\(\int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} (-2+2 x+x^2)+e^{x+x \log (4)} (100-80 x+100 x^2+40 x^3+(60+60 x^2) \log (4))}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} (1+2 x+x^2)+e^{x+x \log (4)} (40+40 x+40 x^2+40 x^3)} \, dx\) [2735]

Optimal result

Integrand size = 125, antiderivative size = 28 \[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=3+x+\frac {3}{1+x+20 e^{-x-x \log (4)} \left (1+x^2\right )} \] Output:

x+3+3/(1+5/exp(x+2*x*ln(2))*(4*x^2+4)+x)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=\int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx \] Input:

Integrate[(400 + 800*x^2 + 400*x^4 + E^(2*x + 2*x*Log[4])*(-2 + 2*x + x^2) 
 + E^(x + x*Log[4])*(100 - 80*x + 100*x^2 + 40*x^3 + (60 + 60*x^2)*Log[4]) 
)/(400 + 800*x^2 + 400*x^4 + E^(2*x + 2*x*Log[4])*(1 + 2*x + x^2) + E^(x + 
 x*Log[4])*(40 + 40*x + 40*x^2 + 40*x^3)),x]
 

Output:

Integrate[(400 + 800*x^2 + 400*x^4 + E^(2*x + 2*x*Log[4])*(-2 + 2*x + x^2) 
 + E^(x + x*Log[4])*(100 - 80*x + 100*x^2 + 40*x^3 + (60 + 60*x^2)*Log[4]) 
)/(400 + 800*x^2 + 400*x^4 + E^(2*x + 2*x*Log[4])*(1 + 2*x + x^2) + E^(x + 
 x*Log[4])*(40 + 40*x + 40*x^2 + 40*x^3)), 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.

\(\displaystyle \int \frac {400 x^4+800 x^2+\left (x^2+2 x-2\right ) e^{2 x+2 x \log (4)}+e^{x+x \log (4)} \left (40 x^3+100 x^2+\left (60 x^2+60\right ) \log (4)-80 x+100\right )+400}{400 x^4+800 x^2+\left (x^2+2 x+1\right ) e^{2 x+2 x \log (4)}+\left (40 x^3+40 x^2+40 x+40\right ) e^{x+x \log (4)}+400} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {400 x^4+800 x^2+\left (x^2+2 x-2\right ) e^{2 x+2 x \log (4)}+e^{x+x \log (4)} \left (40 x^3+100 x^2+\left (60 x^2+60\right ) \log (4)-80 x+100\right )+400}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {(4 e)^{2 x} x^2}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}+\frac {800 x^2}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}-\frac {5\ 4^{x+2} e^x x}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}+\frac {2^{4 x+1} e^{2 x} x}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}-\frac {2^{4 x+1} e^{2 x}}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}+\frac {400}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}+\frac {25\ 4^{x+1} e^x x^2 \left (1+\frac {3 \log (4)}{5}\right )}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}+\frac {25\ 4^{x+1} e^x \left (1+\frac {3 \log (4)}{5}\right )}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}+\frac {400 x^4}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}+\frac {5\ 2^{2 x+3} e^x x^3}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 400 \int \frac {1}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx+800 \int \frac {x^2}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx+\int \frac {(4 e)^{2 x} x^2}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx+20 (5+\log (64)) \int \frac {e^{x (1+\log (4))}}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx-2 \int \frac {e^{2 x (1+\log (4))}}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx-80 \int \frac {e^{x (1+\log (4))} x}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx+2 \int \frac {e^{2 x (1+\log (4))} x}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx+20 (5+\log (64)) \int \frac {e^{x (1+\log (4))} x^2}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx+400 \int \frac {x^4}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx+40 \int \frac {e^{x (1+\log (4))} x^3}{\left (20 x^2+(4 e)^x x+(4 e)^x+20\right )^2}dx\)

Input:

Int[(400 + 800*x^2 + 400*x^4 + E^(2*x + 2*x*Log[4])*(-2 + 2*x + x^2) + E^( 
x + x*Log[4])*(100 - 80*x + 100*x^2 + 40*x^3 + (60 + 60*x^2)*Log[4]))/(400 
 + 800*x^2 + 400*x^4 + E^(2*x + 2*x*Log[4])*(1 + 2*x + x^2) + E^(x + x*Log 
[4])*(40 + 40*x + 40*x^2 + 40*x^3)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57

method	result	size
risch	\(x +\frac {3}{1+x}-\frac {60 \left (x^{2}+1\right )}{\left (1+x \right ) \left ({\mathrm e}^{x} 4^{x} x +20 x^{2}+{\mathrm e}^{x} 4^{x}+20\right )}\)	\(44\)
norman	\(\frac {-20 x^{2}+2 \,{\mathrm e}^{x +2 x \ln \left (2\right )}+{\mathrm e}^{x +2 x \ln \left (2\right )} x^{2}+20 x +20 x^{3}-20}{{\mathrm e}^{x +2 x \ln \left (2\right )} x +20 x^{2}+{\mathrm e}^{x +2 x \ln \left (2\right )}+20}\)	\(66\)
parallelrisch	\(\frac {20 x^{3}+{\mathrm e}^{\left (1+2 \ln \left (2\right )\right ) x} x^{2}-20-20 x^{2}+20 x +2 \,{\mathrm e}^{\left (1+2 \ln \left (2\right )\right ) x}}{20 x^{2}+{\mathrm e}^{\left (1+2 \ln \left (2\right )\right ) x} x +{\mathrm e}^{\left (1+2 \ln \left (2\right )\right ) x}+20}\)	\(70\)

Input:

int(((x^2+2*x-2)*exp(x+2*x*ln(2))^2+(2*(60*x^2+60)*ln(2)+40*x^3+100*x^2-80 
*x+100)*exp(x+2*x*ln(2))+400*x^4+800*x^2+400)/((x^2+2*x+1)*exp(x+2*x*ln(2) 
)^2+(40*x^3+40*x^2+40*x+40)*exp(x+2*x*ln(2))+400*x^4+800*x^2+400),x,method 
=_RETURNVERBOSE)
 

Output:

x+3/(1+x)-60*(x^2+1)/(1+x)/(exp(x)*4^x*x+20*x^2+exp(x)*4^x+20)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=\frac {20 \, x^{3} + {\left (x^{2} + x + 3\right )} e^{\left (2 \, x \log \left (2\right ) + x\right )} + 20 \, x}{20 \, x^{2} + {\left (x + 1\right )} e^{\left (2 \, x \log \left (2\right ) + x\right )} + 20} \] Input:

integrate(((x^2+2*x-2)*exp(x+2*x*log(2))^2+(2*(60*x^2+60)*log(2)+40*x^3+10 
0*x^2-80*x+100)*exp(x+2*x*log(2))+400*x^4+800*x^2+400)/((x^2+2*x+1)*exp(x+ 
2*x*log(2))^2+(40*x^3+40*x^2+40*x+40)*exp(x+2*x*log(2))+400*x^4+800*x^2+40 
0),x, algorithm="fricas")
 

Output:

(20*x^3 + (x^2 + x + 3)*e^(2*x*log(2) + x) + 20*x)/(20*x^2 + (x + 1)*e^(2* 
x*log(2) + x) + 20)
 

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=x + \frac {- 60 x^{2} - 60}{20 x^{3} + 20 x^{2} + 20 x + \left (x^{2} + 2 x + 1\right ) e^{x + 2 x \log {\left (2 \right )}} + 20} + \frac {3}{x + 1} \] Input:

integrate(((x**2+2*x-2)*exp(x+2*x*ln(2))**2+(2*(60*x**2+60)*ln(2)+40*x**3+ 
100*x**2-80*x+100)*exp(x+2*x*ln(2))+400*x**4+800*x**2+400)/((x**2+2*x+1)*e 
xp(x+2*x*ln(2))**2+(40*x**3+40*x**2+40*x+40)*exp(x+2*x*ln(2))+400*x**4+800 
*x**2+400),x)
 

Output:

x + (-60*x**2 - 60)/(20*x**3 + 20*x**2 + 20*x + (x**2 + 2*x + 1)*exp(x + 2 
*x*log(2)) + 20) + 3/(x + 1)
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=\frac {20 \, x^{3} + {\left (x^{2} + x + 3\right )} e^{\left (2 \, x \log \left (2\right ) + x\right )} + 20 \, x}{20 \, x^{2} + {\left (x + 1\right )} e^{\left (2 \, x \log \left (2\right ) + x\right )} + 20} \] Input:

integrate(((x^2+2*x-2)*exp(x+2*x*log(2))^2+(2*(60*x^2+60)*log(2)+40*x^3+10 
0*x^2-80*x+100)*exp(x+2*x*log(2))+400*x^4+800*x^2+400)/((x^2+2*x+1)*exp(x+ 
2*x*log(2))^2+(40*x^3+40*x^2+40*x+40)*exp(x+2*x*log(2))+400*x^4+800*x^2+40 
0),x, algorithm="maxima")
 

Output:

(20*x^3 + (x^2 + x + 3)*e^(2*x*log(2) + x) + 20*x)/(20*x^2 + (x + 1)*e^(2* 
x*log(2) + x) + 20)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (27) = 54\).

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.46 \[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=\frac {20 \, x^{3} + x^{2} e^{\left (2 \, x \log \left (2\right ) + x\right )} + x e^{\left (2 \, x \log \left (2\right ) + x\right )} + 20 \, x + 3 \, e^{\left (2 \, x \log \left (2\right ) + x\right )}}{20 \, x^{2} + x e^{\left (2 \, x \log \left (2\right ) + x\right )} + e^{\left (2 \, x \log \left (2\right ) + x\right )} + 20} \] Input:

integrate(((x^2+2*x-2)*exp(x+2*x*log(2))^2+(2*(60*x^2+60)*log(2)+40*x^3+10 
0*x^2-80*x+100)*exp(x+2*x*log(2))+400*x^4+800*x^2+400)/((x^2+2*x+1)*exp(x+ 
2*x*log(2))^2+(40*x^3+40*x^2+40*x+40)*exp(x+2*x*log(2))+400*x^4+800*x^2+40 
0),x, algorithm="giac")
 

Output:

(20*x^3 + x^2*e^(2*x*log(2) + x) + x*e^(2*x*log(2) + x) + 20*x + 3*e^(2*x* 
log(2) + x))/(20*x^2 + x*e^(2*x*log(2) + x) + e^(2*x*log(2) + x) + 20)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=\int \frac {{\mathrm {e}}^{2\,x+4\,x\,\ln \left (2\right )}\,\left (x^2+2\,x-2\right )+800\,x^2+400\,x^4+{\mathrm {e}}^{x+2\,x\,\ln \left (2\right )}\,\left (2\,\ln \left (2\right )\,\left (60\,x^2+60\right )-80\,x+100\,x^2+40\,x^3+100\right )+400}{{\mathrm {e}}^{2\,x+4\,x\,\ln \left (2\right )}\,\left (x^2+2\,x+1\right )+{\mathrm {e}}^{x+2\,x\,\ln \left (2\right )}\,\left (40\,x^3+40\,x^2+40\,x+40\right )+800\,x^2+400\,x^4+400} \,d x \] Input:

int((exp(2*x + 4*x*log(2))*(2*x + x^2 - 2) + 800*x^2 + 400*x^4 + exp(x + 2 
*x*log(2))*(2*log(2)*(60*x^2 + 60) - 80*x + 100*x^2 + 40*x^3 + 100) + 400) 
/(exp(2*x + 4*x*log(2))*(2*x + x^2 + 1) + exp(x + 2*x*log(2))*(40*x + 40*x 
^2 + 40*x^3 + 40) + 800*x^2 + 400*x^4 + 400),x)
 

Output:

int((exp(2*x + 4*x*log(2))*(2*x + x^2 - 2) + 800*x^2 + 400*x^4 + exp(x + 2 
*x*log(2))*(2*log(2)*(60*x^2 + 60) - 80*x + 100*x^2 + 40*x^3 + 100) + 400) 
/(exp(2*x + 4*x*log(2))*(2*x + x^2 + 1) + exp(x + 2*x*log(2))*(40*x + 40*x 
^2 + 40*x^3 + 40) + 800*x^2 + 400*x^4 + 400), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (-2+2 x+x^2\right )+e^{x+x \log (4)} \left (100-80 x+100 x^2+40 x^3+\left (60+60 x^2\right ) \log (4)\right )}{400+800 x^2+400 x^4+e^{2 x+2 x \log (4)} \left (1+2 x+x^2\right )+e^{x+x \log (4)} \left (40+40 x+40 x^2+40 x^3\right )} \, dx=\frac {e^{x} 2^{2 x} x^{2}-2 e^{x} 2^{2 x} x +20 x^{3}-60 x^{2}+20 x -60}{e^{x} 2^{2 x} x +e^{x} 2^{2 x}+20 x^{2}+20} \] Input:

int(((x^2+2*x-2)*exp(x+2*x*log(2))^2+(2*(60*x^2+60)*log(2)+40*x^3+100*x^2- 
80*x+100)*exp(x+2*x*log(2))+400*x^4+800*x^2+400)/((x^2+2*x+1)*exp(x+2*x*lo 
g(2))^2+(40*x^3+40*x^2+40*x+40)*exp(x+2*x*log(2))+400*x^4+800*x^2+400),x)
 

Output:

(e**x*2**(2*x)*x**2 - 2*e**x*2**(2*x)*x + 20*x**3 - 60*x**2 + 20*x - 60)/( 
e**x*2**(2*x)*x + e**x*2**(2*x) + 20*x**2 + 20)