\(\int \frac {e^{\frac {32-12 e^{\frac {1}{3} (-e^x+3 x)}-4 x-12 x^2}{4+x}} (-48-96 x-12 x^2+e^{\frac {1}{3} (-e^x+3 x)} (-36-12 x+e^x (16+4 x)))}{16+8 x+x^2} \, dx\) [4042]

Optimal result

Integrand size = 86, antiderivative size = 31 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{-4+\frac {12 \left (4-e^{-\frac {e^x}{3}+x}-x^2\right )}{4+x}} \]

[Out]

exp(12*(4-x^2-exp(-1/3*exp(x)+x))/(4+x)-4)

Rubi [F]

\[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=\int \frac {\exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right ) \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx \]

[In]

Int[(E^((32 - 12*E^((-E^x + 3*x)/3) - 4*x - 12*x^2)/(4 + x))*(-48 - 96*x - 12*x^2 + E^((-E^x + 3*x)/3)*(-36 -
12*x + E^x*(16 + 4*x))))/(16 + 8*x + x^2),x]

[Out]

-12*Defer[Int][E^((32 - 12*E^((-E^x + 3*x)/3) - 4*x - 12*x^2)/(4 + x)), x] + 144*Defer[Int][E^((32 - 12*E^((-E
^x + 3*x)/3) - 4*x - 12*x^2)/(4 + x))/(4 + x)^2, x] + 12*Defer[Int][E^(-1/3*E^x + x + (32 - 12*E^((-E^x + 3*x)
/3) - 4*x - 12*x^2)/(4 + x))/(4 + x)^2, x] - 12*Defer[Int][E^(-1/3*E^x + x + (32 - 12*E^((-E^x + 3*x)/3) - 4*x
- 12*x^2)/(4 + x))/(4 + x), x] + 4*Defer[Int][E^((-E^x + 6*x)/3 + (32 - 12*E^((-E^x + 3*x)/3) - 4*x - 12*x^2)
/(4 + x))/(4 + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right ) \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{(4+x)^2} \, dx \\ & = \int \left (-\frac {12 \exp \left (-\frac {e^x}{3}+x+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right ) (3+x)}{(4+x)^2}+\frac {4 \exp \left (\frac {1}{3} \left (-e^x+6 x\right )+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{4+x}-\frac {12 \exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right ) \left (4+8 x+x^2\right )}{(4+x)^2}\right ) \, dx \\ & = 4 \int \frac {\exp \left (\frac {1}{3} \left (-e^x+6 x\right )+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{4+x} \, dx-12 \int \frac {\exp \left (-\frac {e^x}{3}+x+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right ) (3+x)}{(4+x)^2} \, dx-12 \int \frac {\exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right ) \left (4+8 x+x^2\right )}{(4+x)^2} \, dx \\ & = 4 \int \frac {\exp \left (\frac {1}{3} \left (-e^x+6 x\right )+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{4+x} \, dx-12 \int \left (\exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )-\frac {12 \exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{(4+x)^2}\right ) \, dx-12 \int \left (-\frac {\exp \left (-\frac {e^x}{3}+x+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{(4+x)^2}+\frac {\exp \left (-\frac {e^x}{3}+x+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{4+x}\right ) \, dx \\ & = 4 \int \frac {\exp \left (\frac {1}{3} \left (-e^x+6 x\right )+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{4+x} \, dx-12 \int \exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right ) \, dx+12 \int \frac {\exp \left (-\frac {e^x}{3}+x+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{(4+x)^2} \, dx-12 \int \frac {\exp \left (-\frac {e^x}{3}+x+\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{4+x} \, dx+144 \int \frac {\exp \left (\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}\right )}{(4+x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{44-12 x-\frac {144}{4+x}-\frac {12 e^{-\frac {e^x}{3}+x}}{4+x}} \]

[In]

Integrate[(E^((32 - 12*E^((-E^x + 3*x)/3) - 4*x - 12*x^2)/(4 + x))*(-48 - 96*x - 12*x^2 + E^((-E^x + 3*x)/3)*(
-36 - 12*x + E^x*(16 + 4*x))))/(16 + 8*x + x^2),x]

[Out]

E^(44 - 12*x - 144/(4 + x) - (12*E^(-1/3*E^x + x))/(4 + x))

Maple [A] (verified)

Time = 3.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
risch	\({\mathrm e}^{-\frac {4 \left (3 x^{2}+3 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{3}+x}+x -8\right )}{4+x}}\)	\(26\)
parallelrisch	\({\mathrm e}^{-\frac {4 \left (3 x^{2}+3 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{3}+x}+x -8\right )}{4+x}}\)	\(26\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-12 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{3}+x}-12 x^{2}-4 x +32}{4+x}}+4 \,{\mathrm e}^{\frac {-12 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{3}+x}-12 x^{2}-4 x +32}{4+x}}}{4+x}\)	\(64\)

[In]

int((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x^2-96*x-48)*exp((-12*exp(-1/3*exp(x)+x)-12*x^2-4*x+32)/(
4+x))/(x^2+8*x+16),x,method=_RETURNVERBOSE)

[Out]

exp(-4*(3*x^2+3*exp(-1/3*exp(x)+x)+x-8)/(4+x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {4 \, {\left (3 \, x^{2} + x + 3 \, e^{\left (x - \frac {1}{3} \, e^{x}\right )} - 8\right )}}{x + 4}\right )} \]

[In]

integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x^2-96*x-48)*exp((-12*exp(-1/3*exp(x)+x)-12*x^2-4*x
+32)/(4+x))/(x^2+8*x+16),x, algorithm="fricas")

[Out]

e^(-4*(3*x^2 + x + 3*e^(x - 1/3*e^x) - 8)/(x + 4))

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\frac {- 12 x^{2} - 4 x - 12 e^{x - \frac {e^{x}}{3}} + 32}{x + 4}} \]

[In]

integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x**2-96*x-48)*exp((-12*exp(-1/3*exp(x)+x)-12*x**2-4
*x+32)/(4+x))/(x**2+8*x+16),x)

[Out]

exp((-12*x**2 - 4*x - 12*exp(x - exp(x)/3) + 32)/(x + 4))

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\left (-12 \, x - \frac {12 \, e^{\left (x - \frac {1}{3} \, e^{x}\right )}}{x + 4} - \frac {144}{x + 4} + 44\right )} \]

[In]

integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x^2-96*x-48)*exp((-12*exp(-1/3*exp(x)+x)-12*x^2-4*x
+32)/(4+x))/(x^2+8*x+16),x, algorithm="maxima")

[Out]

e^(-12*x - 12*e^(x - 1/3*e^x)/(x + 4) - 144/(x + 4) + 44)

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {12 \, x^{2}}{x + 4} - \frac {4 \, x}{x + 4} - \frac {12 \, e^{\left (x - \frac {1}{3} \, e^{x}\right )}}{x + 4} + \frac {32}{x + 4}\right )} \]

[In]

integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x^2-96*x-48)*exp((-12*exp(-1/3*exp(x)+x)-12*x^2-4*x
+32)/(4+x))/(x^2+8*x+16),x, algorithm="giac")

[Out]

e^(-12*x^2/(x + 4) - 4*x/(x + 4) - 12*e^(x - 1/3*e^x)/(x + 4) + 32/(x + 4))

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx={\mathrm {e}}^{-\frac {4\,x}{x+4}}\,{\mathrm {e}}^{-\frac {12\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{3}}\,{\mathrm {e}}^x}{x+4}}\,{\mathrm {e}}^{-\frac {12\,x^2}{x+4}}\,{\mathrm {e}}^{\frac {32}{x+4}} \]

[In]

int(-(exp(-(4*x + 12*exp(x - exp(x)/3) + 12*x^2 - 32)/(x + 4))*(96*x + exp(x - exp(x)/3)*(12*x - exp(x)*(4*x +
16) + 36) + 12*x^2 + 48))/(8*x + x^2 + 16),x)

[Out]

exp(-(4*x)/(x + 4))*exp(-(12*exp(-exp(x)/3)*exp(x))/(x + 4))*exp(-(12*x^2)/(x + 4))*exp(32/(x + 4))