\(\int \frac {e^{2 x} (200+100 x-50 x^2)+e^x (-4+18 x-6 x^2)+(e^{2 x} (-200 x+50 x^2)+e^x (4 x-9 x^2+2 x^3)) \log (-4 x+9 x^2-2 x^3+e^x (200 x-50 x^2))}{(4 x-9 x^2+2 x^3+e^x (-200 x+50 x^2)) \log ^2(-4 x+9 x^2-2 x^3+e^x (200 x-50 x^2))} \, dx\) [755]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 154, antiderivative size = 27 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=1+\frac {e^x}{\log \left (2 (4-x) x \left (-\frac {1}{2}+25 e^x+x\right )\right )} \] Output:

1+exp(x)/ln(2*(x+25*exp(x)-1/2)*(4-x)*x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^x}{\log \left (-\left ((-4+x) x \left (-1+50 e^x+2 x\right )\right )\right )} \] Input:

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]
 

Output:

E^x/Log[-((-4 + x)*x*(-1 + 50*E^x + 2*x))]
 

Rubi [F]

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

\(\displaystyle -\frac {2}{25} \int \frac {x^2}{\left (2 x+50 e^x-1\right ) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\frac {3}{50} \int \frac {1}{\log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\int \frac {e^x}{\log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\int \frac {e^x}{(x-4) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\int \frac {e^x}{x \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx+\frac {1}{25} \int \frac {x}{\log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx-\frac {3}{50} \int \frac {1}{\left (2 x+50 e^x-1\right ) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx+\frac {4}{25} \int \frac {x}{\left (2 x+50 e^x-1\right ) \log ^2\left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx+\int \frac {e^x}{\log \left (-\left ((x-4) x \left (2 x+50 e^x-1\right )\right )\right )}dx\)

Input:

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]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 0.72 (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 \(\frac {2 i {\mathrm e}^{x}}{\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right ) \operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )-\pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{2}+2 \pi {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right ) {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{2}-\pi {\operatorname {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )}^{3}-2 \pi +2 i \ln \left (x \right )+2 i \ln \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )}\) \(222\)

Input:

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)
 

Output:

exp(x)/ln((-50*x^2+200*x)*exp(x)-2*x^3+9*x^2-4*x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log \left (-2 \, x^{3} + 9 \, x^{2} - 50 \, {\left (x^{2} - 4 \, x\right )} e^{x} - 4 \, x\right )} \] Input:

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="fricas")
 

Output:

e^x/log(-2*x^3 + 9*x^2 - 50*(x^2 - 4*x)*e^x - 4*x)
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log {\left (- 2 x^{3} + 9 x^{2} - 4 x + \left (- 50 x^{2} + 200 x\right ) e^{x} \right )}} \] Input:

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)
 

Output:

exp(x)/log(-2*x**3 + 9*x**2 - 4*x + (-50*x**2 + 200*x)*exp(x))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log \left (x - 4\right ) + \log \left (x\right ) + \log \left (-2 \, x - 50 \, e^{x} + 1\right )} \] Input:

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="maxima")
 

Output:

e^x/(log(x - 4) + log(x) + log(-2*x - 50*e^x + 1))
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\log \left (-2 \, x^{3} - 50 \, x^{2} e^{x} + 9 \, x^{2} + 200 \, x e^{x} - 4 \, x\right )} \] Input:

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="giac")
 

Output:

e^x/log(-2*x^3 - 50*x^2*e^x + 9*x^2 + 200*x*e^x - 4*x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\int -\frac {\ln \left ({\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-4\,x+9\,x^2-2\,x^3\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (200\,x-50\,x^2\right )-{\mathrm {e}}^x\,\left (2\,x^3-9\,x^2+4\,x\right )\right )-{\mathrm {e}}^{2\,x}\,\left (-50\,x^2+100\,x+200\right )+{\mathrm {e}}^x\,\left (6\,x^2-18\,x+4\right )}{{\ln \left ({\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-4\,x+9\,x^2-2\,x^3\right )}^2\,\left (4\,x-{\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-9\,x^2+2\,x^3\right )} \,d x \] Input:

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)
 

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{2 x} \left (200+100 x-50 x^2\right )+e^x \left (-4+18 x-6 x^2\right )+\left (e^{2 x} \left (-200 x+50 x^2\right )+e^x \left (4 x-9 x^2+2 x^3\right )\right ) \log \left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )}{\left (4 x-9 x^2+2 x^3+e^x \left (-200 x+50 x^2\right )\right ) \log ^2\left (-4 x+9 x^2-2 x^3+e^x \left (200 x-50 x^2\right )\right )} \, dx=\frac {e^{x}}{\mathrm {log}\left (-50 e^{x} x^{2}+200 e^{x} x -2 x^{3}+9 x^{2}-4 x \right )} \] Input:

int((((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)
 

Output:

e**x/log( - 50*e**x*x**2 + 200*e**x*x - 2*x**3 + 9*x**2 - 4*x)