\(\int \frac {256\ 25^{108+36 x+4 x^2} (e^{-x} x)^{108+36 x+4 x^2} (108-72 x-32 x^2-4 x^3+(36 x+8 x^2) \log (25 e^{-x} x))}{x} \, dx\) [1662]

Optimal result

Integrand size = 69, antiderivative size = 29 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=3+e^{4 \left (\log (4)+\left (2-x+(5+x)^2\right ) \log \left (25 e^{-x} x\right )\right )} \] Output:

3+exp(4*ln(25*x/exp(x))*(2-x+(5+x)^2)+8*ln(2))
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=\int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx \] Input:

Integrate[(256*25^(108 + 36*x + 4*x^2)*(x/E^x)^(108 + 36*x + 4*x^2)*(108 - 
 72*x - 32*x^2 - 4*x^3 + (36*x + 8*x^2)*Log[(25*x)/E^x]))/x,x]
 

Output:

256*Integrate[(25^(108 + 36*x + 4*x^2)*(x/E^x)^(108 + 36*x + 4*x^2)*(108 - 
 72*x - 32*x^2 - 4*x^3 + (36*x + 8*x^2)*Log[(25*x)/E^x]))/x, 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.

\(\displaystyle \int \frac {256\ 25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108} \left (-4 x^3-32 x^2+\left (8 x^2+36 x\right ) \log \left (25 e^{-x} x\right )-72 x+108\right )}{x} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle 256 \int \frac {4\ 25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108} \left (-x^3-8 x^2-18 x+\left (2 x^2+9 x\right ) \log \left (25 e^{-x} x\right )+27\right )}{x}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 1024 \int \frac {25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108} \left (-x^3-8 x^2-18 x+\left (2 x^2+9 x\right ) \log \left (25 e^{-x} x\right )+27\right )}{x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 1024 \int \left (\frac {25^{4 x^2+36 x+108} \left (-x^3-8 x^2-18 x+27\right ) \left (e^{-x} x\right )^{4 x^2+36 x+108}}{x}+25^{4 x^2+36 x+108} (2 x+9) \log \left (25 e^{-x} x\right ) \left (e^{-x} x\right )^{4 x^2+36 x+108}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 1024 \left (-18 \int 25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108}dx+27 \int \frac {25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108}}{x}dx-8 \int 25^{4 x^2+36 x+108} x \left (e^{-x} x\right )^{4 x^2+36 x+108}dx-\int 25^{4 x^2+36 x+108} x^2 \left (e^{-x} x\right )^{4 x^2+36 x+108}dx+9 \int \int 25^{4 \left (x^2+9 x+27\right )} \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dxdx-9 \int \frac {\int 25^{4 \left (x^2+9 x+27\right )} \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dx}{x}dx+2 \int \int 25^{4 \left (x^2+9 x+27\right )} x \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dxdx-2 \int \frac {\int 25^{4 \left (x^2+9 x+27\right )} x \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dx}{x}dx+9 \log \left (25 e^{-x} x\right ) \int 25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108}dx+2 \log \left (25 e^{-x} x\right ) \int 25^{4 x^2+36 x+108} x \left (e^{-x} x\right )^{4 x^2+36 x+108}dx\right )\)

Input:

Int[(256*25^(108 + 36*x + 4*x^2)*(x/E^x)^(108 + 36*x + 4*x^2)*(108 - 72*x 
- 32*x^2 - 4*x^3 + (36*x + 8*x^2)*Log[(25*x)/E^x]))/x,x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\({\mathrm e}^{\left (4 x^{2}+36 x +108\right ) \ln \left (25 x \,{\mathrm e}^{-x}\right )+8 \ln \left (2\right )}\)	\(26\)
risch	\(256 \,{\mathrm e}^{2 \left (x^{2}+9 x +27\right ) \left (-i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} \pi +i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi +i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )-i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )+4 \ln \left (5\right )+2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x}\right )\right )}\)	\(112\)

Input:

int(((8*x^2+36*x)*ln(25*x/exp(x))-4*x^3-32*x^2-72*x+108)*exp((4*x^2+36*x+1 
08)*ln(25*x/exp(x))+8*ln(2))/x,x,method=_RETURNVERBOSE)
 

Output:

exp((4*x^2+36*x+108)*ln(25*x/exp(x))+8*ln(2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=e^{\left (4 \, {\left (x^{2} + 9 \, x + 27\right )} \log \left (25 \, x e^{\left (-x\right )}\right ) + 8 \, \log \left (2\right )\right )} \] Input:

integrate(((8*x^2+36*x)*log(25*x/exp(x))-4*x^3-32*x^2-72*x+108)*exp((4*x^2 
+36*x+108)*log(25*x/exp(x))+8*log(2))/x,x, algorithm="fricas")
 

Output:

e^(4*(x^2 + 9*x + 27)*log(25*x*e^(-x)) + 8*log(2))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=\text {Timed out} \] Input:

integrate(((8*x**2+36*x)*ln(25*x/exp(x))-4*x**3-32*x**2-72*x+108)*exp((4*x 
**2+36*x+108)*ln(25*x/exp(x))+8*ln(2))/x,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=2430865342914508479353150021007861031480567253406705911367623677652226107045071656712478446533481881623815074044969719579967204481363296508789062500000000 \, x^{108} e^{\left (-4 \, x^{3} + 8 \, x^{2} \log \left (5\right ) + 4 \, x^{2} \log \left (x\right ) - 36 \, x^{2} + 72 \, x \log \left (5\right ) + 36 \, x \log \left (x\right ) - 108 \, x\right )} \] Input:

integrate(((8*x^2+36*x)*log(25*x/exp(x))-4*x^3-32*x^2-72*x+108)*exp((4*x^2 
+36*x+108)*log(25*x/exp(x))+8*log(2))/x,x, algorithm="maxima")
 

Output:

24308653429145084793531500210078610314805672534067059113676236776522261070 
45071656712478446533481881623815074044969719579967204481363296508789062500 
000000*x^108*e^(-4*x^3 + 8*x^2*log(5) + 4*x^2*log(x) - 36*x^2 + 72*x*log(5 
) + 36*x*log(x) - 108*x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=e^{\left (4 \, x^{2} \log \left (25 \, x e^{\left (-x\right )}\right ) + 36 \, x \log \left (25 \, x e^{\left (-x\right )}\right ) + 8 \, \log \left (2\right ) + 108 \, \log \left (25 \, x e^{\left (-x\right )}\right )\right )} \] Input:

integrate(((8*x^2+36*x)*log(25*x/exp(x))-4*x^3-32*x^2-72*x+108)*exp((4*x^2 
+36*x+108)*log(25*x/exp(x))+8*log(2))/x,x, algorithm="giac")
 

Output:

e^(4*x^2*log(25*x*e^(-x)) + 36*x*log(25*x*e^(-x)) + 8*log(2) + 108*log(25* 
x*e^(-x)))
 

Mupad [B] (verification not implemented)

Time = 3.76 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=256\,5^{8\,x^2+72\,x+216}\,x^{4\,x^2+36\,x+108}\,{\mathrm {e}}^{-108\,x}\,{\mathrm {e}}^{-4\,x^3}\,{\mathrm {e}}^{-36\,x^2} \] Input:

int(-(exp(8*log(2) + log(25*x*exp(-x))*(36*x + 4*x^2 + 108))*(72*x - log(2 
5*x*exp(-x))*(36*x + 8*x^2) + 32*x^2 + 4*x^3 - 108))/x,x)
 

Output:

256*5^(72*x + 8*x^2 + 216)*x^(36*x + 4*x^2 + 108)*exp(-108*x)*exp(-4*x^3)* 
exp(-36*x^2)
 

Reduce [F]

\[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=\int \frac {\left (\left (8 x^{2}+36 x \right ) \mathrm {log}\left (\frac {25 x}{{\mathrm e}^{x}}\right )-4 x^{3}-32 x^{2}-72 x +108\right ) {\mathrm e}^{\left (4 x^{2}+36 x +108\right ) \mathrm {log}\left (\frac {25 x}{{\mathrm e}^{x}}\right )+8 \,\mathrm {log}\left (2\right )}}{x}d x \] Input:

int(((8*x^2+36*x)*log(25*x/exp(x))-4*x^3-32*x^2-72*x+108)*exp((4*x^2+36*x+ 
108)*log(25*x/exp(x))+8*log(2))/x,x)
 

Output:

int(((8*x^2+36*x)*log(25*x/exp(x))-4*x^3-32*x^2-72*x+108)*exp((4*x^2+36*x+ 
108)*log(25*x/exp(x))+8*log(2))/x,x)