Integrand size = 154, antiderivative size = 27 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=1+\frac {e^x}{\log \left (2 (4-x) x \left (-\frac {1}{2}+25 e^x+x\right )\right )} \]
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^x}{\log \left (-\left ((-4+x) x \left (-1+50 e^x+2 x\right )\right )\right )} \]
Integrate[(E^(2*x)*(200 + 100*x - 50*x^2) + E^x*(-4 + 18*x - 6*x^2) + (E^( 2*x)*(-200*x + 50*x^2) + E^x*(4*x - 9*x^2 + 2*x^3))*Log[-4*x + 9*x^2 - 2*x ^3 + E^x*(200*x - 50*x^2)])/((4*x - 9*x^2 + 2*x^3 + E^x*(-200*x + 50*x^2)) *Log[-4*x + 9*x^2 - 2*x^3 + E^x*(200*x - 50*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.
| \(\displaystyle \int \frac {e^{2 x} \left (-50 x^2+100 x+200\right )+e^x \left (-6 x^2+18 x-4\right )+\left (e^{2 x} \left (50 x^2-200 x\right )+e^x \left (2 x^3-9 x^2+4 x\right )\right ) \log \left (-2 x^3+9 x^2+e^x \left (200 x-50 x^2\right )-4 x\right )}{\left (2 x^3-9 x^2+e^x \left (50 x^2-200 x\right )+4 x\right ) \log ^2\left (-2 x^3+9 x^2+e^x \left (200 x-50 x^2\right )-4 x\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{2 x} \left (-50 x^2+100 x+200\right )+e^x \left (-6 x^2+18 x-4\right )+\left (e^{2 x} \left (50 x^2-200 x\right )+e^x \left (2 x^3-9 x^2+4 x\right )\right ) \log \left (-2 x^3+9 x^2+e^x \left (200 x-50 x^2\right )-4 x\right )}{\left (-2 x-50 e^x+1\right ) (4-x) x \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^x \left (-x^2+x^2 \log \left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )+2 x-4 x \log \left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )+4\right )}{(x-4) x \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}-\frac {4 x^2-8 x+3}{50 \left (2 x+50 e^x-1\right ) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}+\frac {2 x-3}{50 \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2}{25} \int \frac {x^2}{\left (2 x+50 e^x-1\right ) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\frac {3}{50} \int \frac {1}{\log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\int \frac {e^x}{\log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\int \frac {e^x}{(x-4) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\int \frac {e^x}{x \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx+\frac {1}{25} \int \frac {x}{\log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\frac {3}{50} \int \frac {1}{\left (2 x+50 e^x-1\right ) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx+\frac {4}{25} \int \frac {x}{\left (2 x+50 e^x-1\right ) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx+\int \frac {e^x}{\log \left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx\) |
Int[(E^(2*x)*(200 + 100*x - 50*x^2) + E^x*(-4 + 18*x - 6*x^2) + (E^(2*x)*( -200*x + 50*x^2) + E^x*(4*x - 9*x^2 + 2*x^3))*Log[-4*x + 9*x^2 - 2*x^3 + E ^x*(200*x - 50*x^2)])/((4*x - 9*x^2 + 2*x^3 + E^x*(-200*x + 50*x^2))*Log[- 4*x + 9*x^2 - 2*x^3 + E^x*(200*x - 50*x^2)]^2),x]
3.8.55.3.1 Defintions of rubi rules used
Time = 2.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22
| method | result | size |
| parallelrisch | \(\frac {{\mathrm e}^{x}}{\ln \left (\left (-50 x^{2}+200 x \right ) {\mathrm e}^{x}-2 x^{3}+9 x^{2}-4 x \right )}\) | \(33\) |
| risch | \(\frac {2 i {\mathrm e}^{x}}{\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right ) \operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )-\pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right ) {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{2}-\pi {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{3}+2 \pi {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{2}-2 \pi +2 i \ln \left (x \right )+2 i \ln \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )}\) | \(222\) |
int((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*ln((-50*x^2+200*x) *exp(x)-2*x^3+9*x^2-4*x)+(-50*x^2+100*x+200)*exp(x)^2+(-6*x^2+18*x-4)*exp( x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/ln((-50*x^2+200*x)*exp(x)-2*x^ 3+9*x^2-4*x)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log \left (-2 \, x^{3} + 9 \, x^{2} - 50 \, {\left (x^{2} - 4 \, x\right )} e^{x} - 4 \, x\right )} \]
integrate((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*log((-50*x^2 +200*x)*exp(x)-2*x^3+9*x^2-4*x)+(-50*x^2+100*x+200)*exp(x)^2+(-6*x^2+18*x- 4)*exp(x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/log((-50*x^2+200*x)*exp (x)-2*x^3+9*x^2-4*x)^2,x, algorithm=\
Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log {\left (- 2 x^{3} + 9 x^{2} - 4 x + \left (- 50 x^{2} + 200 x\right ) e^{x} \right )}} \]
integrate((((50*x**2-200*x)*exp(x)**2+(2*x**3-9*x**2+4*x)*exp(x))*ln((-50* x**2+200*x)*exp(x)-2*x**3+9*x**2-4*x)+(-50*x**2+100*x+200)*exp(x)**2+(-6*x **2+18*x-4)*exp(x))/((50*x**2-200*x)*exp(x)+2*x**3-9*x**2+4*x)/ln((-50*x** 2+200*x)*exp(x)-2*x**3+9*x**2-4*x)**2,x)
Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log \left (x - 4\right ) + \log \left (x\right ) + \log \left (-2 \, x - 50 \, e^{x} + 1\right )} \]
integrate((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*log((-50*x^2 +200*x)*exp(x)-2*x^3+9*x^2-4*x)+(-50*x^2+100*x+200)*exp(x)^2+(-6*x^2+18*x- 4)*exp(x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/log((-50*x^2+200*x)*exp (x)-2*x^3+9*x^2-4*x)^2,x, algorithm=\
Time = 0.38 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log \left (-2 \, x^{3} - 50 \, x^{2} e^{x} + 9 \, x^{2} + 200 \, x e^{x} - 4 \, x\right )} \]
integrate((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*log((-50*x^2 +200*x)*exp(x)-2*x^3+9*x^2-4*x)+(-50*x^2+100*x+200)*exp(x)^2+(-6*x^2+18*x- 4)*exp(x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/log((-50*x^2+200*x)*exp (x)-2*x^3+9*x^2-4*x)^2,x, algorithm=\
Timed out. \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\int -\frac {\ln \left ({\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-4\,x+9\,x^2-2\,x^3\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (200\,x-50\,x^2\right )-{\mathrm {e}}^x\,\left (2\,x^3-9\,x^2+4\,x\right )\right )-{\mathrm {e}}^{2\,x}\,\left (-50\,x^2+100\,x+200\right )+{\mathrm {e}}^x\,\left (6\,x^2-18\,x+4\right )}{{\ln \left ({\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-4\,x+9\,x^2-2\,x^3\right )}^2\,\left (4\,x-{\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-9\,x^2+2\,x^3\right )} \,d x \]
int(-(log(exp(x)*(200*x - 50*x^2) - 4*x + 9*x^2 - 2*x^3)*(exp(2*x)*(200*x - 50*x^2) - exp(x)*(4*x - 9*x^2 + 2*x^3)) - exp(2*x)*(100*x - 50*x^2 + 200 ) + exp(x)*(6*x^2 - 18*x + 4))/(log(exp(x)*(200*x - 50*x^2) - 4*x + 9*x^2 - 2*x^3)^2*(4*x - exp(x)*(200*x - 50*x^2) - 9*x^2 + 2*x^3)),x)