Integrand size = 75, antiderivative size = 25 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\frac {e^{-4+\frac {x}{4}} \left (e^{20}+\frac {\log (x)}{x}\right )}{x}} \] Output:
exp(exp(1/4*x)/exp(4)/x*(exp(20)+ln(x)/x))
Time = 1.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\frac {e^{16+\frac {x}{4}}}{x}} x^{\frac {e^{-4+\frac {x}{4}}}{x^2}} \] Input:
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]
Output:
E^(E^(16 + x/4)/x)*x^(E^(-4 + x/4)/x^2)
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 |
\(\displaystyle \frac {1}{4} \left (-4 \int \frac {e^{\frac {x}{4}+16+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}}}{x^2}dx+\int \frac {e^{\frac {x}{4}+16+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}}}{x}dx+\int \frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}} \log (x)}{x^2}dx+4 \int \frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}}}{x^3}dx-8 \int \frac {e^{\frac {x}{4}-4+\frac {e^{\frac {x}{4}+20} x+e^{x/4} \log (x)}{e^4 x^2}} \log (x)}{x^3}dx\right )\) |
Input:
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]
Output:
$Aborted
Time = 4.12 (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 | \(x^{\frac {{\mathrm e}^{-4+\frac {x}{4}}}{x^{2}}} {\mathrm e}^{\frac {{\mathrm e}^{16+\frac {x}{4}}}{x}}\) | \(25\) |
Input:
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)
Output:
exp(exp(1/4*x)*(x*exp(20)+ln(x))/x^2/exp(4))
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\left (-\frac {{\left (4 \, x^{2} e^{24} - x e^{\left (\frac {1}{4} \, x + 40\right )} - e^{\left (\frac {1}{4} \, x + 20\right )} \log \left (x\right )\right )} e^{\left (-24\right )}}{x^{2}} + 4\right )} \] Input:
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="fricas")
Output:
e^(-(4*x^2*e^24 - x*e^(1/4*x + 40) - e^(1/4*x + 20)*log(x))*e^(-24)/x^2 + 4)
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\frac {x e^{20} e^{\frac {x}{4}} + e^{\frac {x}{4}} \log {\left (x \right )}}{x^{2} e^{4}}} \] Input:
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)
Output:
exp((x*exp(20)*exp(x/4) + exp(x/4)*log(x))*exp(-4)/x**2)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\left (\frac {e^{\left (\frac {1}{4} \, x + 16\right )}}{x} + \frac {e^{\left (\frac {1}{4} \, x - 4\right )} \log \left (x\right )}{x^{2}}\right )} \] Input:
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="maxima")
Output:
e^(e^(1/4*x + 16)/x + e^(1/4*x - 4)*log(x)/x^2)
\[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=\int { \frac {{\left ({\left (x - 8\right )} e^{\left (\frac {1}{4} \, x\right )} \log \left (x\right ) + {\left ({\left (x^{2} - 4 \, x\right )} e^{20} + 4\right )} e^{\left (\frac {1}{4} \, x\right )}\right )} e^{\left (\frac {{\left (x e^{\left (\frac {1}{4} \, x + 20\right )} + e^{\left (\frac {1}{4} \, x\right )} \log \left (x\right )\right )} e^{\left (-4\right )}}{x^{2}} - 4\right )}}{4 \, x^{3}} \,d x } \] Input:
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="giac")
Output:
integrate(1/4*((x - 8)*e^(1/4*x)*log(x) + ((x^2 - 4*x)*e^20 + 4)*e^(1/4*x) )*e^((x*e^(1/4*x + 20) + e^(1/4*x)*log(x))*e^(-4)/x^2 - 4)/x^3, x)
Time = 3.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=x^{\frac {{\mathrm {e}}^{\frac {x}{4}-4}}{x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x/4}\,{\mathrm {e}}^{16}}{x}} \] Input:
int(-(exp((exp(-4)*(exp(x/4)*log(x) + x*exp(x/4)*exp(20)))/x^2)*exp(-4)*(e xp(x/4)*(exp(20)*(4*x - x^2) - 4) - exp(x/4)*log(x)*(x - 8)))/(4*x^3),x)
Output:
x^(exp(x/4 - 4)/x^2)*exp((exp(x/4)*exp(16))/x)
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-4+\frac {e^{20+\frac {x}{4}} x+e^{x/4} \log (x)}{e^4 x^2}} \left (e^{x/4} \left (4+e^{20} \left (-4 x+x^2\right )\right )+e^{x/4} (-8+x) \log (x)\right )}{4 x^3} \, dx=e^{\frac {e^{\frac {x}{4}} \mathrm {log}\left (x \right )+e^{\frac {x}{4}} e^{20} x}{e^{4} x^{2}}} \] Input:
int(1/4*((-8+x)*exp(1/4*x)*log(x)+((x^2-4*x)*exp(20)+4)*exp(1/4*x))*exp((e xp(1/4*x)*log(x)+x*exp(20)*exp(1/4*x))/x^2/exp(4))/x^3/exp(4),x)
Output:
e**((e**(x/4)*log(x) + e**(x/4)*e**20*x)/(e**4*x**2))