Integrand size = 93, antiderivative size = 24 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=5^{\frac {1}{2+x-\frac {x}{\log \left (\frac {x}{3}\right )}}}-e^5 \]
\[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=\int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx \]
Integrate[(E^((Log[5]*Log[x/3])/(-x + (2 + x)*Log[x/3]))*(-Log[5] + Log[5] *Log[x/3] - Log[5]*Log[x/3]^2))/(x^2 + (-4*x - 2*x^2)*Log[x/3] + (4 + 4*x + x^2)*Log[x/3]^2),x]
Integrate[(E^((Log[5]*Log[x/3])/(-x + (2 + x)*Log[x/3]))*(-Log[5] + Log[5] *Log[x/3] - Log[5]*Log[x/3]^2))/(x^2 + (-4*x - 2*x^2)*Log[x/3] + (4 + 4*x + x^2)*Log[x/3]^2), 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.
\(\displaystyle \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{(x+2) \log \left (\frac {x}{3}\right )-x}} \left (-\log (5) \log ^2\left (\frac {x}{3}\right )+\log (5) \log \left (\frac {x}{3}\right )-\log (5)\right )}{x^2+\left (x^2+4 x+4\right ) \log ^2\left (\frac {x}{3}\right )+\left (-2 x^2-4 x\right ) \log \left (\frac {x}{3}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\log (5) e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{(x+2) \log \left (\frac {x}{3}\right )-x}} \left (-\log ^2\left (\frac {x}{3}\right )+\log \left (\frac {x}{3}\right )-1\right )}{\left (x+x \left (-\log \left (\frac {x}{3}\right )\right )-2 \log \left (\frac {x}{3}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (5) \int -\frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}} \left (\log ^2\left (\frac {x}{3}\right )-\log \left (\frac {x}{3}\right )+1\right )}{\left (-\log \left (\frac {x}{3}\right ) x+x-2 \log \left (\frac {x}{3}\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\log (5) \int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}} \left (\log ^2\left (\frac {x}{3}\right )-\log \left (\frac {x}{3}\right )+1\right )}{\left (-\log \left (\frac {x}{3}\right ) x+x-2 \log \left (\frac {x}{3}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\log (5) \int \left (\frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}} (x-2)}{(x+2)^2 \left (\log \left (\frac {x}{3}\right ) x-x+2 \log \left (\frac {x}{3}\right )\right )}+\frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}}}{(x+2)^2}+\frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}} \left (x^2+2 x+4\right )}{(x+2)^2 \left (\log \left (\frac {x}{3}\right ) x-x+2 \log \left (\frac {x}{3}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log (5) \left (\int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}}}{(x+2)^2}dx+\int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}}}{\left (\log \left (\frac {x}{3}\right ) x-x+2 \log \left (\frac {x}{3}\right )\right )^2}dx+4 \int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}}}{(x+2)^2 \left (\log \left (\frac {x}{3}\right ) x-x+2 \log \left (\frac {x}{3}\right )\right )^2}dx-2 \int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}}}{(x+2) \left (\log \left (\frac {x}{3}\right ) x-x+2 \log \left (\frac {x}{3}\right )\right )^2}dx-4 \int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}}}{(x+2)^2 \left (\log \left (\frac {x}{3}\right ) x-x+2 \log \left (\frac {x}{3}\right )\right )}dx+\int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}}}{(x+2) \left (\log \left (\frac {x}{3}\right ) x-x+2 \log \left (\frac {x}{3}\right )\right )}dx\right )\) |
Int[(E^((Log[5]*Log[x/3])/(-x + (2 + x)*Log[x/3]))*(-Log[5] + Log[5]*Log[x /3] - Log[5]*Log[x/3]^2))/(x^2 + (-4*x - 2*x^2)*Log[x/3] + (4 + 4*x + x^2) *Log[x/3]^2),x]
3.9.33.3.1 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 = 6.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{x \ln \left (\frac {x}{3}\right )+2 \ln \left (\frac {x}{3}\right )-x}}\) | \(27\) |
risch | \(\left (\frac {x}{3}\right )^{\frac {\ln \left (5\right )}{x \left (\ln \left (x \right )-\ln \left (3\right )\right )+2 \ln \left (x \right )-2 \ln \left (3\right )-x}}\) | \(31\) |
norman | \(\frac {x \ln \left (\frac {x}{3}\right ) {\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}-x \,{\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}+2 \ln \left (\frac {x}{3}\right ) {\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}}{x \ln \left (\frac {x}{3}\right )+2 \ln \left (\frac {x}{3}\right )-x}\) | \(102\) |
int((-ln(5)*ln(1/3*x)^2+ln(5)*ln(1/3*x)-ln(5))*exp(ln(5)*ln(1/3*x)/((2+x)* ln(1/3*x)-x))/((x^2+4*x+4)*ln(1/3*x)^2+(-2*x^2-4*x)*ln(1/3*x)+x^2),x,metho d=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\left (\frac {\log \left (5\right ) \log \left (\frac {1}{3} \, x\right )}{{\left (x + 2\right )} \log \left (\frac {1}{3} \, x\right ) - x}\right )} \]
integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1 /3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3 *x)+x^2),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\frac {\log {\left (5 \right )} \log {\left (\frac {x}{3} \right )}}{- x + \left (x + 2\right ) \log {\left (\frac {x}{3} \right )}}} \]
integrate((-ln(5)*ln(1/3*x)**2+ln(5)*ln(1/3*x)-ln(5))*exp(ln(5)*ln(1/3*x)/ ((2+x)*ln(1/3*x)-x))/((x**2+4*x+4)*ln(1/3*x)**2+(-2*x**2-4*x)*ln(1/3*x)+x* *2),x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (21) = 42\).
Time = 0.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\left (\frac {\log \left (5\right ) \log \left (3\right )}{x {\left (\log \left (3\right ) + 1\right )} - {\left (x + 2\right )} \log \left (x\right ) + 2 \, \log \left (3\right )} - \frac {\log \left (5\right ) \log \left (x\right )}{x {\left (\log \left (3\right ) + 1\right )} - {\left (x + 2\right )} \log \left (x\right ) + 2 \, \log \left (3\right )}\right )} \]
integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1 /3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3 *x)+x^2),x, algorithm=\
e^(log(5)*log(3)/(x*(log(3) + 1) - (x + 2)*log(x) + 2*log(3)) - log(5)*log (x)/(x*(log(3) + 1) - (x + 2)*log(x) + 2*log(3)))
Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\left (\frac {\log \left (5\right ) \log \left (\frac {1}{3} \, x\right )}{x \log \left (\frac {1}{3} \, x\right ) - x + 2 \, \log \left (\frac {1}{3} \, x\right )}\right )} \]
integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1 /3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3 *x)+x^2),x, algorithm=\
Time = 9.82 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx={\mathrm {e}}^{-\frac {\ln \left (5\right )\,\ln \left (x\right )}{x+2\,\ln \left (3\right )-2\,\ln \left (x\right )+x\,\ln \left (3\right )-x\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {\ln \left (3\right )\,\ln \left (5\right )}{x+2\,\ln \left (3\right )-2\,\ln \left (x\right )+x\,\ln \left (3\right )-x\,\ln \left (x\right )}} \]
int(-(exp(-(log(x/3)*log(5))/(x - log(x/3)*(x + 2)))*(log(5) - log(x/3)*lo g(5) + log(x/3)^2*log(5)))/(log(x/3)^2*(4*x + x^2 + 4) - log(x/3)*(4*x + 2 *x^2) + x^2),x)