8*x^2+30*x-32)*exp(x)+16*x^4-120*x^3+353*x ^2-480*x+256),x, algorithm="giac")
Output:
e^(3*x/(4*x^2 - 15*x - e^x + 16) - 3*e^x/(4*x^2 - 15*x - e^x + 16))
Time = 4.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx={\mathrm {e}}^{\frac {3\,{\mathrm {e}}^x}{15\,x+{\mathrm {e}}^x-4\,x^2-16}}\,{\mathrm {e}}^{-\frac {3\,x}{15\,x+{\mathrm {e}}^x-4\,x^2-16}} \] Input:
int(-(exp(-(2*(3*x - 3*exp(x)))/(30*x + 2*exp(x) - 8*x^2 - 32))*(exp(x)*(1 2*x^2 - 72*x + 96) + 12*x^2 - 48))/(exp(2*x) - 480*x - exp(x)*(8*x^2 - 30* x + 32) + 353*x^2 - 120*x^3 + 16*x^4 + 256),x)
Output:
exp((3*exp(x))/(15*x + exp(x) - 4*x^2 - 16))*exp(-(3*x)/(15*x + exp(x) - 4 *x^2 - 16))
Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=\frac {e^{\frac {12 x^{2}+48}{e^{x}-4 x^{2}+15 x -16}} e^{3}}{e^{\frac {48 x}{e^{x}-4 x^{2}+15 x -16}}} \] Input:
int(((-12*x^2+72*x-96)*exp(x)-12*x^2+48)*exp((3*exp(x)-3*x)/(2*exp(x)-8*x^ 2+30*x-32))^2/(exp(x)^2+(-8*x^2+30*x-32)*exp(x)+16*x^4-120*x^3+353*x^2-480 *x+256),x)
Output:
(e**((12*x**2 + 48)/(e**x - 4*x**2 + 15*x - 16))*e**3)/e**((48*x)/(e**x - 4*x**2 + 15*x - 16))