Integrand size = 59, antiderivative size = 27 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\log \left (3+e^2+x+\left (4-e^x\right ) x-\frac {4 x}{4-\log (4)}\right ) \] Output:
ln(exp(2)-4*x/(4-2*ln(2))+3+x+x*(-exp(x)+4))
Time = 5.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\log \left (12+4 e^2+16 x-4 e^x x-3 \log (4)-e^2 \log (4)-5 x \log (4)+e^x x \log (4)\right ) \] Input:
Integrate[(16 - 5*Log[4] + E^x*(-4 - 4*x + (1 + x)*Log[4]))/(12 + 4*E^2 + 16*x + (-3 - E^2 - 5*x)*Log[4] + E^x*(-4*x + x*Log[4])),x]
Output:
Log[12 + 4*E^2 + 16*x - 4*E^x*x - 3*Log[4] - E^2*Log[4] - 5*x*Log[4] + E^x *x*Log[4]]
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.
\(\displaystyle \int \frac {e^x (-4 x+(x+1) \log (4)-4)+16-5 \log (4)}{16 x+\left (-5 x-e^2-3\right ) \log (4)+e^x (x \log (4)-4 x)+4 e^2+12} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x (-4 x+(x+1) \log (4)-4)+16 \left (1-\frac {5 \log (2)}{8}\right )}{16 x+\left (-5 x-e^2-3\right ) \log (4)+e^x (x \log (4)-4 x)+12 \left (1+\frac {e^2}{3}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-\left (x^2 (16-5 \log (4))\right )-x \left (12+e^2 (4-\log (4))-\log (64)\right )-12+\log (64)-e^2 (4-\log (4))}{x \left (-4 e^x x \left (1-\frac {\log (2)}{2}\right )+16 x \left (1-\frac {5 \log (2)}{8}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (\log (4)-4)-3 \log (4)\right )\right )\right )}+\frac {x+1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\left (12+e^2 (4-\log (4))-\log (64)\right ) \int \frac {1}{x \left (-4 e^x \left (1-\frac {\log (2)}{2}\right ) x+16 \left (1-\frac {5 \log (2)}{8}\right ) x+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )}dx-(16-5 \log (4)) \int \frac {x}{-4 e^x \left (1-\frac {\log (2)}{2}\right ) x+16 \left (1-\frac {5 \log (2)}{8}\right ) x+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )}dx+\left (12+e^2 (4-\log (4))-\log (64)\right ) \int \frac {1}{4 e^x \left (1-\frac {\log (2)}{2}\right ) x-16 \left (1-\frac {5 \log (2)}{8}\right ) x-12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )}dx+x+\log (x)\) |
Input:
Int[(16 - 5*Log[4] + E^x*(-4 - 4*x + (1 + x)*Log[4]))/(12 + 4*E^2 + 16*x + (-3 - E^2 - 5*x)*Log[4] + E^x*(-4*x + x*Log[4])),x]
Output:
$Aborted
Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37
method | result | size |
norman | \(\ln \left (-x \ln \left (2\right ) {\mathrm e}^{x}+{\mathrm e}^{2} \ln \left (2\right )+5 x \ln \left (2\right )+2 \,{\mathrm e}^{x} x +3 \ln \left (2\right )-2 \,{\mathrm e}^{2}-8 x -6\right )\) | \(37\) |
risch | \(\ln \left (x \right )+\ln \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2} \ln \left (2\right )+5 x \ln \left (2\right )+3 \ln \left (2\right )-2 \,{\mathrm e}^{2}-8 x -6}{x \left (\ln \left (2\right )-2\right )}\right )\) | \(42\) |
parallelrisch | \(\ln \left (-\frac {-x \ln \left (2\right ) {\mathrm e}^{x}+{\mathrm e}^{2} \ln \left (2\right )+5 x \ln \left (2\right )+2 \,{\mathrm e}^{x} x +3 \ln \left (2\right )-2 \,{\mathrm e}^{2}-8 x -6}{\ln \left (2\right )-2}\right )\) | \(45\) |
Input:
int(((2*ln(2)*(1+x)-4*x-4)*exp(x)-10*ln(2)+16)/((2*x*ln(2)-4*x)*exp(x)+2*( -exp(2)-5*x-3)*ln(2)+4*exp(2)+16*x+12),x,method=_RETURNVERBOSE)
Output:
ln(-x*ln(2)*exp(x)+exp(2)*ln(2)+5*x*ln(2)+2*exp(x)*x+3*ln(2)-2*exp(2)-8*x- 6)
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\log \left (x\right ) + \log \left (\frac {{\left (x \log \left (2\right ) - 2 \, x\right )} e^{x} - {\left (5 \, x + e^{2} + 3\right )} \log \left (2\right ) + 8 \, x + 2 \, e^{2} + 6}{x}\right ) \] Input:
integrate(((2*log(2)*(1+x)-4*x-4)*exp(x)-10*log(2)+16)/((2*x*log(2)-4*x)*e xp(x)+2*(-exp(2)-5*x-3)*log(2)+4*exp(2)+16*x+12),x, algorithm="fricas")
Output:
log(x) + log(((x*log(2) - 2*x)*e^x - (5*x + e^2 + 3)*log(2) + 8*x + 2*e^2 + 6)/x)
Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\log {\left (x \right )} + \log {\left (e^{x} + \frac {- 5 x \log {\left (2 \right )} + 8 x - e^{2} \log {\left (2 \right )} - 3 \log {\left (2 \right )} + 6 + 2 e^{2}}{- 2 x + x \log {\left (2 \right )}} \right )} \] Input:
integrate(((2*ln(2)*(1+x)-4*x-4)*exp(x)-10*ln(2)+16)/((2*x*ln(2)-4*x)*exp( x)+2*(-exp(2)-5*x-3)*ln(2)+4*exp(2)+16*x+12),x)
Output:
log(x) + log(exp(x) + (-5*x*log(2) + 8*x - exp(2)*log(2) - 3*log(2) + 6 + 2*exp(2))/(-2*x + x*log(2)))
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\log \left (x\right ) + \log \left (\frac {x {\left (\log \left (2\right ) - 2\right )} e^{x} - x {\left (5 \, \log \left (2\right ) - 8\right )} - {\left (\log \left (2\right ) - 2\right )} e^{2} - 3 \, \log \left (2\right ) + 6}{x {\left (\log \left (2\right ) - 2\right )}}\right ) \] Input:
integrate(((2*log(2)*(1+x)-4*x-4)*exp(x)-10*log(2)+16)/((2*x*log(2)-4*x)*e xp(x)+2*(-exp(2)-5*x-3)*log(2)+4*exp(2)+16*x+12),x, algorithm="maxima")
Output:
log(x) + log((x*(log(2) - 2)*e^x - x*(5*log(2) - 8) - (log(2) - 2)*e^2 - 3 *log(2) + 6)/(x*(log(2) - 2)))
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\log \left (x e^{x} \log \left (2\right ) - 2 \, x e^{x} - 5 \, x \log \left (2\right ) - e^{2} \log \left (2\right ) + 8 \, x + 2 \, e^{2} - 3 \, \log \left (2\right ) + 6\right ) \] Input:
integrate(((2*log(2)*(1+x)-4*x-4)*exp(x)-10*log(2)+16)/((2*x*log(2)-4*x)*e xp(x)+2*(-exp(2)-5*x-3)*log(2)+4*exp(2)+16*x+12),x, algorithm="giac")
Output:
log(x*e^x*log(2) - 2*x*e^x - 5*x*log(2) - e^2*log(2) + 8*x + 2*e^2 - 3*log (2) + 6)
Time = 7.96 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\ln \left (16\,x+4\,{\mathrm {e}}^2-2\,\ln \left (2\right )\,\left (5\,x+{\mathrm {e}}^2+3\right )-{\mathrm {e}}^x\,\left (4\,x-2\,x\,\ln \left (2\right )\right )+12\right ) \] Input:
int(-(10*log(2) + exp(x)*(4*x - 2*log(2)*(x + 1) + 4) - 16)/(16*x + 4*exp( 2) - 2*log(2)*(5*x + exp(2) + 3) - exp(x)*(4*x - 2*x*log(2)) + 12),x)
Output:
log(16*x + 4*exp(2) - 2*log(2)*(5*x + exp(2) + 3) - exp(x)*(4*x - 2*x*log( 2)) + 12)
Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx=\mathrm {log}\left (e^{x} \mathrm {log}\left (2\right ) x -2 e^{x} x -\mathrm {log}\left (2\right ) e^{2}-5 \,\mathrm {log}\left (2\right ) x -3 \,\mathrm {log}\left (2\right )+2 e^{2}+8 x +6\right ) \] Input:
int(((2*log(2)*(1+x)-4*x-4)*exp(x)-10*log(2)+16)/((2*x*log(2)-4*x)*exp(x)+ 2*(-exp(2)-5*x-3)*log(2)+4*exp(2)+16*x+12),x)
Output:
log(e**x*log(2)*x - 2*e**x*x - log(2)*e**2 - 5*log(2)*x - 3*log(2) + 2*e** 2 + 8*x + 6)