Integrand size = 104, antiderivative size = 29 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=-x-\log (x)+\frac {2 x+\log (x)}{4-e^{2+2 x}+x} \] Output:
(2*x+ln(x))/(x+4-exp(1)^2*exp(x)^2)-x-ln(x)
Time = 0.37 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=-x-\frac {2 x}{-4+e^{2+2 x}-x}-\log (x)-\frac {\log (x)}{-4+e^{2+2 x}-x} \] Input:
Integrate[(-12 + E^(4 + 4*x)*(-1 - x) - 15*x - 9*x^2 - x^3 + E^(2 + 2*x)*( 7 + 8*x + 6*x^2) + (-x + 2*E^(2 + 2*x)*x)*Log[x])/(16*x + E^(4 + 4*x)*x + 8*x^2 + x^3 + E^(2 + 2*x)*(-8*x - 2*x^2)),x]
Output:
-x - (2*x)/(-4 + E^(2 + 2*x) - x) - Log[x] - Log[x]/(-4 + E^(2 + 2*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.
\(\displaystyle \int \frac {-x^3-9 x^2+e^{2 x+2} \left (6 x^2+8 x+7\right )-15 x+e^{4 x+4} (-x-1)+\left (2 e^{2 x+2} x-x\right ) \log (x)-12}{x^3+8 x^2+e^{2 x+2} \left (-2 x^2-8 x\right )+e^{4 x+4} x+16 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^3-9 x^2+e^{2 x+2} \left (6 x^2+8 x+7\right )-15 x+e^{4 x+4} (-x-1)+\left (2 e^{2 x+2} x-x\right ) \log (x)-12}{x \left (x-e^{2 x+2}+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {4 x^2-2 x+2 x \log (x)-1}{x \left (x-e^{2 x+2}+4\right )}+\frac {-x-1}{x}+\frac {(2 x+7) (2 x+\log (x))}{\left (-x+e^{2 x+2}-4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {x^2}{\left (-x+e^{2 x+2}-4\right )^2}dx-2 \int \frac {1}{-x+e^{2 x+2}-4}dx+14 \int \frac {x}{\left (-x+e^{2 x+2}-4\right )^2}dx+\int \frac {1}{x \left (x-e^{2 x+2}+4\right )}dx-4 \int \frac {x}{x-e^{2 x+2}+4}dx-2 \int \frac {\int \frac {1}{-x+e^{2 x+2}-4}dx}{x}dx-7 \int \frac {\int \frac {1}{\left (x-e^{2 x+2}+4\right )^2}dx}{x}dx-2 \int \frac {\int \frac {x}{\left (x-e^{2 x+2}+4\right )^2}dx}{x}dx+7 \log (x) \int \frac {1}{\left (-x+e^{2 x+2}-4\right )^2}dx+2 \log (x) \int \frac {1}{-x+e^{2 x+2}-4}dx+2 \log (x) \int \frac {x}{\left (-x+e^{2 x+2}-4\right )^2}dx-x-\log (x)\) |
Input:
Int[(-12 + E^(4 + 4*x)*(-1 - x) - 15*x - 9*x^2 - x^3 + E^(2 + 2*x)*(7 + 8* x + 6*x^2) + (-x + 2*E^(2 + 2*x)*x)*Log[x])/(16*x + E^(4 + 4*x)*x + 8*x^2 + x^3 + E^(2 + 2*x)*(-8*x - 2*x^2)),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(63\) vs. \(2(30)=60\).
Time = 1.39 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21
method | result | size |
parallelrisch | \(\frac {-x \,{\mathrm e}^{2} {\mathrm e}^{2 x}-{\mathrm e}^{2} \ln \left (x \right ) {\mathrm e}^{2 x}-8+2 \,{\mathrm e}^{2} {\mathrm e}^{2 x}+x^{2}+x \ln \left (x \right )+3 \ln \left (x \right )}{{\mathrm e}^{2} {\mathrm e}^{2 x}-x -4}\) | \(64\) |
risch | \(-\frac {\ln \left (x \right )}{{\mathrm e}^{2+2 x}-x -4}-\frac {\ln \left (x \right ) {\mathrm e}^{2+2 x}+x \,{\mathrm e}^{2+2 x}-x \ln \left (x \right )-x^{2}-4 \ln \left (x \right )-2 x}{{\mathrm e}^{2+2 x}-x -4}\) | \(69\) |
Input:
int(((2*x*exp(1)^2*exp(x)^2-x)*ln(x)+(-1-x)*exp(1)^4*exp(x)^4+(6*x^2+8*x+7 )*exp(1)^2*exp(x)^2-x^3-9*x^2-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2-8*x)*e xp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x,method=_RETURNVERBOSE)
Output:
(-x*exp(1)^2*exp(x)^2-exp(1)^2*ln(x)*exp(x)^2-8+2*exp(1)^2*exp(x)^2+x^2+x* ln(x)+3*ln(x))/(exp(1)^2*exp(x)^2-x-4)
Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=-\frac {x^{2} - x e^{\left (2 \, x + 2\right )} + {\left (x - e^{\left (2 \, x + 2\right )} + 3\right )} \log \left (x\right ) + 2 \, x}{x - e^{\left (2 \, x + 2\right )} + 4} \] Input:
integrate(((2*x*exp(1)^2*exp(x)^2-x)*log(x)+(-1-x)*exp(1)^4*exp(x)^4+(6*x^ 2+8*x+7)*exp(1)^2*exp(x)^2-x^3-9*x^2-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2 -8*x)*exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x, algorithm="fricas")
Output:
-(x^2 - x*e^(2*x + 2) + (x - e^(2*x + 2) + 3)*log(x) + 2*x)/(x - e^(2*x + 2) + 4)
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=- x + \frac {- 2 x - \log {\left (x \right )}}{- x + e^{2} e^{2 x} - 4} - \log {\left (x \right )} \] Input:
integrate(((2*x*exp(1)**2*exp(x)**2-x)*ln(x)+(-1-x)*exp(1)**4*exp(x)**4+(6 *x**2+8*x+7)*exp(1)**2*exp(x)**2-x**3-9*x**2-15*x-12)/(x*exp(1)**4*exp(x)* *4+(-2*x**2-8*x)*exp(1)**2*exp(x)**2+x**3+8*x**2+16*x),x)
Output:
-x + (-2*x - log(x))/(-x + exp(2)*exp(2*x) - 4) - log(x)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=-\frac {x^{2} - x e^{\left (2 \, x + 2\right )} + 2 \, x - \log \left (x\right )}{x - e^{\left (2 \, x + 2\right )} + 4} - \log \left (x\right ) \] Input:
integrate(((2*x*exp(1)^2*exp(x)^2-x)*log(x)+(-1-x)*exp(1)^4*exp(x)^4+(6*x^ 2+8*x+7)*exp(1)^2*exp(x)^2-x^3-9*x^2-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2 -8*x)*exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x, algorithm="maxima")
Output:
-(x^2 - x*e^(2*x + 2) + 2*x - log(x))/(x - e^(2*x + 2) + 4) - log(x)
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=-\frac {x^{2} - x e^{\left (2 \, x + 2\right )} + x \log \left (x\right ) - e^{\left (2 \, x + 2\right )} \log \left (x\right ) + 2 \, x + 3 \, \log \left (x\right )}{x - e^{\left (2 \, x + 2\right )} + 4} \] Input:
integrate(((2*x*exp(1)^2*exp(x)^2-x)*log(x)+(-1-x)*exp(1)^4*exp(x)^4+(6*x^ 2+8*x+7)*exp(1)^2*exp(x)^2-x^3-9*x^2-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2 -8*x)*exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x, algorithm="giac")
Output:
-(x^2 - x*e^(2*x + 2) + x*log(x) - e^(2*x + 2)*log(x) + 2*x + 3*log(x))/(x - e^(2*x + 2) + 4)
Time = 2.84 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.10 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=\frac {\ln \left (x\right )}{x-{\mathrm {e}}^{2\,x+2}+4}-\ln \left (x\right )-x-\frac {2\,\left (2\,x^2+7\,x\right )}{\left ({\mathrm {e}}^{2\,x}-{\mathrm {e}}^{-2}\,\left (x+4\right )\right )\,\left (7\,{\mathrm {e}}^2+2\,x\,{\mathrm {e}}^2\right )} \] Input:
int(-(15*x + log(x)*(x - 2*x*exp(2*x)*exp(2)) + 9*x^2 + x^3 - exp(2*x)*exp (2)*(8*x + 6*x^2 + 7) + exp(4*x)*exp(4)*(x + 1) + 12)/(16*x + 8*x^2 + x^3 - exp(2*x)*exp(2)*(8*x + 2*x^2) + x*exp(4*x)*exp(4)),x)
Output:
log(x)/(x - exp(2*x + 2) + 4) - log(x) - x - (2*(7*x + 2*x^2))/((exp(2*x) - exp(-2)*(x + 4))*(7*exp(2) + 2*x*exp(2)))
Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx=\frac {-e^{2 x} \mathrm {log}\left (x \right ) e^{2}-e^{2 x} e^{2} x +2 e^{2 x} e^{2}+\mathrm {log}\left (x \right ) x +3 \,\mathrm {log}\left (x \right )+x^{2}-8}{e^{2 x} e^{2}-x -4} \] Input:
int(((2*x*exp(1)^2*exp(x)^2-x)*log(x)+(-1-x)*exp(1)^4*exp(x)^4+(6*x^2+8*x+ 7)*exp(1)^2*exp(x)^2-x^3-9*x^2-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2-8*x)* exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x)
Output:
( - e**(2*x)*log(x)*e**2 - e**(2*x)*e**2*x + 2*e**(2*x)*e**2 + log(x)*x + 3*log(x) + x**2 - 8)/(e**(2*x)*e**2 - x - 4)