Integrand size = 221, antiderivative size = 32 \[ \int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx=\frac {x}{-e^{\frac {2 x}{4+x-x^2}}+\log \left (\frac {10}{3-x}\right )} \]
Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx=\frac {e^{\frac {2 x}{-4-x+x^2}} x}{-1+e^{\frac {2 x}{-4-x+x^2}} \log \left (-\frac {10}{-3+x}\right )} \]
Integrate[(16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + (48 - 16*x - 21*x^2 - 5*x^ 3 + 7*x^4 - x^5)/E^((2*x)/(-4 - x + x^2)) + (-48 - 8*x + 29*x^2 - x^3 - 5* x^4 + x^5)*Log[-10/(-3 + x)])/((-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)/E^ ((4*x)/(-4 - x + x^2)) + ((96 + 16*x - 58*x^2 + 2*x^3 + 10*x^4 - 2*x^5)*Lo g[-10/(-3 + x)])/E^((2*x)/(-4 - x + x^2)) + (-48 - 8*x + 29*x^2 - x^3 - 5* x^4 + x^5)*Log[-10/(-3 + x)]^2),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.
| \(\displaystyle \int \frac {x^5-2 x^4-7 x^3+8 x^2+e^{-\frac {2 x}{x^2-x-4}} \left (-x^5+7 x^4-5 x^3-21 x^2-16 x+48\right )+\left (x^5-5 x^4-x^3+29 x^2-8 x-48\right ) \log \left (-\frac {10}{x-3}\right )+16 x}{e^{-\frac {4 x}{x^2-x-4}} \left (x^5-5 x^4-x^3+29 x^2-8 x-48\right )+\left (x^5-5 x^4-x^3+29 x^2-8 x-48\right ) \log ^2\left (-\frac {10}{x-3}\right )+e^{-\frac {2 x}{x^2-x-4}} \left (-2 x^5+10 x^4+2 x^3-58 x^2+16 x+96\right ) \log \left (-\frac {10}{x-3}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-x^5+2 x^4+7 x^3-8 x^2-e^{-\frac {2 x}{x^2-x-4}} \left (-x^5+7 x^4-5 x^3-21 x^2-16 x+48\right )-\left (x^5-5 x^4-x^3+29 x^2-8 x-48\right ) \log \left (-\frac {10}{x-3}\right )-16 x}{(3-x) \left (-x^2+x+4\right )^2 \left (e^{\frac {2 x}{-x^2+x+4}}-\log \left (-\frac {10}{x-3}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x \left (x^4-2 x^3+2 x^3 \log \left (-\frac {10}{x-3}\right )-7 x^2-6 x^2 \log \left (-\frac {10}{x-3}\right )+8 x+8 x \log \left (-\frac {10}{x-3}\right )-24 \log \left (-\frac {10}{x-3}\right )+16\right )}{(x-3) \left (x^2-x-4\right )^2 \left (e^{\frac {2 x}{-x^2+x+4}}-\log \left (-\frac {10}{x-3}\right )\right )^2}-\frac {x^4-4 x^3-7 x^2+16}{\left (x^2-x-4\right )^2 \left (e^{\frac {2 x}{-x^2+x+4}}-\log \left (-\frac {10}{x-3}\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {x \left (x^4-2 x^3+2 x^3 \log \left (-\frac {10}{x-3}\right )-7 x^2-6 x^2 \log \left (-\frac {10}{x-3}\right )+8 x+8 x \log \left (-\frac {10}{x-3}\right )-24 \log \left (-\frac {10}{x-3}\right )+16\right )}{(x-3) \left (x^2-x-4\right )^2 \left (e^{\frac {2 x}{-x^2+x+4}}-\log \left (-\frac {10}{x-3}\right )\right )^2}-\frac {x^4-4 x^3-7 x^2+16}{\left (x^2-x-4\right )^2 \left (e^{\frac {2 x}{-x^2+x+4}}-\log \left (-\frac {10}{x-3}\right )\right )}\right )dx\) |
Int[(16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + (48 - 16*x - 21*x^2 - 5*x^3 + 7* x^4 - x^5)/E^((2*x)/(-4 - x + x^2)) + (-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)*Log[-10/(-3 + x)])/((-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)/E^((4*x) /(-4 - x + x^2)) + ((96 + 16*x - 58*x^2 + 2*x^3 + 10*x^4 - 2*x^5)*Log[-10/ (-3 + x)])/E^((2*x)/(-4 - x + x^2)) + (-48 - 8*x + 29*x^2 - x^3 - 5*x^4 + x^5)*Log[-10/(-3 + x)]^2),x]
3.20.36.3.1 Defintions of rubi rules used
Time = 3.60 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94
| method | result | size |
| parallelrisch | \(\frac {x}{\ln \left (-\frac {10}{-3+x}\right )-{\mathrm e}^{-\frac {2 x}{x^{2}-x -4}}}\) | \(30\) |
| risch | \(\frac {2 x}{-2 i \pi \operatorname {csgn}\left (\frac {i}{-3+x}\right )^{2}+2 i \pi \operatorname {csgn}\left (\frac {i}{-3+x}\right )^{3}+2 i \pi +2 \ln \left (2\right )+2 \ln \left (5\right )-2 \,{\mathrm e}^{-\frac {2 x}{x^{2}-x -4}}-2 \ln \left (-3+x \right )}\) | \(71\) |
int(((x^5-5*x^4-x^3+29*x^2-8*x-48)*ln(-10/(-3+x))+(-x^5+7*x^4-5*x^3-21*x^2 -16*x+48)*exp(-2*x/(x^2-x-4))+x^5-2*x^4-7*x^3+8*x^2+16*x)/((x^5-5*x^4-x^3+ 29*x^2-8*x-48)*ln(-10/(-3+x))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+96)*exp(- 2*x/(x^2-x-4))*ln(-10/(-3+x))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(-2*x/(x^2- x-4))^2),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx=-\frac {x}{e^{\left (-\frac {2 \, x}{x^{2} - x - 4}\right )} - \log \left (-\frac {10}{x - 3}\right )} \]
integrate(((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(-3+x))+(-x^5+7*x^4-5*x^3 -21*x^2-16*x+48)*exp(-2*x/(x^2-x-4))+x^5-2*x^4-7*x^3+8*x^2+16*x)/((x^5-5*x ^4-x^3+29*x^2-8*x-48)*log(-10/(-3+x))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+9 6)*exp(-2*x/(x^2-x-4))*log(-10/(-3+x))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(- 2*x/(x^2-x-4))^2),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx=- \frac {x}{- \log {\left (- \frac {10}{x - 3} \right )} + e^{- \frac {2 x}{x^{2} - x - 4}}} \]
integrate(((x**5-5*x**4-x**3+29*x**2-8*x-48)*ln(-10/(-3+x))+(-x**5+7*x**4- 5*x**3-21*x**2-16*x+48)*exp(-2*x/(x**2-x-4))+x**5-2*x**4-7*x**3+8*x**2+16* x)/((x**5-5*x**4-x**3+29*x**2-8*x-48)*ln(-10/(-3+x))**2+(-2*x**5+10*x**4+2 *x**3-58*x**2+16*x+96)*exp(-2*x/(x**2-x-4))*ln(-10/(-3+x))+(x**5-5*x**4-x* *3+29*x**2-8*x-48)*exp(-2*x/(x**2-x-4))**2),x)
Time = 0.63 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx=\frac {x e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )}}{{\left (\log \left (5\right ) + \log \left (2\right ) - \log \left (-x + 3\right )\right )} e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )} - 1} \]
integrate(((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(-3+x))+(-x^5+7*x^4-5*x^3 -21*x^2-16*x+48)*exp(-2*x/(x^2-x-4))+x^5-2*x^4-7*x^3+8*x^2+16*x)/((x^5-5*x ^4-x^3+29*x^2-8*x-48)*log(-10/(-3+x))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+9 6)*exp(-2*x/(x^2-x-4))*log(-10/(-3+x))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(- 2*x/(x^2-x-4))^2),x, algorithm=\
Time = 0.39 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx=\frac {x e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )}}{e^{\left (\frac {2 \, x}{x^{2} - x - 4}\right )} \log \left (-\frac {10}{x - 3}\right ) - 1} \]
integrate(((x^5-5*x^4-x^3+29*x^2-8*x-48)*log(-10/(-3+x))+(-x^5+7*x^4-5*x^3 -21*x^2-16*x+48)*exp(-2*x/(x^2-x-4))+x^5-2*x^4-7*x^3+8*x^2+16*x)/((x^5-5*x ^4-x^3+29*x^2-8*x-48)*log(-10/(-3+x))^2+(-2*x^5+10*x^4+2*x^3-58*x^2+16*x+9 6)*exp(-2*x/(x^2-x-4))*log(-10/(-3+x))+(x^5-5*x^4-x^3+29*x^2-8*x-48)*exp(- 2*x/(x^2-x-4))^2),x, algorithm=\
Time = 1.00 (sec) , antiderivative size = 293, normalized size of antiderivative = 9.16 \[ \int \frac {16 x+8 x^2-7 x^3-2 x^4+x^5+e^{-\frac {2 x}{-4-x+x^2}} \left (48-16 x-21 x^2-5 x^3+7 x^4-x^5\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log \left (-\frac {10}{-3+x}\right )}{e^{-\frac {4 x}{-4-x+x^2}} \left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right )+e^{-\frac {2 x}{-4-x+x^2}} \left (96+16 x-58 x^2+2 x^3+10 x^4-2 x^5\right ) \log \left (-\frac {10}{-3+x}\right )+\left (-48-8 x+29 x^2-x^3-5 x^4+x^5\right ) \log ^2\left (-\frac {10}{-3+x}\right )} \, dx=-\frac {\left (x^5-2\,x^4-7\,x^3+8\,x^2+16\,x\right )\,{\left (-x^5+5\,x^4+x^3-29\,x^2+8\,x+48\right )}^2-\ln \left (-\frac {10}{x-3}\right )\,\left (-2\,x^4+6\,x^3-8\,x^2+24\,x\right )\,{\left (-x^5+5\,x^4+x^3-29\,x^2+8\,x+48\right )}^2}{\left (\ln \left (-\frac {10}{x-3}\right )-{\mathrm {e}}^{\frac {2\,x}{-x^2+x+4}}\right )\,\left (x-3\right )\,{\left (-x^2+x+4\right )}^2\,\left (512\,x-1152\,\ln \left (-\frac {10}{x-3}\right )+472\,x^2\,\ln \left (-\frac {10}{x-3}\right )-208\,x^3\,\ln \left (-\frac {10}{x-3}\right )+78\,x^4\,\ln \left (-\frac {10}{x-3}\right )-36\,x^6\,\ln \left (-\frac {10}{x-3}\right )+16\,x^7\,\ln \left (-\frac {10}{x-3}\right )-2\,x^8\,\ln \left (-\frac {10}{x-3}\right )-736\,x^2-368\,x^3+323\,x^4+83\,x^5-74\,x^6-2\,x^7+7\,x^8-x^9+192\,x\,\ln \left (-\frac {10}{x-3}\right )+768\right )} \]
int((log(-10/(x - 3))*(8*x - 29*x^2 + x^3 + 5*x^4 - x^5 + 48) - 16*x + exp ((2*x)/(x - x^2 + 4))*(16*x + 21*x^2 + 5*x^3 - 7*x^4 + x^5 - 48) - 8*x^2 + 7*x^3 + 2*x^4 - x^5)/(log(-10/(x - 3))^2*(8*x - 29*x^2 + x^3 + 5*x^4 - x^ 5 + 48) + exp((4*x)/(x - x^2 + 4))*(8*x - 29*x^2 + x^3 + 5*x^4 - x^5 + 48) - log(-10/(x - 3))*exp((2*x)/(x - x^2 + 4))*(16*x - 58*x^2 + 2*x^3 + 10*x ^4 - 2*x^5 + 96)),x)
-((16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5)*(8*x - 29*x^2 + x^3 + 5*x^4 - x^5 + 48)^2 - log(-10/(x - 3))*(24*x - 8*x^2 + 6*x^3 - 2*x^4)*(8*x - 29*x^2 + x ^3 + 5*x^4 - x^5 + 48)^2)/((log(-10/(x - 3)) - exp((2*x)/(x - x^2 + 4)))*( x - 3)*(x - x^2 + 4)^2*(512*x - 1152*log(-10/(x - 3)) + 472*x^2*log(-10/(x - 3)) - 208*x^3*log(-10/(x - 3)) + 78*x^4*log(-10/(x - 3)) - 36*x^6*log(- 10/(x - 3)) + 16*x^7*log(-10/(x - 3)) - 2*x^8*log(-10/(x - 3)) - 736*x^2 - 368*x^3 + 323*x^4 + 83*x^5 - 74*x^6 - 2*x^7 + 7*x^8 - x^9 + 192*x*log(-10 /(x - 3)) + 768))