Integrand size = 89, antiderivative size = 33 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=1+\frac {4 e^{-2 x-\frac {x}{4 \left (-2+\frac {x^2}{3}\right )}}}{-2+2 x} \] Output:
exp(2*ln(2)-2*x-1/4*x/(1/3*x^2-2))/(-2+2*x)+1
Time = 1.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=\frac {2 e^{-x \left (2+\frac {3}{4 \left (-6+x^2\right )}\right )}}{-1+x} \] Input:
Integrate[(E^((45*x - 8*x^3 + (-24 + 4*x^2)*Log[4])/(-24 + 4*x^2))*(126 - 270*x - 51*x^2 + 99*x^3 + 4*x^4 - 8*x^5))/(288 - 576*x + 192*x^2 + 192*x^3 - 88*x^4 - 16*x^5 + 8*x^6),x]
Output:
2/(E^(x*(2 + 3/(4*(-6 + x^2))))*(-1 + 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.
\(\displaystyle \int \frac {\left (-8 x^5+4 x^4+99 x^3-51 x^2-270 x+126\right ) \exp \left (\frac {-8 x^3+\left (4 x^2-24\right ) \log (4)+45 x}{4 x^2-24}\right )}{8 x^6-16 x^5-88 x^4+192 x^3+192 x^2-576 x+288} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-8 x^5+4 x^4+99 x^3-51 x^2-270 x+126\right ) \exp \left (\frac {-8 x^3+\left (4 x^2-24\right ) \log (4)+45 x}{4 x^2-24}\right )}{250 (x-1)}+\frac {(-4 x-9) \left (-8 x^5+4 x^4+99 x^3-51 x^2-270 x+126\right ) \exp \left (\frac {-8 x^3+\left (4 x^2-24\right ) \log (4)+45 x}{4 x^2-24}\right )}{1000 \left (x^2-6\right )}+\frac {\left (-8 x^5+4 x^4+99 x^3-51 x^2-270 x+126\right ) \exp \left (\frac {-8 x^3+\left (4 x^2-24\right ) \log (4)+45 x}{4 x^2-24}\right )}{200 (x-1)^2}+\frac {(2 x+7) \left (-8 x^5+4 x^4+99 x^3-51 x^2-270 x+126\right ) \exp \left (\frac {-8 x^3+\left (4 x^2-24\right ) \log (4)+45 x}{4 x^2-24}\right )}{200 \left (x^2-6\right )^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {x \left (45-8 x^2\right )}{4 \left (x^2-6\right )}} \left (-8 x^5+4 x^4+99 x^3-51 x^2-270 x+126\right )}{2 (1-x)^2 \left (6-x^2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}} \left (-8 x^5+4 x^4+99 x^3-51 x^2-270 x+126\right )}{(1-x)^2 \left (6-x^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {21 e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}} (x+1)}{25 \left (x^2-6\right )}+\frac {36 e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}} (x+1)}{5 \left (x^2-6\right )^2}-\frac {179 e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{25 (x-1)}-\frac {4 e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{(x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{10} \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{\left (\sqrt {6}-x\right )^2}dx+\frac {7}{100} \left (6+\sqrt {6}\right ) \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{\sqrt {6}-x}dx+\frac {1}{10} \sqrt {\frac {3}{2}} \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{\sqrt {6}-x}dx-4 \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{(x-1)^2}dx-\frac {179}{25} \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{x-1}dx+\frac {3}{10} \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{\left (x+\sqrt {6}\right )^2}dx-\frac {7}{100} \left (6-\sqrt {6}\right ) \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{x+\sqrt {6}}dx+\frac {1}{10} \sqrt {\frac {3}{2}} \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}}}{x+\sqrt {6}}dx+\frac {36}{5} \int \frac {e^{-\frac {x \left (45-8 x^2\right )}{4 \left (6-x^2\right )}} x}{\left (x^2-6\right )^2}dx\right )\) |
Input:
Int[(E^((45*x - 8*x^3 + (-24 + 4*x^2)*Log[4])/(-24 + 4*x^2))*(126 - 270*x - 51*x^2 + 99*x^3 + 4*x^4 - 8*x^5))/(288 - 576*x + 192*x^2 + 192*x^3 - 88* x^4 - 16*x^5 + 8*x^6),x]
Output:
$Aborted
Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15
method | result | size |
gosper | \(\frac {{\mathrm e}^{\frac {8 x^{2} \ln \left (2\right )-8 x^{3}-48 \ln \left (2\right )+45 x}{4 x^{2}-24}}}{-2+2 x}\) | \(38\) |
risch | \(\frac {{\mathrm e}^{\frac {8 x^{2} \ln \left (2\right )-8 x^{3}-48 \ln \left (2\right )+45 x}{4 x^{2}-24}}}{-2+2 x}\) | \(38\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {8 x^{2} \ln \left (2\right )-8 x^{3}-48 \ln \left (2\right )+45 x}{4 x^{2}-24}}}{-2+2 x}\) | \(38\) |
norman | \(\frac {\frac {x^{2} {\mathrm e}^{\frac {2 \left (4 x^{2}-24\right ) \ln \left (2\right )-8 x^{3}+45 x}{4 x^{2}-24}}}{2}-3 \,{\mathrm e}^{\frac {2 \left (4 x^{2}-24\right ) \ln \left (2\right )-8 x^{3}+45 x}{4 x^{2}-24}}}{x^{3}-x^{2}-6 x +6}\) | \(87\) |
orering | \(-\frac {4 \left (x^{2}-6\right )^{2} \left (-1+x \right ) \left (-8 x^{5}+4 x^{4}+99 x^{3}-51 x^{2}-270 x +126\right ) {\mathrm e}^{\frac {2 \left (4 x^{2}-24\right ) \ln \left (2\right )-8 x^{3}+45 x}{4 x^{2}-24}}}{\left (8 x^{5}-4 x^{4}-99 x^{3}+51 x^{2}+270 x -126\right ) \left (8 x^{6}-16 x^{5}-88 x^{4}+192 x^{3}+192 x^{2}-576 x +288\right )}\) | \(128\) |
Input:
int((-8*x^5+4*x^4+99*x^3-51*x^2-270*x+126)*exp((2*(4*x^2-24)*ln(2)-8*x^3+4 5*x)/(4*x^2-24))/(8*x^6-16*x^5-88*x^4+192*x^3+192*x^2-576*x+288),x,method= _RETURNVERBOSE)
Output:
1/2*exp(1/4*(8*x^2*ln(2)-8*x^3-48*ln(2)+45*x)/(x^2-6))/(-1+x)
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=\frac {e^{\left (-\frac {8 \, x^{3} - 8 \, {\left (x^{2} - 6\right )} \log \left (2\right ) - 45 \, x}{4 \, {\left (x^{2} - 6\right )}}\right )}}{2 \, {\left (x - 1\right )}} \] Input:
integrate((-8*x^5+4*x^4+99*x^3-51*x^2-270*x+126)*exp((2*(4*x^2-24)*log(2)- 8*x^3+45*x)/(4*x^2-24))/(8*x^6-16*x^5-88*x^4+192*x^3+192*x^2-576*x+288),x, algorithm="fricas")
Output:
1/2*e^(-1/4*(8*x^3 - 8*(x^2 - 6)*log(2) - 45*x)/(x^2 - 6))/(x - 1)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=\frac {e^{\frac {- 8 x^{3} + 45 x + \left (8 x^{2} - 48\right ) \log {\left (2 \right )}}{4 x^{2} - 24}}}{2 x - 2} \] Input:
integrate((-8*x**5+4*x**4+99*x**3-51*x**2-270*x+126)*exp((2*(4*x**2-24)*ln (2)-8*x**3+45*x)/(4*x**2-24))/(8*x**6-16*x**5-88*x**4+192*x**3+192*x**2-57 6*x+288),x)
Output:
exp((-8*x**3 + 45*x + (8*x**2 - 48)*log(2))/(4*x**2 - 24))/(2*x - 2)
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=\frac {2 \, e^{\left (-2 \, x - \frac {3 \, x}{4 \, {\left (x^{2} - 6\right )}}\right )}}{x - 1} \] Input:
integrate((-8*x^5+4*x^4+99*x^3-51*x^2-270*x+126)*exp((2*(4*x^2-24)*log(2)- 8*x^3+45*x)/(4*x^2-24))/(8*x^6-16*x^5-88*x^4+192*x^3+192*x^2-576*x+288),x, algorithm="maxima")
Output:
2*e^(-2*x - 3/4*x/(x^2 - 6))/(x - 1)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=\frac {2 \, e^{\left (-\frac {8 \, x^{3} - 45 \, x}{4 \, {\left (x^{2} - 6\right )}}\right )}}{x - 1} \] Input:
integrate((-8*x^5+4*x^4+99*x^3-51*x^2-270*x+126)*exp((2*(4*x^2-24)*log(2)- 8*x^3+45*x)/(4*x^2-24))/(8*x^6-16*x^5-88*x^4+192*x^3+192*x^2-576*x+288),x, algorithm="giac")
Output:
2*e^(-1/4*(8*x^3 - 45*x)/(x^2 - 6))/(x - 1)
Time = 3.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=\frac {2\,{\mathrm {e}}^{-\frac {2\,x^3}{x^2-6}}\,{\mathrm {e}}^{\frac {45\,x}{4\,\left (x^2-6\right )}}}{x-1} \] Input:
int(-(exp((45*x + 2*log(2)*(4*x^2 - 24) - 8*x^3)/(4*x^2 - 24))*(270*x + 51 *x^2 - 99*x^3 - 4*x^4 + 8*x^5 - 126))/(192*x^2 - 576*x + 192*x^3 - 88*x^4 - 16*x^5 + 8*x^6 + 288),x)
Output:
(2*exp(-(2*x^3)/(x^2 - 6))*exp((45*x)/(4*(x^2 - 6))))/(x - 1)
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {45 x-8 x^3+\left (-24+4 x^2\right ) \log (4)}{-24+4 x^2}} \left (126-270 x-51 x^2+99 x^3+4 x^4-8 x^5\right )}{288-576 x+192 x^2+192 x^3-88 x^4-16 x^5+8 x^6} \, dx=\frac {2}{e^{\frac {8 x^{3}-45 x}{4 x^{2}-24}} \left (x -1\right )} \] Input:
int((-8*x^5+4*x^4+99*x^3-51*x^2-270*x+126)*exp((2*(4*x^2-24)*log(2)-8*x^3+ 45*x)/(4*x^2-24))/(8*x^6-16*x^5-88*x^4+192*x^3+192*x^2-576*x+288),x)
Output:
2/(e**((8*x**3 - 45*x)/(4*x**2 - 24))*(x - 1))