Integrand size = 137, antiderivative size = 31 \[ \int \frac {4-2 x+4 x^3+e^x \left (2-2 x^2\right )+e^{3 x} \left (4+e^x (2-8 x)-12 x+8 x^2+6 x^3\right )+\left (-4-2 x^2+e^x (-2+2 x)\right ) \log (x)}{-2 x^2-e^x x^2+x^3+x^4+e^{3 x} \left (-2 x-e^x x+x^2+x^3\right )+\left (2 x+e^x x-x^2-x^3\right ) \log (x)} \, dx=\log \left (\frac {\left (-2-e^x+x+x^2\right )^2 \left (e^{3 x}+x-\log (x)\right )^2}{x^2}\right ) \]
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {4-2 x+4 x^3+e^x \left (2-2 x^2\right )+e^{3 x} \left (4+e^x (2-8 x)-12 x+8 x^2+6 x^3\right )+\left (-4-2 x^2+e^x (-2+2 x)\right ) \log (x)}{-2 x^2-e^x x^2+x^3+x^4+e^{3 x} \left (-2 x-e^x x+x^2+x^3\right )+\left (2 x+e^x x-x^2-x^3\right ) \log (x)} \, dx=2 \left (-\log (x)+\log \left (2+e^x-x-x^2\right )+\log \left (e^{3 x}+x-\log (x)\right )\right ) \]
Integrate[(4 - 2*x + 4*x^3 + E^x*(2 - 2*x^2) + E^(3*x)*(4 + E^x*(2 - 8*x) - 12*x + 8*x^2 + 6*x^3) + (-4 - 2*x^2 + E^x*(-2 + 2*x))*Log[x])/(-2*x^2 - E^x*x^2 + x^3 + x^4 + E^(3*x)*(-2*x - E^x*x + x^2 + x^3) + (2*x + E^x*x - x^2 - x^3)*Log[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.
\(\displaystyle \int \frac {4 x^3+e^x \left (2-2 x^2\right )+\left (-2 x^2+e^x (2 x-2)-4\right ) \log (x)+e^{3 x} \left (6 x^3+8 x^2-12 x+e^x (2-8 x)+4\right )-2 x+4}{x^4+x^3-e^x x^2-2 x^2+e^{3 x} \left (x^3+x^2-e^x x-2 x\right )+\left (-x^3-x^2+e^x x+2 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-4 x^3-e^x \left (2-2 x^2\right )-\left (-2 x^2+e^x (2 x-2)-4\right ) \log (x)-e^{3 x} \left (6 x^3+8 x^2-12 x+e^x (2-8 x)+4\right )+2 x-4}{x \left (-x^2-x+e^x+2\right ) \left (x+e^{3 x}-\log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (x^2-x-3\right )}{-x^2-x+e^x+2}-\frac {2 \left (3 x^2-x-3 x \log (x)+1\right )}{x \left (x+e^{3 x}-\log (x)\right )}+\frac {2 (4 x-1)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -6 \int \frac {1}{-x^2-x+e^x+2}dx+2 \int \frac {x}{x^2+x-e^x-2}dx-2 \int \frac {x^2}{x^2+x-e^x-2}dx+2 \int \frac {1}{x+e^{3 x}-\log (x)}dx-2 \int \frac {1}{x \left (x+e^{3 x}-\log (x)\right )}dx-6 \int \frac {x}{x+e^{3 x}-\log (x)}dx+6 \int \frac {\log (x)}{x+e^{3 x}-\log (x)}dx+8 x-2 \log (x)\) |
Int[(4 - 2*x + 4*x^3 + E^x*(2 - 2*x^2) + E^(3*x)*(4 + E^x*(2 - 8*x) - 12*x + 8*x^2 + 6*x^3) + (-4 - 2*x^2 + E^x*(-2 + 2*x))*Log[x])/(-2*x^2 - E^x*x^ 2 + x^3 + x^4 + E^(3*x)*(-2*x - E^x*x + x^2 + x^3) + (2*x + E^x*x - x^2 - x^3)*Log[x]),x]
3.25.92.3.1 Defintions of rubi rules used
Time = 0.72 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(2 \ln \left (x -\ln \left (x \right )+{\mathrm e}^{3 x}\right )+2 \ln \left (x -{\mathrm e}^{x}+x^{2}-2\right )-2 \ln \left (x \right )\) | \(32\) |
risch | \(-2 \ln \left (x \right )+2 \ln \left (2-x -x^{2}+{\mathrm e}^{x}\right )+2 \ln \left (-{\mathrm e}^{3 x}-x +\ln \left (x \right )\right )\) | \(36\) |
int((((-2+2*x)*exp(x)-2*x^2-4)*ln(x)+((-8*x+2)*exp(x)+6*x^3+8*x^2-12*x+4)* exp(3*x)+(-2*x^2+2)*exp(x)+4*x^3-2*x+4)/((exp(x)*x-x^3-x^2+2*x)*ln(x)+(-ex p(x)*x+x^3+x^2-2*x)*exp(3*x)-exp(x)*x^2+x^4+x^3-2*x^2),x,method=_RETURNVER BOSE)
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {4-2 x+4 x^3+e^x \left (2-2 x^2\right )+e^{3 x} \left (4+e^x (2-8 x)-12 x+8 x^2+6 x^3\right )+\left (-4-2 x^2+e^x (-2+2 x)\right ) \log (x)}{-2 x^2-e^x x^2+x^3+x^4+e^{3 x} \left (-2 x-e^x x+x^2+x^3\right )+\left (2 x+e^x x-x^2-x^3\right ) \log (x)} \, dx=2 \, \log \left (-x^{2} - x + e^{x} + 2\right ) - 2 \, \log \left (x\right ) + 2 \, \log \left (-x - e^{\left (3 \, x\right )} + \log \left (x\right )\right ) \]
integrate((((-2+2*x)*exp(x)-2*x^2-4)*log(x)+((-8*x+2)*exp(x)+6*x^3+8*x^2-1 2*x+4)*exp(3*x)+(-2*x^2+2)*exp(x)+4*x^3-2*x+4)/((exp(x)*x-x^3-x^2+2*x)*log (x)+(-exp(x)*x+x^3+x^2-2*x)*exp(3*x)-exp(x)*x^2+x^4+x^3-2*x^2),x, algorith m=\
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 1.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.94 \[ \int \frac {4-2 x+4 x^3+e^x \left (2-2 x^2\right )+e^{3 x} \left (4+e^x (2-8 x)-12 x+8 x^2+6 x^3\right )+\left (-4-2 x^2+e^x (-2+2 x)\right ) \log (x)}{-2 x^2-e^x x^2+x^3+x^4+e^{3 x} \left (-2 x-e^x x+x^2+x^3\right )+\left (2 x+e^x x-x^2-x^3\right ) \log (x)} \, dx=- 2 \log {\left (x \right )} + 2 \log {\left (- x^{3} + x^{2} \log {\left (x \right )} - x^{2} + x \log {\left (x \right )} + 2 x + \left (x - \log {\left (x \right )}\right ) e^{x} + \left (- x^{2} - x + 2\right ) e^{3 x} + e^{4 x} - 2 \log {\left (x \right )} \right )} \]
integrate((((-2+2*x)*exp(x)-2*x**2-4)*ln(x)+((-8*x+2)*exp(x)+6*x**3+8*x**2 -12*x+4)*exp(3*x)+(-2*x**2+2)*exp(x)+4*x**3-2*x+4)/((exp(x)*x-x**3-x**2+2* x)*ln(x)+(-exp(x)*x+x**3+x**2-2*x)*exp(3*x)-exp(x)*x**2+x**4+x**3-2*x**2), x)
-2*log(x) + 2*log(-x**3 + x**2*log(x) - x**2 + x*log(x) + 2*x + (x - log(x ))*exp(x) + (-x**2 - x + 2)*exp(3*x) + exp(4*x) - 2*log(x))
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {4-2 x+4 x^3+e^x \left (2-2 x^2\right )+e^{3 x} \left (4+e^x (2-8 x)-12 x+8 x^2+6 x^3\right )+\left (-4-2 x^2+e^x (-2+2 x)\right ) \log (x)}{-2 x^2-e^x x^2+x^3+x^4+e^{3 x} \left (-2 x-e^x x+x^2+x^3\right )+\left (2 x+e^x x-x^2-x^3\right ) \log (x)} \, dx=2 \, \log \left (-x^{2} - x + e^{x} + 2\right ) + 2 \, \log \left (x + e^{\left (3 \, x\right )} - \log \left (x\right )\right ) - 2 \, \log \left (x\right ) \]
integrate((((-2+2*x)*exp(x)-2*x^2-4)*log(x)+((-8*x+2)*exp(x)+6*x^3+8*x^2-1 2*x+4)*exp(3*x)+(-2*x^2+2)*exp(x)+4*x^3-2*x+4)/((exp(x)*x-x^3-x^2+2*x)*log (x)+(-exp(x)*x+x^3+x^2-2*x)*exp(3*x)-exp(x)*x^2+x^4+x^3-2*x^2),x, algorith m=\
Timed out. \[ \int \frac {4-2 x+4 x^3+e^x \left (2-2 x^2\right )+e^{3 x} \left (4+e^x (2-8 x)-12 x+8 x^2+6 x^3\right )+\left (-4-2 x^2+e^x (-2+2 x)\right ) \log (x)}{-2 x^2-e^x x^2+x^3+x^4+e^{3 x} \left (-2 x-e^x x+x^2+x^3\right )+\left (2 x+e^x x-x^2-x^3\right ) \log (x)} \, dx=\text {Timed out} \]
integrate((((-2+2*x)*exp(x)-2*x^2-4)*log(x)+((-8*x+2)*exp(x)+6*x^3+8*x^2-1 2*x+4)*exp(3*x)+(-2*x^2+2)*exp(x)+4*x^3-2*x+4)/((exp(x)*x-x^3-x^2+2*x)*log (x)+(-exp(x)*x+x^3+x^2-2*x)*exp(3*x)-exp(x)*x^2+x^4+x^3-2*x^2),x, algorith m=\
Time = 9.68 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {4-2 x+4 x^3+e^x \left (2-2 x^2\right )+e^{3 x} \left (4+e^x (2-8 x)-12 x+8 x^2+6 x^3\right )+\left (-4-2 x^2+e^x (-2+2 x)\right ) \log (x)}{-2 x^2-e^x x^2+x^3+x^4+e^{3 x} \left (-2 x-e^x x+x^2+x^3\right )+\left (2 x+e^x x-x^2-x^3\right ) \log (x)} \, dx=2\,\ln \left (x-{\mathrm {e}}^x+x^2-2\right )+2\,\ln \left (x+{\mathrm {e}}^{3\,x}-\ln \left (x\right )\right )-2\,\ln \left (x\right ) \]