Integrand size = 74, antiderivative size = 24 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx=e^5 \log \left (\log \left (\frac {5+2 e^{-x/3}-x}{x}\right )\right ) \] Output:
ln(ln((5-x+2/exp(1/3*x))/x))*exp(5)
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx=e^5 \log \left (\log \left (\frac {5+2 e^{-x/3}-x}{x}\right )\right ) \] Input:
Integrate[(15*E^(5 + x/3) + E^5*(6 + 2*x))/((-6*x + E^(x/3)*(-15*x + 3*x^2 ))*Log[(2 + E^(x/3)*(5 - x))/(E^(x/3)*x)]),x]
Output:
E^5*Log[Log[(5 + 2/E^(x/3) - 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.
\(\displaystyle \int \frac {e^5 (2 x+6)+15 e^{\frac {x}{3}+5}}{\left (e^{x/3} \left (3 x^2-15 x\right )-6 x\right ) \log \left (\frac {e^{-x/3} \left (e^{x/3} (5-x)+2\right )}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^5 \left (-2 x-15 e^{x/3}-6\right )}{3 x \left (-e^{x/3} x+5 e^{x/3}+2\right ) \log \left (\frac {e^{-x/3} \left (e^{x/3} (5-x)+2\right )}{x}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} e^5 \int -\frac {2 x+15 e^{x/3}+6}{x \left (-e^{x/3} x+5 e^{x/3}+2\right ) \log \left (\frac {e^{-x/3} \left (e^{x/3} (5-x)+2\right )}{x}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} e^5 \int \frac {2 x+15 e^{x/3}+6}{x \left (-e^{x/3} x+5 e^{x/3}+2\right ) \log \left (\frac {e^{-x/3} \left (e^{x/3} (5-x)+2\right )}{x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{3} e^5 \int \left (\frac {2 (2-x)}{(5-x) \left (-e^{x/3} x+5 e^{x/3}+2\right ) \log \left (-1+\frac {2 e^{-x/3}}{x}+\frac {5}{x}\right )}+\frac {15}{(5-x) x \log \left (-1+\frac {2 e^{-x/3}}{x}+\frac {5}{x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{3} e^5 \left (3 \int \frac {1}{(5-x) \log \left (-1+\frac {2 e^{-x/3}}{x}+\frac {5}{x}\right )}dx+3 \int \frac {1}{x \log \left (-1+\frac {2 e^{-x/3}}{x}+\frac {5}{x}\right )}dx+2 \int \frac {1}{\left (-e^{x/3} x+5 e^{x/3}+2\right ) \log \left (-1+\frac {2 e^{-x/3}}{x}+\frac {5}{x}\right )}dx+6 \int \frac {1}{(x-5) \left (-e^{x/3} x+5 e^{x/3}+2\right ) \log \left (-1+\frac {2 e^{-x/3}}{x}+\frac {5}{x}\right )}dx\right )\) |
Input:
Int[(15*E^(5 + x/3) + E^5*(6 + 2*x))/((-6*x + E^(x/3)*(-15*x + 3*x^2))*Log [(2 + E^(x/3)*(5 - x))/(E^(x/3)*x)]),x]
Output:
$Aborted
Time = 0.98 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
norman | \({\mathrm e}^{5} \ln \left (\ln \left (\frac {\left (\left (5-x \right ) {\mathrm e}^{\frac {x}{3}}+2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )\right )\) | \(28\) |
parallelrisch | \({\mathrm e}^{5} \ln \left (\ln \left (\frac {\left (\left (5-x \right ) {\mathrm e}^{\frac {x}{3}}+2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )\right )\) | \(28\) |
risch | \({\mathrm e}^{5} \ln \left (\ln \left ({\mathrm e}^{\frac {x}{3}}\right )+\frac {i \left (\pi \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )-\pi \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-\pi {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )}^{3}+2 \pi {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )-2 i \ln \left (x \right )+2 i \ln \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )-2 \pi \right )}{2}\right )\) | \(402\) |
Input:
int((15*exp(5)*exp(1/3*x)+(2*x+6)*exp(5))/((3*x^2-15*x)*exp(1/3*x)-6*x)/ln (((5-x)*exp(1/3*x)+2)/x/exp(1/3*x)),x,method=_RETURNVERBOSE)
Output:
exp(5)*ln(ln(((5-x)*exp(1/3*x)+2)/x/exp(1/3*x)))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx=e^{5} \log \left (\log \left (-\frac {{\left ({\left (x - 5\right )} e^{\left (\frac {1}{3} \, x + 5\right )} - 2 \, e^{5}\right )} e^{\left (-\frac {1}{3} \, x - 5\right )}}{x}\right )\right ) \] Input:
integrate((15*exp(5)*exp(1/3*x)+(2*x+6)*exp(5))/((3*x^2-15*x)*exp(1/3*x)-6 *x)/log(((5-x)*exp(1/3*x)+2)/x/exp(1/3*x)),x, algorithm="fricas")
Output:
e^5*log(log(-((x - 5)*e^(1/3*x + 5) - 2*e^5)*e^(-1/3*x - 5)/x))
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx=e^{5} \log {\left (\log {\left (\frac {\left (\left (5 - x\right ) e^{\frac {x}{3}} + 2\right ) e^{- \frac {x}{3}}}{x} \right )} \right )} \] Input:
integrate((15*exp(5)*exp(1/3*x)+(2*x+6)*exp(5))/((3*x**2-15*x)*exp(1/3*x)- 6*x)/ln(((5-x)*exp(1/3*x)+2)/x/exp(1/3*x)),x)
Output:
exp(5)*log(log(((5 - x)*exp(x/3) + 2)*exp(-x/3)/x))
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx=e^{5} \log \left (-\frac {1}{3} \, x + \log \left (-{\left (x - 5\right )} e^{\left (\frac {1}{3} \, x\right )} + 2\right ) - \log \left (x\right )\right ) \] Input:
integrate((15*exp(5)*exp(1/3*x)+(2*x+6)*exp(5))/((3*x^2-15*x)*exp(1/3*x)-6 *x)/log(((5-x)*exp(1/3*x)+2)/x/exp(1/3*x)),x, algorithm="maxima")
Output:
e^5*log(-1/3*x + log(-(x - 5)*e^(1/3*x) + 2) - log(x))
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx=e^{5} \log \left (x - 3 \, \log \left (-x e^{\left (\frac {1}{3} \, x\right )} + 5 \, e^{\left (\frac {1}{3} \, x\right )} + 2\right ) + 3 \, \log \left (x\right )\right ) \] Input:
integrate((15*exp(5)*exp(1/3*x)+(2*x+6)*exp(5))/((3*x^2-15*x)*exp(1/3*x)-6 *x)/log(((5-x)*exp(1/3*x)+2)/x/exp(1/3*x)),x, algorithm="giac")
Output:
e^5*log(x - 3*log(-x*e^(1/3*x) + 5*e^(1/3*x) + 2) + 3*log(x))
Time = 2.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx={\mathrm {e}}^5\,\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{-\frac {x}{3}}-x+5}{x}\right )\right ) \] Input:
int(-(15*exp(x/3)*exp(5) + exp(5)*(2*x + 6))/(log(-(exp(-x/3)*(exp(x/3)*(x - 5) - 2))/x)*(6*x + exp(x/3)*(15*x - 3*x^2))),x)
Output:
exp(5)*log(log((2*exp(-x/3) - x + 5)/x))
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {15 e^{5+\frac {x}{3}}+e^5 (6+2 x)}{\left (-6 x+e^{x/3} \left (-15 x+3 x^2\right )\right ) \log \left (\frac {e^{-x/3} \left (2+e^{x/3} (5-x)\right )}{x}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {-e^{\frac {x}{3}} x +5 e^{\frac {x}{3}}+2}{e^{\frac {x}{3}} x}\right )\right ) e^{5} \] Input:
int((15*exp(5)*exp(1/3*x)+(2*x+6)*exp(5))/((3*x^2-15*x)*exp(1/3*x)-6*x)/lo g(((5-x)*exp(1/3*x)+2)/x/exp(1/3*x)),x)
Output:
log(log(( - e**(x/3)*x + 5*e**(x/3) + 2)/(e**(x/3)*x)))*e**5