Integrand size = 135, antiderivative size = 27 \[ \int \frac {-8-33 x+17 x^2+e^x \left (-15+5 x+5 x^2\right )+\left (28+14 x-14 x^2+e^x \left (10+5 x-5 x^2\right )\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )}{-10-5 x+5 x^2+\left (20+10 x-10 x^2\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )} \, dx=x-\frac {-2+e^x+\frac {4 x}{5}}{-1+\log \left (-2-x+x^2\right )} \]
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {-8-33 x+17 x^2+e^x \left (-15+5 x+5 x^2\right )+\left (28+14 x-14 x^2+e^x \left (10+5 x-5 x^2\right )\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )}{-10-5 x+5 x^2+\left (20+10 x-10 x^2\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )} \, dx=\frac {1}{5} \left (5 x-\frac {-10+5 e^x+4 x}{-1+\log \left (-2-x+x^2\right )}\right ) \]
Integrate[(-8 - 33*x + 17*x^2 + E^x*(-15 + 5*x + 5*x^2) + (28 + 14*x - 14* x^2 + E^x*(10 + 5*x - 5*x^2))*Log[-2 - x + x^2] + (-10 - 5*x + 5*x^2)*Log[ -2 - x + x^2]^2)/(-10 - 5*x + 5*x^2 + (20 + 10*x - 10*x^2)*Log[-2 - x + x^ 2] + (-10 - 5*x + 5*x^2)*Log[-2 - x + x^2]^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.
| \(\displaystyle \int \frac {17 x^2+e^x \left (5 x^2+5 x-15\right )+\left (5 x^2-5 x-10\right ) \log ^2\left (x^2-x-2\right )+\left (-14 x^2+e^x \left (-5 x^2+5 x+10\right )+14 x+28\right ) \log \left (x^2-x-2\right )-33 x-8}{5 x^2+\left (5 x^2-5 x-10\right ) \log ^2\left (x^2-x-2\right )+\left (-10 x^2+10 x+20\right ) \log \left (x^2-x-2\right )-5 x-10} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-17 x^2-e^x \left (5 x^2+5 x-15\right )-\left (5 x^2-5 x-10\right ) \log ^2\left (x^2-x-2\right )-\left (-14 x^2+e^x \left (-5 x^2+5 x+10\right )+14 x+28\right ) \log \left (x^2-x-2\right )+33 x+8}{5 \left (-x^2+x+2\right ) \left (1-\log \left (x^2-x-2\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {-17 x^2+33 x+5 \left (-x^2+x+2\right ) \log ^2\left (x^2-x-2\right )+5 e^x \left (-x^2-x+3\right )-\left (-14 x^2+14 x+5 e^x \left (-x^2+x+2\right )+28\right ) \log \left (x^2-x-2\right )+8}{\left (-x^2+x+2\right ) \left (1-\log \left (x^2-x-2\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {1}{5} \int \left (-\frac {14 \log \left (x^2-x-2\right ) x^2}{(x-2) (x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}+\frac {17 x^2}{(x-2) (x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}+\frac {14 \log \left (x^2-x-2\right ) x}{(x-2) (x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}-\frac {33 x}{(x-2) (x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}+\frac {5 \log ^2\left (x^2-x-2\right )}{\left (\log \left (x^2-x-2\right )-1\right )^2}+\frac {28 \log \left (x^2-x-2\right )}{(x-2) (x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}-\frac {5 e^x \left (\log \left (x^2-x-2\right ) x^2-x^2-\log \left (x^2-x-2\right ) x-x-2 \log \left (x^2-x-2\right )+3\right )}{(x-2) (x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}-\frac {8}{(x-2) (x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (8 \int \frac {1}{\left (\log \left (x^2-x-2\right )-1\right )^2}dx-2 \int \frac {1}{(x-2) \left (\log \left (x^2-x-2\right )-1\right )^2}dx+5 \int \frac {e^x}{(x-2) \left (\log \left (x^2-x-2\right )-1\right )^2}dx-14 \int \frac {1}{(x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}dx+5 \int \frac {e^x}{(x+1) \left (\log \left (x^2-x-2\right )-1\right )^2}dx-4 \int \frac {1}{\log \left (x^2-x-2\right )-1}dx-5 \int \frac {e^x}{\log \left (x^2-x-2\right )-1}dx+5 x\right )\) |
Int[(-8 - 33*x + 17*x^2 + E^x*(-15 + 5*x + 5*x^2) + (28 + 14*x - 14*x^2 + E^x*(10 + 5*x - 5*x^2))*Log[-2 - x + x^2] + (-10 - 5*x + 5*x^2)*Log[-2 - x + x^2]^2)/(-10 - 5*x + 5*x^2 + (20 + 10*x - 10*x^2)*Log[-2 - x + x^2] + ( -10 - 5*x + 5*x^2)*Log[-2 - x + x^2]^2),x]
3.8.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.44 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(x -\frac {4 x +5 \,{\mathrm e}^{x}-10}{5 \left (\ln \left (x^{2}-x -2\right )-1\right )}\) | \(27\) |
| norman | \(\frac {\ln \left (x^{2}-x -2\right ) x -\frac {9 x}{5}+2-{\mathrm e}^{x}}{\ln \left (x^{2}-x -2\right )-1}\) | \(35\) |
| parallelrisch | \(\frac {5 \ln \left (x^{2}-x -2\right ) x -9 x -5 \,{\mathrm e}^{x}+10 \ln \left (x^{2}-x -2\right )}{5 \ln \left (x^{2}-x -2\right )-5}\) | \(47\) |
| default | \(\frac {\ln \left (x^{2}-x -2\right ) x -\frac {9 x}{5}+2}{\ln \left (x^{2}-x -2\right )-1}-\frac {{\mathrm e}^{x}}{\ln \left (x^{2}-x -2\right )-1}\) | \(49\) |
| parts | \(\frac {\ln \left (x^{2}-x -2\right ) x -\frac {9 x}{5}+2}{\ln \left (x^{2}-x -2\right )-1}-\frac {{\mathrm e}^{x}}{\ln \left (x^{2}-x -2\right )-1}\) | \(49\) |
int(((5*x^2-5*x-10)*ln(x^2-x-2)^2+((-5*x^2+5*x+10)*exp(x)-14*x^2+14*x+28)* ln(x^2-x-2)+(5*x^2+5*x-15)*exp(x)+17*x^2-33*x-8)/((5*x^2-5*x-10)*ln(x^2-x- 2)^2+(-10*x^2+10*x+20)*ln(x^2-x-2)+5*x^2-5*x-10),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-8-33 x+17 x^2+e^x \left (-15+5 x+5 x^2\right )+\left (28+14 x-14 x^2+e^x \left (10+5 x-5 x^2\right )\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )}{-10-5 x+5 x^2+\left (20+10 x-10 x^2\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )} \, dx=\frac {5 \, x \log \left (x^{2} - x - 2\right ) - 9 \, x - 5 \, e^{x} + 10}{5 \, {\left (\log \left (x^{2} - x - 2\right ) - 1\right )}} \]
integrate(((5*x^2-5*x-10)*log(x^2-x-2)^2+((-5*x^2+5*x+10)*exp(x)-14*x^2+14 *x+28)*log(x^2-x-2)+(5*x^2+5*x-15)*exp(x)+17*x^2-33*x-8)/((5*x^2-5*x-10)*l og(x^2-x-2)^2+(-10*x^2+10*x+20)*log(x^2-x-2)+5*x^2-5*x-10),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-8-33 x+17 x^2+e^x \left (-15+5 x+5 x^2\right )+\left (28+14 x-14 x^2+e^x \left (10+5 x-5 x^2\right )\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )}{-10-5 x+5 x^2+\left (20+10 x-10 x^2\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )} \, dx=x + \frac {10 - 4 x}{5 \log {\left (x^{2} - x - 2 \right )} - 5} - \frac {e^{x}}{\log {\left (x^{2} - x - 2 \right )} - 1} \]
integrate(((5*x**2-5*x-10)*ln(x**2-x-2)**2+((-5*x**2+5*x+10)*exp(x)-14*x** 2+14*x+28)*ln(x**2-x-2)+(5*x**2+5*x-15)*exp(x)+17*x**2-33*x-8)/((5*x**2-5* x-10)*ln(x**2-x-2)**2+(-10*x**2+10*x+20)*ln(x**2-x-2)+5*x**2-5*x-10),x)
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-8-33 x+17 x^2+e^x \left (-15+5 x+5 x^2\right )+\left (28+14 x-14 x^2+e^x \left (10+5 x-5 x^2\right )\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )}{-10-5 x+5 x^2+\left (20+10 x-10 x^2\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )} \, dx=\frac {5 \, x \log \left (x + 1\right ) + 5 \, x \log \left (x - 2\right ) - 9 \, x - 5 \, e^{x} + 10}{5 \, {\left (\log \left (x + 1\right ) + \log \left (x - 2\right ) - 1\right )}} \]
integrate(((5*x^2-5*x-10)*log(x^2-x-2)^2+((-5*x^2+5*x+10)*exp(x)-14*x^2+14 *x+28)*log(x^2-x-2)+(5*x^2+5*x-15)*exp(x)+17*x^2-33*x-8)/((5*x^2-5*x-10)*l og(x^2-x-2)^2+(-10*x^2+10*x+20)*log(x^2-x-2)+5*x^2-5*x-10),x, algorithm=\
Time = 0.36 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-8-33 x+17 x^2+e^x \left (-15+5 x+5 x^2\right )+\left (28+14 x-14 x^2+e^x \left (10+5 x-5 x^2\right )\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )}{-10-5 x+5 x^2+\left (20+10 x-10 x^2\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )} \, dx=\frac {5 \, x \log \left (x^{2} - x - 2\right ) - 9 \, x - 5 \, e^{x} + 10}{5 \, {\left (\log \left (x^{2} - x - 2\right ) - 1\right )}} \]
integrate(((5*x^2-5*x-10)*log(x^2-x-2)^2+((-5*x^2+5*x+10)*exp(x)-14*x^2+14 *x+28)*log(x^2-x-2)+(5*x^2+5*x-15)*exp(x)+17*x^2-33*x-8)/((5*x^2-5*x-10)*l og(x^2-x-2)^2+(-10*x^2+10*x+20)*log(x^2-x-2)+5*x^2-5*x-10),x, algorithm=\
Time = 8.38 (sec) , antiderivative size = 112, normalized size of antiderivative = 4.15 \[ \int \frac {-8-33 x+17 x^2+e^x \left (-15+5 x+5 x^2\right )+\left (28+14 x-14 x^2+e^x \left (10+5 x-5 x^2\right )\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )}{-10-5 x+5 x^2+\left (20+10 x-10 x^2\right ) \log \left (-2-x+x^2\right )+\left (-10-5 x+5 x^2\right ) \log ^2\left (-2-x+x^2\right )} \, dx=\frac {3\,x}{5}+\frac {9}{10\,\left (x-\frac {1}{2}\right )}-\frac {\frac {5\,x^2\,{\mathrm {e}}^x-15\,{\mathrm {e}}^x-28\,x+5\,x\,{\mathrm {e}}^x+12\,x^2+2}{5\,\left (2\,x-1\right )}+\frac {\ln \left (x^2-x-2\right )\,\left (5\,{\mathrm {e}}^x+4\right )\,\left (-x^2+x+2\right )}{5\,\left (2\,x-1\right )}}{\ln \left (x^2-x-2\right )-1}+\frac {{\mathrm {e}}^x\,\left (-\frac {x^2}{2}+\frac {x}{2}+1\right )}{x-\frac {1}{2}} \]
int((33*x + log(x^2 - x - 2)^2*(5*x - 5*x^2 + 10) - exp(x)*(5*x + 5*x^2 - 15) - 17*x^2 - log(x^2 - x - 2)*(14*x + exp(x)*(5*x - 5*x^2 + 10) - 14*x^2 + 28) + 8)/(5*x - log(x^2 - x - 2)*(10*x - 10*x^2 + 20) + log(x^2 - x - 2 )^2*(5*x - 5*x^2 + 10) - 5*x^2 + 10),x)