Integrand size = 69, antiderivative size = 29 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=3+e^{4 \left (\log (4)+\left (2-x+(5+x)^2\right ) \log \left (25 e^{-x} x\right )\right )} \]
\[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=\int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx \]
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]
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]
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 |
| \(\displaystyle 1024 \left (-18 \int 25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108}dx+27 \int \frac {25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108}}{x}dx-8 \int 25^{4 x^2+36 x+108} x \left (e^{-x} x\right )^{4 x^2+36 x+108}dx-\int 25^{4 x^2+36 x+108} x^2 \left (e^{-x} x\right )^{4 x^2+36 x+108}dx+9 \int \int 25^{4 \left (x^2+9 x+27\right )} \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dxdx-9 \int \frac {\int 25^{4 \left (x^2+9 x+27\right )} \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dx}{x}dx+2 \int \int 25^{4 \left (x^2+9 x+27\right )} x \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dxdx-2 \int \frac {\int 25^{4 \left (x^2+9 x+27\right )} x \left (e^{-x} x\right )^{4 \left (x^2+9 x+27\right )}dx}{x}dx+9 \log \left (25 e^{-x} x\right ) \int 25^{4 x^2+36 x+108} \left (e^{-x} x\right )^{4 x^2+36 x+108}dx+2 \log \left (25 e^{-x} x\right ) \int 25^{4 x^2+36 x+108} x \left (e^{-x} x\right )^{4 x^2+36 x+108}dx\right )\) |
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]
3.17.62.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.16 (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 | \(256 \,{\mathrm e}^{2 \left (x^{2}+9 x +27\right ) \left (i \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \pi -i \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right ) \pi -i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} \pi +i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi +2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x}\right )+4 \ln \left (5\right )\right )}\) | \(112\) |
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)
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=e^{\left (4 \, {\left (x^{2} + 9 \, x + 27\right )} \log \left (25 \, x e^{\left (-x\right )}\right ) + 8 \, \log \left (2\right )\right )} \]
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=\
Timed out. \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=\text {Timed out} \]
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)
Time = 0.37 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=2430865342914508479353150021007861031480567253406705911367623677652226107045071656712478446533481881623815074044969719579967204481363296508789062500000000 \, x^{108} e^{\left (-4 \, x^{3} + 8 \, x^{2} \log \left (5\right ) + 4 \, x^{2} \log \left (x\right ) - 36 \, x^{2} + 72 \, x \log \left (5\right ) + 36 \, x \log \left (x\right ) - 108 \, x\right )} \]
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=\
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)
Time = 0.37 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=e^{\left (4 \, x^{2} \log \left (25 \, x e^{\left (-x\right )}\right ) + 36 \, x \log \left (25 \, x e^{\left (-x\right )}\right ) + 8 \, \log \left (2\right ) + 108 \, \log \left (25 \, x e^{\left (-x\right )}\right )\right )} \]
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=\
Time = 10.51 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {256\ 25^{108+36 x+4 x^2} \left (e^{-x} x\right )^{108+36 x+4 x^2} \left (108-72 x-32 x^2-4 x^3+\left (36 x+8 x^2\right ) \log \left (25 e^{-x} x\right )\right )}{x} \, dx=256\,5^{8\,x^2+72\,x+216}\,x^{4\,x^2+36\,x+108}\,{\mathrm {e}}^{-108\,x}\,{\mathrm {e}}^{-4\,x^3}\,{\mathrm {e}}^{-36\,x^2} \]