\(\int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} (16 e^x x^3 \log (2)+(-16 x^2-16 x^3) \log (2))}{40000+64 e^{10+2 e^x-2 x} x^2+(400 e x+800 x^2) \log (2)+(e^2 x^2+4 e x^3+4 x^4) \log ^2(2)+e^{5+e^x-x} (-3200 x+(-16 e x^2-32 x^3) \log (2))} \, dx\) [2477]

Optimal result

Integrand size = 141, antiderivative size = 35 \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=\frac {x}{\frac {e}{2}+x+\frac {4 \left (-e^{5+e^x-x}+\frac {25}{x}\right )}{\log (2)}} \] Output:

x/(4/ln(2)*(25/x-exp(exp(x)+5-x))+1/2*exp(1)+x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=\frac {2 e^x x^2 \log (2)}{-8 e^{5+e^x} x+e^{1+x} x \log (2)+e^x \left (200+x^2 \log (4)\right )} \] Input:

Integrate[(800*x*Log[2] + 2*E*x^2*Log[2]^2 + E^(5 + E^x - x)*(16*E^x*x^3*L 
og[2] + (-16*x^2 - 16*x^3)*Log[2]))/(40000 + 64*E^(10 + 2*E^x - 2*x)*x^2 + 
 (400*E*x + 800*x^2)*Log[2] + (E^2*x^2 + 4*E*x^3 + 4*x^4)*Log[2]^2 + E^(5 
+ E^x - x)*(-3200*x + (-16*E*x^2 - 32*x^3)*Log[2])),x]
 

Output:

(2*E^x*x^2*Log[2])/(-8*E^(5 + E^x)*x + E^(1 + x)*x*Log[2] + E^x*(200 + x^2 
*Log[4]))
 

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.

\(\displaystyle \int \frac {2 e x^2 \log ^2(2)+e^{-x+e^x+5} \left (16 e^x x^3 \log (2)+\left (-16 x^3-16 x^2\right ) \log (2)\right )+800 x \log (2)}{64 e^{-2 x+2 e^x+10} x^2+\left (800 x^2+400 e x\right ) \log (2)+e^{-x+e^x+5} \left (\left (-32 x^3-16 e x^2\right ) \log (2)-3200 x\right )+\left (4 x^4+4 e x^3+e^2 x^2\right ) \log ^2(2)+40000} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{2 x} \left (2 e x^2 \log ^2(2)+e^{-x+e^x+5} \left (16 e^x x^3 \log (2)+\left (-16 x^3-16 x^2\right ) \log (2)\right )+800 x \log (2)\right )}{\left (e^x x^2 \log (4)-8 e^{e^x+5} x+200 e^x+e^{x+1} x \log (2)\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {e^{2 x-e^x-5} (x+1) \log (2) \left (x^2 \log (4)+e x \log (2)+200\right )}{4 \left (-e^x x^2 \log (4)+8 e^{e^x+5} x-200 e^x-e^{x+1} x \log (2)\right )}+\frac {2 e^{2 x} x \log (2) \left (x^3 (-\log (4))+8 e^{e^x+5} x^2-e x^2 \log (2) \left (1+\frac {\log (4)}{e \log (2)}\right )-200 x+200\right )}{\left (e^x x^2 \log (4)-8 e^{e^x+5} x+200 e^x+e^{x+1} x \log (2)\right )^2}-\frac {1}{4} e^{x-e^x-5} (x+1) \log (2)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} (2+e) \log ^2(2) \int \frac {e^{2 x-e^x-5} x^2}{e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x}dx+400 \log (2) \int \frac {e^{2 x} x}{\left (e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x\right )^2}dx-400 \log (2) \int \frac {e^{2 x} x^2}{\left (e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x\right )^2}dx+50 \log (2) \int \frac {e^{2 x-e^x-5}}{e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x}dx+\frac {1}{4} \log (2) (200+e \log (2)) \int \frac {e^{2 x-e^x-5} x}{e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x}dx-2 \log (2) \log (4) \int \frac {e^{2 x} x^4}{\left (e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x\right )^2}dx-2 (2+e) \log ^2(2) \int \frac {e^{2 x} x^3}{\left (e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x\right )^2}dx+16 \log (2) \int \frac {e^{2 x+e^x+5} x^3}{\left (-e^x \log (4) x^2+8 e^{5+e^x} x-e^{x+1} \log (2) x-200 e^x\right )^2}dx+\frac {1}{4} \log (2) \log (4) \int \frac {e^{2 x-e^x-5} x^3}{e^x \log (4) x^2-8 e^{5+e^x} x+e^{x+1} \log (2) x+200 e^x}dx-\frac {1}{4} \log (2) \int e^{x-e^x-5} xdx+\frac {1}{4} e^{-e^x-5} \log (2)\)

Input:

Int[(800*x*Log[2] + 2*E*x^2*Log[2]^2 + E^(5 + E^x - x)*(16*E^x*x^3*Log[2] 
+ (-16*x^2 - 16*x^3)*Log[2]))/(40000 + 64*E^(10 + 2*E^x - 2*x)*x^2 + (400* 
E*x + 800*x^2)*Log[2] + (E^2*x^2 + 4*E*x^3 + 4*x^4)*Log[2]^2 + E^(5 + E^x 
- x)*(-3200*x + (-16*E*x^2 - 32*x^3)*Log[2])),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03

method	result	size
norman	\(\frac {2 x^{2} \ln \left (2\right )}{x \,{\mathrm e} \ln \left (2\right )+2 x^{2} \ln \left (2\right )-8 \,{\mathrm e}^{{\mathrm e}^{x}+5-x} x +200}\)	\(36\)
risch	\(\frac {2 x^{2} \ln \left (2\right )}{x \,{\mathrm e} \ln \left (2\right )+2 x^{2} \ln \left (2\right )-8 \,{\mathrm e}^{{\mathrm e}^{x}+5-x} x +200}\)	\(36\)
parallelrisch	\(\frac {2 x^{2} \ln \left (2\right )}{x \,{\mathrm e} \ln \left (2\right )+2 x^{2} \ln \left (2\right )-8 \,{\mathrm e}^{{\mathrm e}^{x}+5-x} x +200}\)	\(36\)

Input:

int(((16*x^3*ln(2)*exp(x)+(-16*x^3-16*x^2)*ln(2))*exp(exp(x)+5-x)+2*x^2*ex 
p(1)*ln(2)^2+800*x*ln(2))/(64*x^2*exp(exp(x)+5-x)^2+((-16*x^2*exp(1)-32*x^ 
3)*ln(2)-3200*x)*exp(exp(x)+5-x)+(x^2*exp(1)^2+4*x^3*exp(1)+4*x^4)*ln(2)^2 
+(400*x*exp(1)+800*x^2)*ln(2)+40000),x,method=_RETURNVERBOSE)
 

Output:

2*x^2*ln(2)/(x*exp(1)*ln(2)+2*x^2*ln(2)-8*exp(exp(x)+5-x)*x+200)
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=-\frac {2 \, x^{2} \log \left (2\right )}{8 \, x e^{\left (-x + e^{x} + 5\right )} - {\left (2 \, x^{2} + x e\right )} \log \left (2\right ) - 200} \] Input:

integrate(((16*x^3*log(2)*exp(x)+(-16*x^3-16*x^2)*log(2))*exp(exp(x)+5-x)+ 
2*x^2*exp(1)*log(2)^2+800*x*log(2))/(64*x^2*exp(exp(x)+5-x)^2+((-16*x^2*ex 
p(1)-32*x^3)*log(2)-3200*x)*exp(exp(x)+5-x)+(x^2*exp(1)^2+4*x^3*exp(1)+4*x 
^4)*log(2)^2+(400*exp(1)*x+800*x^2)*log(2)+40000),x, algorithm="fricas")
 

Output:

-2*x^2*log(2)/(8*x*e^(-x + e^x + 5) - (2*x^2 + x*e)*log(2) - 200)
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=- \frac {2 x^{2} \log {\left (2 \right )}}{- 2 x^{2} \log {\left (2 \right )} + 8 x e^{- x + e^{x} + 5} - e x \log {\left (2 \right )} - 200} \] Input:

integrate(((16*x**3*ln(2)*exp(x)+(-16*x**3-16*x**2)*ln(2))*exp(exp(x)+5-x) 
+2*x**2*exp(1)*ln(2)**2+800*x*ln(2))/(64*x**2*exp(exp(x)+5-x)**2+((-16*x** 
2*exp(1)-32*x**3)*ln(2)-3200*x)*exp(exp(x)+5-x)+(x**2*exp(1)**2+4*x**3*exp 
(1)+4*x**4)*ln(2)**2+(400*exp(1)*x+800*x**2)*ln(2)+40000),x)
 

Output:

-2*x**2*log(2)/(-2*x**2*log(2) + 8*x*exp(-x + exp(x) + 5) - E*x*log(2) - 2 
00)
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=\frac {2 \, x^{2} e^{x} \log \left (2\right )}{{\left (2 \, x^{2} \log \left (2\right ) + x e \log \left (2\right ) + 200\right )} e^{x} - 8 \, x e^{\left (e^{x} + 5\right )}} \] Input:

integrate(((16*x^3*log(2)*exp(x)+(-16*x^3-16*x^2)*log(2))*exp(exp(x)+5-x)+ 
2*x^2*exp(1)*log(2)^2+800*x*log(2))/(64*x^2*exp(exp(x)+5-x)^2+((-16*x^2*ex 
p(1)-32*x^3)*log(2)-3200*x)*exp(exp(x)+5-x)+(x^2*exp(1)^2+4*x^3*exp(1)+4*x 
^4)*log(2)^2+(400*exp(1)*x+800*x^2)*log(2)+40000),x, algorithm="maxima")
 

Output:

2*x^2*e^x*log(2)/((2*x^2*log(2) + x*e*log(2) + 200)*e^x - 8*x*e^(e^x + 5))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1093 vs. \(2 (33) = 66\).

Time = 0.23 (sec) , antiderivative size = 1093, normalized size of antiderivative = 31.23 \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=\text {Too large to display} \] Input:

integrate(((16*x^3*log(2)*exp(x)+(-16*x^3-16*x^2)*log(2))*exp(exp(x)+5-x)+ 
2*x^2*exp(1)*log(2)^2+800*x*log(2))/(64*x^2*exp(exp(x)+5-x)^2+((-16*x^2*ex 
p(1)-32*x^3)*log(2)-3200*x)*exp(exp(x)+5-x)+(x^2*exp(1)^2+4*x^3*exp(1)+4*x 
^4)*log(2)^2+(400*exp(1)*x+800*x^2)*log(2)+40000),x, algorithm="giac")
 

Output:

2*(4*x^7*e^(3/2*x)*log(2)^3 - 4*x^7*e^(1/2*x)*log(2)^3 - 4*x^6*e^(1/2*x)*l 
og(2)^3 + 4*x^6*e^(3/2*x + 1)*log(2)^3 - 4*x^6*e^(1/2*x + 1)*log(2)^3 - 16 
*x^6*e^(1/2*x + e^x + 5)*log(2)^2 + 16*x^6*e^(-1/2*x + e^x + 5)*log(2)^2 + 
 x^5*e^(3/2*x + 2)*log(2)^3 - x^5*e^(1/2*x + 2)*log(2)^3 - 2*x^5*e^(1/2*x 
+ 1)*log(2)^3 + 800*x^5*e^(3/2*x)*log(2)^2 - 800*x^5*e^(1/2*x)*log(2)^2 - 
8*x^5*e^(1/2*x + e^x + 6)*log(2)^2 + 8*x^5*e^(-1/2*x + e^x + 6)*log(2)^2 + 
 16*x^5*e^(-1/2*x + e^x + 5)*log(2)^2 + 400*x^4*e^(3/2*x + 1)*log(2)^2 - 4 
00*x^4*e^(1/2*x + 1)*log(2)^2 - 1600*x^4*e^(1/2*x + e^x + 5)*log(2) + 1600 
*x^4*e^(-1/2*x + e^x + 5)*log(2) + 200*x^3*e^(1/2*x + 1)*log(2)^2 + 40000* 
x^3*e^(3/2*x)*log(2) - 40000*x^3*e^(1/2*x)*log(2) - 1600*x^3*e^(-1/2*x + e 
^x + 5)*log(2) + 40000*x^2*e^(1/2*x)*log(2))/(8*x^7*e^(3/2*x)*log(2)^3 - 8 
*x^7*e^(1/2*x)*log(2)^3 - 8*x^6*e^(1/2*x)*log(2)^3 + 12*x^6*e^(3/2*x + 1)* 
log(2)^3 - 12*x^6*e^(1/2*x + 1)*log(2)^3 - 64*x^6*e^(1/2*x + e^x + 5)*log( 
2)^2 + 64*x^6*e^(-1/2*x + e^x + 5)*log(2)^2 + 6*x^5*e^(3/2*x + 2)*log(2)^3 
 - 6*x^5*e^(1/2*x + 2)*log(2)^3 - 8*x^5*e^(1/2*x + 1)*log(2)^3 + 2400*x^5* 
e^(3/2*x)*log(2)^2 - 2400*x^5*e^(1/2*x)*log(2)^2 - 64*x^5*e^(1/2*x + e^x + 
 6)*log(2)^2 + 64*x^5*e^(-1/2*x + e^x + 6)*log(2)^2 + 64*x^5*e^(-1/2*x + e 
^x + 5)*log(2)^2 + x^4*e^(3/2*x + 3)*log(2)^3 - x^4*e^(1/2*x + 3)*log(2)^3 
 - 2*x^4*e^(1/2*x + 2)*log(2)^3 + 128*x^5*e^(-1/2*x + 2*e^x + 10)*log(2) - 
 128*x^5*e^(-3/2*x + 2*e^x + 10)*log(2) - 800*x^4*e^(1/2*x)*log(2)^2 + ...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=\int \frac {800\,x\,\ln \left (2\right )-{\mathrm {e}}^{{\mathrm {e}}^x-x+5}\,\left (\ln \left (2\right )\,\left (16\,x^3+16\,x^2\right )-16\,x^3\,{\mathrm {e}}^x\,\ln \left (2\right )\right )+2\,x^2\,\mathrm {e}\,{\ln \left (2\right )}^2}{64\,x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x-2\,x+10}-{\mathrm {e}}^{{\mathrm {e}}^x-x+5}\,\left (3200\,x+\ln \left (2\right )\,\left (32\,x^3+16\,\mathrm {e}\,x^2\right )\right )+\ln \left (2\right )\,\left (800\,x^2+400\,\mathrm {e}\,x\right )+{\ln \left (2\right )}^2\,\left (4\,x^4+4\,\mathrm {e}\,x^3+{\mathrm {e}}^2\,x^2\right )+40000} \,d x \] Input:

int((800*x*log(2) - exp(exp(x) - x + 5)*(log(2)*(16*x^2 + 16*x^3) - 16*x^3 
*exp(x)*log(2)) + 2*x^2*exp(1)*log(2)^2)/(64*x^2*exp(2*exp(x) - 2*x + 10) 
- exp(exp(x) - x + 5)*(3200*x + log(2)*(16*x^2*exp(1) + 32*x^3)) + log(2)* 
(400*x*exp(1) + 800*x^2) + log(2)^2*(x^2*exp(2) + 4*x^3*exp(1) + 4*x^4) + 
40000),x)
 

Output:

int((800*x*log(2) - exp(exp(x) - x + 5)*(log(2)*(16*x^2 + 16*x^3) - 16*x^3 
*exp(x)*log(2)) + 2*x^2*exp(1)*log(2)^2)/(64*x^2*exp(2*exp(x) - 2*x + 10) 
- exp(exp(x) - x + 5)*(3200*x + log(2)*(16*x^2*exp(1) + 32*x^3)) + log(2)* 
(400*x*exp(1) + 800*x^2) + log(2)^2*(x^2*exp(2) + 4*x^3*exp(1) + 4*x^4) + 
40000), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {800 x \log (2)+2 e x^2 \log ^2(2)+e^{5+e^x-x} \left (16 e^x x^3 \log (2)+\left (-16 x^2-16 x^3\right ) \log (2)\right )}{40000+64 e^{10+2 e^x-2 x} x^2+\left (400 e x+800 x^2\right ) \log (2)+\left (e^2 x^2+4 e x^3+4 x^4\right ) \log ^2(2)+e^{5+e^x-x} \left (-3200 x+\left (-16 e x^2-32 x^3\right ) \log (2)\right )} \, dx=-\frac {2 e^{x} \mathrm {log}\left (2\right ) x^{2}}{8 e^{e^{x}} e^{5} x -e^{x} \mathrm {log}\left (2\right ) e x -2 e^{x} \mathrm {log}\left (2\right ) x^{2}-200 e^{x}} \] Input:

int(((16*x^3*log(2)*exp(x)+(-16*x^3-16*x^2)*log(2))*exp(exp(x)+5-x)+2*x^2* 
exp(1)*log(2)^2+800*x*log(2))/(64*x^2*exp(exp(x)+5-x)^2+((-16*x^2*exp(1)-3 
2*x^3)*log(2)-3200*x)*exp(exp(x)+5-x)+(x^2*exp(1)^2+4*x^3*exp(1)+4*x^4)*lo 
g(2)^2+(400*exp(1)*x+800*x^2)*log(2)+40000),x)
 

Output:

( - 2*e**x*log(2)*x**2)/(8*e**(e**x)*e**5*x - e**x*log(2)*e*x - 2*e**x*log 
(2)*x**2 - 200*e**x)