Integrand size = 95, antiderivative size = 23 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=-5+\frac {4 e^x \log \left (3+e^x\right )}{\log \left (\frac {1}{3} (-5+x)\right )} \] Output:
4*ln(3+exp(x))/ln(1/3*x-5/3)*exp(x)-5
Time = 5.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\frac {4 e^x \log \left (3+e^x\right )}{\log \left (\frac {1}{3} (-5+x)\right )} \] Input:
Integrate[(E^(2*x)*(-20 + 4*x)*Log[(-5 + x)/3] + Log[3 + E^x]*(-12*E^x - 4 *E^(2*x) + (E^(2*x)*(-20 + 4*x) + E^x*(-60 + 12*x))*Log[(-5 + x)/3]))/((-1 5 + E^x*(-5 + x) + 3*x)*Log[(-5 + x)/3]^2),x]
Output:
(4*E^x*Log[3 + E^x])/Log[(-5 + x)/3]
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.
\(\displaystyle \int \frac {e^{2 x} (4 x-20) \log \left (\frac {x-5}{3}\right )+\log \left (e^x+3\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (4 x-20)+e^x (12 x-60)\right ) \log \left (\frac {x-5}{3}\right )\right )}{\left (e^x (x-5)+3 x-15\right ) \log ^2\left (\frac {x-5}{3}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^{2 x} (4 x-20) \log \left (\frac {x-5}{3}\right )-\log \left (e^x+3\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (4 x-20)+e^x (12 x-60)\right ) \log \left (\frac {x-5}{3}\right )\right )}{\left (e^x+3\right ) (5-x) \log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 e^x \left (\frac {e^x \log \left (\frac {x-5}{3}\right )}{e^x+3}+\frac {\log \left (e^x+3\right ) \left ((x-5) \log \left (\frac {x-5}{3}\right )-1\right )}{x-5}\right )}{\log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {e^x \left (\frac {e^x \log \left (\frac {x-5}{3}\right )}{3+e^x}+\frac {\log \left (3+e^x\right ) \left ((5-x) \log \left (\frac {x-5}{3}\right )+1\right )}{5-x}\right )}{\log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 4 \int \left (\frac {e^x \left (-x \log \left (\frac {x-5}{3}\right ) \log \left (3+e^x\right )+5 \log \left (\frac {x-5}{3}\right ) \log \left (3+e^x\right )+\log \left (3+e^x\right )-x \log \left (\frac {x-5}{3}\right )+5 \log \left (\frac {x-5}{3}\right )\right )}{(5-x) \log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}+\frac {e^x (\log (27)-3 \log (x-5))}{\left (3+e^x\right ) \log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (\log (27) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}dx+\int \frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}dx+3 \int \frac {e^x \log (x-5)}{\left (-3-e^x\right ) \log ^2\left (\frac {x}{3}-\frac {5}{3}\right )}dx+\int \frac {e^x}{\log \left (\frac {x}{3}-\frac {5}{3}\right )}dx+\int \frac {e^x \log \left (3+e^x\right )}{\log \left (\frac {x}{3}-\frac {5}{3}\right )}dx\right )\) |
Input:
Int[(E^(2*x)*(-20 + 4*x)*Log[(-5 + x)/3] + Log[3 + E^x]*(-12*E^x - 4*E^(2* x) + (E^(2*x)*(-20 + 4*x) + E^x*(-60 + 12*x))*Log[(-5 + x)/3]))/((-15 + E^ x*(-5 + x) + 3*x)*Log[(-5 + x)/3]^2),x]
Output:
$Aborted
Time = 2.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {4 \ln \left (3+{\mathrm e}^{x}\right ) {\mathrm e}^{x}}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )}\) | \(18\) |
parallelrisch | \(\frac {4 \ln \left (3+{\mathrm e}^{x}\right ) {\mathrm e}^{x}}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )}\) | \(18\) |
Input:
int(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*ln(1/3*x-5/3)-4*exp(x)^2-12*exp (x))*ln(3+exp(x))+(4*x-20)*exp(x)^2*ln(1/3*x-5/3))/((-5+x)*exp(x)+3*x-15)/ ln(1/3*x-5/3)^2,x,method=_RETURNVERBOSE)
Output:
4*ln(3+exp(x))/ln(1/3*x-5/3)*exp(x)
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \left (\frac {1}{3} \, x - \frac {5}{3}\right )} \] Input:
integrate(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*log(1/3*x-5/3)-4*exp(x)^2 -12*exp(x))*log(3+exp(x))+(4*x-20)*exp(x)^2*log(1/3*x-5/3))/((-5+x)*exp(x) +3*x-15)/log(1/3*x-5/3)^2,x, algorithm="fricas")
Output:
4*e^x*log(e^x + 3)/log(1/3*x - 5/3)
Timed out. \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\text {Timed out} \] Input:
integrate(((((4*x-20)*exp(x)**2+(12*x-60)*exp(x))*ln(1/3*x-5/3)-4*exp(x)** 2-12*exp(x))*ln(3+exp(x))+(4*x-20)*exp(x)**2*ln(1/3*x-5/3))/((-5+x)*exp(x) +3*x-15)/ln(1/3*x-5/3)**2,x)
Output:
Timed out
Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=-\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \left (3\right ) - \log \left (x - 5\right )} \] Input:
integrate(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*log(1/3*x-5/3)-4*exp(x)^2 -12*exp(x))*log(3+exp(x))+(4*x-20)*exp(x)^2*log(1/3*x-5/3))/((-5+x)*exp(x) +3*x-15)/log(1/3*x-5/3)^2,x, algorithm="maxima")
Output:
-4*e^x*log(e^x + 3)/(log(3) - log(x - 5))
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=-\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \left (3\right ) - \log \left (x - 5\right )} \] Input:
integrate(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*log(1/3*x-5/3)-4*exp(x)^2 -12*exp(x))*log(3+exp(x))+(4*x-20)*exp(x)^2*log(1/3*x-5/3))/((-5+x)*exp(x) +3*x-15)/log(1/3*x-5/3)^2,x, algorithm="giac")
Output:
-4*e^x*log(e^x + 3)/(log(3) - log(x - 5))
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\frac {4\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+3\right )}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )} \] Input:
int(-(log(exp(x) + 3)*(4*exp(2*x) + 12*exp(x) - log(x/3 - 5/3)*(exp(x)*(12 *x - 60) + exp(2*x)*(4*x - 20))) - exp(2*x)*log(x/3 - 5/3)*(4*x - 20))/(lo g(x/3 - 5/3)^2*(3*x + exp(x)*(x - 5) - 15)),x)
Output:
(4*exp(x)*log(exp(x) + 3))/log(x/3 - 5/3)
Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\frac {4 e^{x} \mathrm {log}\left (e^{x}+3\right )}{\mathrm {log}\left (-\frac {5}{3}+\frac {x}{3}\right )} \] Input:
int(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*log(1/3*x-5/3)-4*exp(x)^2-12*ex p(x))*log(3+exp(x))+(4*x-20)*exp(x)^2*log(1/3*x-5/3))/((-5+x)*exp(x)+3*x-1 5)/log(1/3*x-5/3)^2,x)
Output:
(4*e**x*log(e**x + 3))/log((x - 5)/3)