Integrand size = 86, antiderivative size = 31 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{-4+\frac {12 \left (4-e^{-\frac {e^x}{3}+x}-x^2\right )}{4+x}} \] Output:
exp(12*(4-x^2-exp(-1/3*exp(x)+x))/(4+x)-4)
Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{44-12 x-\frac {144}{4+x}-\frac {12 e^{-\frac {e^x}{3}+x}}{4+x}} \] Input:
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]
Output:
E^(44 - 12*x - 144/(4 + x) - (12*E^(-1/3*E^x + x))/(4 + x))
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.
\(\displaystyle \int \frac {\left (-12 x^2-96 x+e^{\frac {1}{3} \left (3 x-e^x\right )} \left (-12 x+e^x (4 x+16)-36\right )-48\right ) \exp \left (\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )}{x^2+8 x+16} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (-12 x^2-96 x+e^{\frac {1}{3} \left (3 x-e^x\right )} \left (-12 x+e^x (4 x+16)-36\right )-48\right ) \exp \left (\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )}{(x+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {12 (x+3) \exp \left (\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}+x-\frac {e^x}{3}\right )}{(x+4)^2}-\frac {12 \left (x^2+8 x+4\right ) \exp \left (\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )}{(x+4)^2}+\frac {4 \exp \left (\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}+\frac {1}{3} \left (6 x-e^x\right )\right )}{x+4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -12 \int \exp \left (\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )dx+144 \int \frac {\exp \left (\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )}{(x+4)^2}dx+12 \int \frac {\exp \left (x-\frac {e^x}{3}+\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )}{(x+4)^2}dx-12 \int \frac {\exp \left (x-\frac {e^x}{3}+\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )}{x+4}dx+4 \int \frac {\exp \left (\frac {1}{3} \left (6 x-e^x\right )+\frac {-12 x^2-4 x-12 e^{\frac {1}{3} \left (3 x-e^x\right )}+32}{x+4}\right )}{x+4}dx\) |
Input:
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]
Output:
$Aborted
Time = 0.75 (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 | \(\frac {x \,{\mathrm e}^{\frac {-12 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{3}+x}-12 x^{2}-4 x +32}{4+x}}+4 \,{\mathrm e}^{\frac {-12 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{3}+x}-12 x^{2}-4 x +32}{4+x}}}{4+x}\) | \(64\) |
Input:
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=_RETURNVER BOSE)
Output:
exp(-4*(3*x^2+3*exp(-1/3*exp(x)+x)+x-8)/(4+x))
Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {4 \, {\left (3 \, x^{2} + x + 3 \, e^{\left (x - \frac {1}{3} \, e^{x}\right )} - 8\right )}}{x + 4}\right )} \] Input:
integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x^2-96*x-48)*ex p((-12*exp(-1/3*exp(x)+x)-12*x^2-4*x+32)/(4+x))/(x^2+8*x+16),x, algorithm= "fricas")
Output:
e^(-4*(3*x^2 + x + 3*e^(x - 1/3*e^x) - 8)/(x + 4))
Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\frac {- 12 x^{2} - 4 x - 12 e^{x - \frac {e^{x}}{3}} + 32}{x + 4}} \] Input:
integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x**2-96*x-48)*e xp((-12*exp(-1/3*exp(x)+x)-12*x**2-4*x+32)/(4+x))/(x**2+8*x+16),x)
Output:
exp((-12*x**2 - 4*x - 12*exp(x - exp(x)/3) + 32)/(x + 4))
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\left (-12 \, x - \frac {12 \, e^{\left (x - \frac {1}{3} \, e^{x}\right )}}{x + 4} - \frac {144}{x + 4} + 44\right )} \] Input:
integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x^2-96*x-48)*ex p((-12*exp(-1/3*exp(x)+x)-12*x^2-4*x+32)/(4+x))/(x^2+8*x+16),x, algorithm= "maxima")
Output:
e^(-12*x - 12*e^(x - 1/3*e^x)/(x + 4) - 144/(x + 4) + 44)
Time = 0.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {12 \, x^{2}}{x + 4} - \frac {4 \, x}{x + 4} - \frac {12 \, e^{\left (x - \frac {1}{3} \, e^{x}\right )}}{x + 4} + \frac {32}{x + 4}\right )} \] Input:
integrate((((4*x+16)*exp(x)-12*x-36)*exp(-1/3*exp(x)+x)-12*x^2-96*x-48)*ex p((-12*exp(-1/3*exp(x)+x)-12*x^2-4*x+32)/(4+x))/(x^2+8*x+16),x, algorithm= "giac")
Output:
e^(-12*x^2/(x + 4) - 4*x/(x + 4) - 12*e^(x - 1/3*e^x)/(x + 4) + 32/(x + 4) )
Time = 3.43 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx={\mathrm {e}}^{-\frac {4\,x}{x+4}}\,{\mathrm {e}}^{-\frac {12\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{3}}\,{\mathrm {e}}^x}{x+4}}\,{\mathrm {e}}^{-\frac {12\,x^2}{x+4}}\,{\mathrm {e}}^{\frac {32}{x+4}} \] Input:
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)
Output:
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))
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.16 \[ \int \frac {e^{\frac {32-12 e^{\frac {1}{3} \left (-e^x+3 x\right )}-4 x-12 x^2}{4+x}} \left (-48-96 x-12 x^2+e^{\frac {1}{3} \left (-e^x+3 x\right )} \left (-36-12 x+e^x (16+4 x)\right )\right )}{16+8 x+x^2} \, dx=\frac {e^{44}}{e^{\frac {12 e^{\frac {e^{x}}{3}} x^{2}+48 e^{\frac {e^{x}}{3}} x +144 e^{\frac {e^{x}}{3}}+12 e^{x}}{e^{\frac {e^{x}}{3}} x +4 e^{\frac {e^{x}}{3}}}}} \] Input:
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)
Output:
e**44/e**((12*e**(e**x/3)*x**2 + 48*e**(e**x/3)*x + 144*e**(e**x/3) + 12*e **x)/(e**(e**x/3)*x + 4*e**(e**x/3)))