\(\int \frac {e^{\frac {2 (3 e^x-3 x)}{-32+2 e^x+30 x-8 x^2}} (48-12 x^2+e^x (-96+72 x-12 x^2))}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x (-32+30 x-8 x^2)} \, dx\) [4902]

Optimal result

Integrand size = 93, antiderivative size = 26 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {6}{2-\frac {8 (2-x)^2}{e^x-x}}} \]

[Out]

exp(3/(2-4*(2-x)*(4-2*x)/(exp(x)-x)))^2

Rubi [F]

\[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=\int \frac {\exp \left (\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}\right ) \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx \]

[In]

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]

[Out]

-1488*Defer[Int][E^((3*(E^x - x))/(-16 + E^x + 15*x - 4*x^2))/(-16 + E^x + 15*x - 4*x^2)^2, x] - 96*Defer[Int]
[E^((3*(E^x - x))/(-16 + E^x + 15*x - 4*x^2))/(-16 + E^x + 15*x - 4*x^2), x] + 2592*Defer[Int][(E^((3*(E^x - x
))/(-16 + E^x + 15*x - 4*x^2))*x)/(16 - E^x - 15*x + 4*x^2)^2, x] - 1668*Defer[Int][(E^((3*(E^x - x))/(-16 + E
^x + 15*x - 4*x^2))*x^2)/(16 - E^x - 15*x + 4*x^2)^2, x] + 468*Defer[Int][(E^((3*(E^x - x))/(-16 + E^x + 15*x
- 4*x^2))*x^3)/(16 - E^x - 15*x + 4*x^2)^2, x] - 48*Defer[Int][(E^((3*(E^x - x))/(-16 + E^x + 15*x - 4*x^2))*x
^4)/(16 - E^x - 15*x + 4*x^2)^2, x] - 72*Defer[Int][(E^((3*(E^x - x))/(-16 + E^x + 15*x - 4*x^2))*x)/(16 - E^x
- 15*x + 4*x^2), x] + 12*Defer[Int][(E^((3*(E^x - x))/(-16 + E^x + 15*x - 4*x^2))*x^2)/(16 - E^x - 15*x + 4*x
^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {12 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (2-x) \left (2+e^x (-4+x)+x\right )}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx \\ & = 12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (2-x) \left (2+e^x (-4+x)+x\right )}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx \\ & = 12 \int \left (-\frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (-2+x)^2 \left (31-23 x+4 x^2\right )}{\left (16-e^x-15 x+4 x^2\right )^2}+\frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} \left (8-6 x+x^2\right )}{16-e^x-15 x+4 x^2}\right ) \, dx \\ & = -\left (12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (-2+x)^2 \left (31-23 x+4 x^2\right )}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx\right )+12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} \left (8-6 x+x^2\right )}{16-e^x-15 x+4 x^2} \, dx \\ & = -\left (12 \int \left (\frac {124 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{\left (-16+e^x+15 x-4 x^2\right )^2}-\frac {216 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{\left (16-e^x-15 x+4 x^2\right )^2}+\frac {139 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{\left (16-e^x-15 x+4 x^2\right )^2}-\frac {39 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^3}{\left (16-e^x-15 x+4 x^2\right )^2}+\frac {4 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^4}{\left (16-e^x-15 x+4 x^2\right )^2}\right ) \, dx\right )+12 \int \left (-\frac {8 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{-16+e^x+15 x-4 x^2}-\frac {6 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{16-e^x-15 x+4 x^2}+\frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{16-e^x-15 x+4 x^2}\right ) \, dx \\ & = 12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{16-e^x-15 x+4 x^2} \, dx-48 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^4}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx-72 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{16-e^x-15 x+4 x^2} \, dx-96 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{-16+e^x+15 x-4 x^2} \, dx+468 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^3}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx-1488 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{\left (-16+e^x+15 x-4 x^2\right )^2} \, dx-1668 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx+2592 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} \]

[In]

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]

[Out]

E^((3*(E^x - x))/(-16 + E^x + 15*x - 4*x^2))

Maple [A] (verified)

Time = 0.54 (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	\({\mathrm e}^{\frac {3 \,{\mathrm e}^{x}-3 x}{-4 x^{2}+{\mathrm e}^{x}+15 x -16}}\)	\(26\)

[In]

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)

[Out]

exp(3*(exp(x)-x)/(-4*x^2+exp(x)+15*x-16))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, {\left (x - e^{x}\right )}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \]

[In]

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")

[Out]

e^(3*(x - e^x)/(4*x^2 - 15*x - e^x + 16))

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {2 \left (- 3 x + 3 e^{x}\right )}{- 8 x^{2} + 30 x + 2 e^{x} - 32}} \]

[In]

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)

[Out]

exp(2*(-3*x + 3*exp(x))/(-8*x**2 + 30*x + 2*exp(x) - 32))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, x}{4 \, x^{2} - 15 \, x - e^{x} + 16} - \frac {3 \, e^{x}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \]

[In]

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")

[Out]

e^(3*x/(4*x^2 - 15*x - e^x + 16) - 3*e^x/(4*x^2 - 15*x - e^x + 16))

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, x}{4 \, x^{2} - 15 \, x - e^{x} + 16} - \frac {3 \, e^{x}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \]

[In]

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")

[Out]

e^(3*x/(4*x^2 - 15*x - e^x + 16) - 3*e^x/(4*x^2 - 15*x - e^x + 16))

Mupad [B] (verification not implemented)

Time = 10.95 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx={\mathrm {e}}^{\frac {3\,{\mathrm {e}}^x}{15\,x+{\mathrm {e}}^x-4\,x^2-16}}\,{\mathrm {e}}^{-\frac {3\,x}{15\,x+{\mathrm {e}}^x-4\,x^2-16}} \]

[In]

int(-(exp(-(2*(3*x - 3*exp(x)))/(30*x + 2*exp(x) - 8*x^2 - 32))*(exp(x)*(12*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)

[Out]

exp((3*exp(x))/(15*x + exp(x) - 4*x^2 - 16))*exp(-(3*x)/(15*x + exp(x) - 4*x^2 - 16))