Integrand size = 75, antiderivative size = 25 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\frac {e^{-4+\frac {x}{4}} \left (e^{20}+\frac {\log (x)}{x}\right )}{x}} \]
Time = 2.58 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\frac {e^{16+\frac {x}{4}}}{x}} x^{\frac {e^{-4+\frac {x}{4}}}{x^2}} \]
Integrate[(E^(-4 + (E^(20 + x/4)*x + E^(x/4)*Log[x])/(E^4*x^2))*(E^(x/4)*( 4 + E^20*(-4*x + x^2)) + E^(x/4)*(-8 + x)*Log[x]))/(4*x^3),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.
| \(\displaystyle \int \frac {e^{\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}-4} \left (e^{x/4} \left (e^{20} \left (x^2-4 x\right )+4\right )+e^{x/4} (x-8) \log (x)\right )}{4 x^3} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {e^{\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}-4} \left (e^{x/4} \left (4-e^{20} \left (4 x-x^2\right )\right )-e^{x/4} (8-x) \log (x)\right )}{x^3}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \frac {1}{4} \int \frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{20} x^2+\log (x) x-4 e^{20} x-8 \log (x)+4\right )}{x^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{4} \int \left (\frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{20} x^2-4 e^{20} x+4\right )}{x^3}+\frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}} (x-8) \log (x)}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (-4 \int \frac {e^{\frac {x}{4}+16+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}}}{x^2}dx+\int \frac {e^{\frac {x}{4}+16+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}}}{x}dx+\int \frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}} \log (x)}{x^2}dx+4 \int \frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}}}{x^3}dx-8 \int \frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}} \log (x)}{x^3}dx\right )\) |
Int[(E^(-4 + (E^(20 + x/4)*x + E^(x/4)*Log[x])/(E^4*x^2))*(E^(x/4)*(4 + E^ 20*(-4*x + x^2)) + E^(x/4)*(-8 + x)*Log[x]))/(4*x^3),x]
3.16.66.3.1 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 = 4.62 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}} \left (x \,{\mathrm e}^{20}+\ln \left (x \right )\right ) {\mathrm e}^{-4}}{x^{2}}}\) | \(21\) |
| risch | \(x^{\frac {{\mathrm e}^{-4+\frac {x}{4}}}{x^{2}}} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}+16}}{x}}\) | \(25\) |
int(1/4*((-8+x)*exp(1/4*x)*ln(x)+((x^2-4*x)*exp(20)+4)*exp(1/4*x))*exp((ex p(1/4*x)*ln(x)+x*exp(20)*exp(1/4*x))/x^2/exp(4))/x^3/exp(4),x,method=_RETU RNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\left (-\frac {{\left (4 \, x^{2} e^{24} - x e^{\left (\frac {1}{4} \, x + 40\right )} - e^{\left (\frac {1}{4} \, x + 20\right )} \log \left (x\right )\right )} e^{\left (-24\right )}}{x^{2}} + 4\right )} \]
integrate(1/4*((-8+x)*exp(1/4*x)*log(x)+((x^2-4*x)*exp(20)+4)*exp(1/4*x))* exp((exp(1/4*x)*log(x)+x*exp(20)*exp(1/4*x))/x^2/exp(4))/x^3/exp(4),x, alg orithm=\
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\frac {x e^{20} e^{\frac {x}{4}} + e^{\frac {x}{4}} \log {\left (x \right )}}{x^{2} e^{4}}} \]
integrate(1/4*((-8+x)*exp(1/4*x)*ln(x)+((x**2-4*x)*exp(20)+4)*exp(1/4*x))* exp((exp(1/4*x)*ln(x)+x*exp(20)*exp(1/4*x))/x**2/exp(4))/x**3/exp(4),x)
Time = 0.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\left (\frac {e^{\left (\frac {1}{4} \, x + 16\right )}}{x} + \frac {e^{\left (\frac {1}{4} \, x - 4\right )} \log \left (x\right )}{x^{2}}\right )} \]
integrate(1/4*((-8+x)*exp(1/4*x)*log(x)+((x^2-4*x)*exp(20)+4)*exp(1/4*x))* exp((exp(1/4*x)*log(x)+x*exp(20)*exp(1/4*x))/x^2/exp(4))/x^3/exp(4),x, alg orithm=\
Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\left (\frac {e^{\left (\frac {1}{4} \, x + 16\right )}}{x} + \frac {e^{\left (\frac {1}{4} \, x - 4\right )} \log \left (x\right )}{x^{2}}\right )} \]
integrate(1/4*((-8+x)*exp(1/4*x)*log(x)+((x^2-4*x)*exp(20)+4)*exp(1/4*x))* exp((exp(1/4*x)*log(x)+x*exp(20)*exp(1/4*x))/x^2/exp(4))/x^3/exp(4),x, alg orithm=\
Time = 12.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=x^{\frac {{\mathrm {e}}^{\frac {x}{4}-4}}{x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x/4}\,{\mathrm {e}}^{16}}{x}} \]