Integrand size = 77, antiderivative size = 26 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=e^3+\log (x)+\log \left (3+e+e^x+x-\frac {x}{5-e^2}\right ) \] Output:
exp(3)+ln(x)+ln(x+3+exp(1)+exp(x)-x/(5-exp(2)))
Time = 1.59 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log (x)+\log \left (15+5 e-3 e^2-e^3+5 e^x-e^{2+x}+4 x-e^2 x\right ) \] Input:
Integrate[(-15 - 5*E - 8*x + E^2*(3 + E + 2*x) + E^x*(-5 - 5*x + E^2*(1 + x)))/(-15*x - 5*E*x - 4*x^2 + E^x*(-5*x + E^2*x) + E^2*(3*x + E*x + x^2)), x]
Output:
Log[x] + Log[15 + 5*E - 3*E^2 - E^3 + 5*E^x - E^(2 + x) + 4*x - E^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.
\(\displaystyle \int \frac {-8 x+e^2 (2 x+e+3)+e^x \left (-5 x+e^2 (x+1)-5\right )-5 e-15}{-4 x^2+e^2 \left (x^2+e x+3 x\right )-5 e x-15 x+e^x \left (e^2 x-5 x\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-8 x+e^2 (2 x+e+3)+e^x \left (-5 x+e^2 (x+1)-5\right )-5 e-15}{-4 x^2+e^2 \left (x^2+e x+3 x\right )+(-15-5 e) x+e^x \left (e^2 x-5 x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {8 x-e^2 (2 x+e+3)-e^x \left (-5 x+e^2 (x+1)-5\right )+15 \left (1+\frac {e}{3}\right )}{4 x^2-e^2 \left (x^2+e x+3 x\right )-(-15-5 e) x-e^x \left (e^2 x-5 x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x+1}{x}+\frac {-\left (\left (4-e^2\right ) x\right )+e^3+2 e^2-5 e-11}{4 \left (1-\frac {e^2}{4}\right ) x+5 \left (1-\frac {e^2}{5}\right ) e^x+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (11+e \left (5-2 e-e^2\right )\right ) \int \frac {1}{-4 \left (1-\frac {e^2}{4}\right ) x-15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )-5 e^x \left (1-\frac {e^2}{5}\right )}dx-(2-e) (2+e) \int \frac {x}{4 \left (1-\frac {e^2}{4}\right ) x+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+5 e^x \left (1-\frac {e^2}{5}\right )}dx+x+\log (x)\) |
Input:
Int[(-15 - 5*E - 8*x + E^2*(3 + E + 2*x) + E^x*(-5 - 5*x + E^2*(1 + x)))/( -15*x - 5*E*x - 4*x^2 + E^x*(-5*x + E^2*x) + E^2*(3*x + E*x + x^2)),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\ln \left (x \right )+\ln \left ({\mathrm e}^{x}+\frac {{\mathrm e}^{3}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-4 x -15}{{\mathrm e}^{2}-5}\right )\) | \(34\) |
norman | \(\ln \left (x \right )+\ln \left ({\mathrm e}^{2} {\mathrm e}+{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-5 \,{\mathrm e}^{x}-4 x -15\right )\) | \(36\) |
parallelrisch | \(\ln \left (x \right )+\ln \left (\frac {{\mathrm e}^{2} {\mathrm e}+{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-5 \,{\mathrm e}^{x}-4 x -15}{{\mathrm e}^{2}-4}\right )\) | \(43\) |
Input:
int((((1+x)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x-15)/(( exp(2)*x-5*x)*exp(x)+(x*exp(1)+x^2+3*x)*exp(2)-5*x*exp(1)-4*x^2-15*x),x,me thod=_RETURNVERBOSE)
Output:
ln(x)+ln(exp(x)+(exp(3)+exp(2)*x+3*exp(2)-5*exp(1)-4*x-15)/(exp(2)-5))
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log \left ({\left (x + 3\right )} e^{2} + {\left (e^{2} - 5\right )} e^{x} - 4 \, x + e^{3} - 5 \, e - 15\right ) + \log \left (x\right ) \] Input:
integrate((((1+x)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x- 15)/((exp(2)*x-5*x)*exp(x)+(exp(1)*x+x^2+3*x)*exp(2)-5*exp(1)*x-4*x^2-15*x ),x, algorithm="fricas")
Output:
log((x + 3)*e^2 + (e^2 - 5)*e^x - 4*x + e^3 - 5*e - 15) + log(x)
Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log {\left (x \right )} + \log {\left (\frac {- 4 x + x e^{2} - 15 - 5 e + e^{3} + 3 e^{2}}{-5 + e^{2}} + e^{x} \right )} \] Input:
integrate((((1+x)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x- 15)/((exp(2)*x-5*x)*exp(x)+(exp(1)*x+x**2+3*x)*exp(2)-5*exp(1)*x-4*x**2-15 *x),x)
Output:
log(x) + log((-4*x + x*exp(2) - 15 - 5*E + exp(3) + 3*exp(2))/(-5 + exp(2) ) + exp(x))
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x {\left (e^{2} - 4\right )} + {\left (e^{2} - 5\right )} e^{x} + e^{3} + 3 \, e^{2} - 5 \, e - 15}{e^{2} - 5}\right ) \] Input:
integrate((((1+x)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x- 15)/((exp(2)*x-5*x)*exp(x)+(exp(1)*x+x^2+3*x)*exp(2)-5*exp(1)*x-4*x^2-15*x ),x, algorithm="maxima")
Output:
log(x) + log((x*(e^2 - 4) + (e^2 - 5)*e^x + e^3 + 3*e^2 - 5*e - 15)/(e^2 - 5))
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log \left (-x e^{2} + 4 \, x - e^{3} - 3 \, e^{2} + 5 \, e - e^{\left (x + 2\right )} + 5 \, e^{x} + 15\right ) + \log \left (x\right ) \] Input:
integrate((((1+x)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x- 15)/((exp(2)*x-5*x)*exp(x)+(exp(1)*x+x^2+3*x)*exp(2)-5*exp(1)*x-4*x^2-15*x ),x, algorithm="giac")
Output:
log(-x*e^2 + 4*x - e^3 - 3*e^2 + 5*e - e^(x + 2) + 5*e^x + 15) + log(x)
Time = 1.68 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\ln \left (4\,x-{\mathrm {e}}^{x+2}+5\,\mathrm {e}-3\,{\mathrm {e}}^2-{\mathrm {e}}^3+5\,{\mathrm {e}}^x-x\,{\mathrm {e}}^2+15\right )+\ln \left (x\right ) \] Input:
int((8*x + 5*exp(1) + exp(x)*(5*x - exp(2)*(x + 1) + 5) - exp(2)*(2*x + ex p(1) + 3) + 15)/(15*x - exp(2)*(3*x + x*exp(1) + x^2) + 5*x*exp(1) + exp(x )*(5*x - x*exp(2)) + 4*x^2),x)
Output:
log(4*x - exp(x + 2) + 5*exp(1) - 3*exp(2) - exp(3) + 5*exp(x) - x*exp(2) + 15) + log(x)
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\mathrm {log}\left (e^{x} e^{2}-5 e^{x}+e^{3}+e^{2} x +3 e^{2}-5 e -4 x -15\right )+\mathrm {log}\left (x \right ) \] Input:
int((((1+x)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x-15)/(( exp(2)*x-5*x)*exp(x)+(exp(1)*x+x^2+3*x)*exp(2)-5*exp(1)*x-4*x^2-15*x),x)
Output:
log(e**x*e**2 - 5*e**x + e**3 + e**2*x + 3*e**2 - 5*e - 4*x - 15) + log(x)