Integrand size = 257, antiderivative size = 38 \[ \int \frac {4 x+6 x^2-4 x^3+\left (-2 x+x^2\right ) \log (2)+e^x \left (6 x-x^2-x^3\right ) \log (2)+\left (4 x+4 x^2+(-4-2 x) \log (2)+e^x (4+2 x) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )}{\left (8 x-6 x^3+2 x^4+\left (-8+4 x+2 x^2-x^3\right ) \log (2)+e^x \left (8-4 x-2 x^2+x^3\right ) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \, dx=\frac {x}{(2-x) \log \left (\log \left (\frac {4}{x^2+(2+x) \left (x-\left (1-e^x\right ) \log (2)\right )}\right )\right )} \]
Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {4 x+6 x^2-4 x^3+\left (-2 x+x^2\right ) \log (2)+e^x \left (6 x-x^2-x^3\right ) \log (2)+\left (4 x+4 x^2+(-4-2 x) \log (2)+e^x (4+2 x) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )}{\left (8 x-6 x^3+2 x^4+\left (-8+4 x+2 x^2-x^3\right ) \log (2)+e^x \left (8-4 x-2 x^2+x^3\right ) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \, dx=-\frac {x}{(-2+x) \log \left (\log \left (\frac {4}{2 x+2 x^2-(2+x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \]
Integrate[(4*x + 6*x^2 - 4*x^3 + (-2*x + x^2)*Log[2] + E^x*(6*x - x^2 - x^ 3)*Log[2] + (4*x + 4*x^2 + (-4 - 2*x)*Log[2] + E^x*(4 + 2*x)*Log[2])*Log[4 /(2*x + 2*x^2 + (-2 - x)*Log[2] + E^x*(2 + x)*Log[2])]*Log[Log[4/(2*x + 2* x^2 + (-2 - x)*Log[2] + E^x*(2 + x)*Log[2])]])/((8*x - 6*x^3 + 2*x^4 + (-8 + 4*x + 2*x^2 - x^3)*Log[2] + E^x*(8 - 4*x - 2*x^2 + x^3)*Log[2])*Log[4/( 2*x + 2*x^2 + (-2 - x)*Log[2] + E^x*(2 + x)*Log[2])]*Log[Log[4/(2*x + 2*x^ 2 + (-2 - x)*Log[2] + E^x*(2 + x)*Log[2])]]^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.
| \(\displaystyle \int \frac {-4 x^3+6 x^2+\left (4 x^2+4 x+(-2 x-4) \log (2)+e^x (2 x+4) \log (2)\right ) \log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right )\right )+\left (x^2-2 x\right ) \log (2)+e^x \left (-x^3-x^2+6 x\right ) \log (2)+4 x}{\left (2 x^4-6 x^3+\left (-x^3+2 x^2+4 x-8\right ) \log (2)+e^x \left (x^3-2 x^2-4 x+8\right ) \log (2)+8 x\right ) \log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-4 x^3+6 x^2+\left (4 x^2+4 x+(-2 x-4) \log (2)+e^x (2 x+4) \log (2)\right ) \log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right )\right )+\left (x^2-2 x\right ) \log (2)+e^x \left (-x^3-x^2+6 x\right ) \log (2)+4 x}{(2-x)^2 \left (2 x^2+e^x x \log (2)+2 x \left (1-\frac {\log (2)}{2}\right )+e^x \log (4)-\log (4)\right ) \log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x^2+2 x+(-x-2) \log (2)+e^x (x+2) \log (2)}\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x \left (x^4 \log (4)-x^3 \log ^2(2)-x^2 \log (2) (12+\log (4))+x \left (4 \log ^2(2)+\log (16)\right )+\log (256)+\log (4) \log (16)\right )}{(2-x)^2 (x \log (2)+\log (4)) \left (2 x^2+e^x x \log (2)+2 x \left (1-\frac {\log (2)}{2}\right )+e^x \log (4)-\log (4)\right ) \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )}-\frac {\log (2) \left (x^3+x^2-2 x \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )-4 \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )-6 x\right )}{(x-2)^2 (x \log (2)+\log (4)) \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {x \left (x^4 \log (4)-x^3 \log ^2(2)-x^2 \log (2) (12+\log (4))+x \left (4 \log ^2(2)+\log (16)\right )+\log (256)+\log (4) \log (16)\right )}{(2-x)^2 (x \log (2)+\log (4)) \left (2 x^2+e^x x \log (2)+2 x \left (1-\frac {\log (2)}{2}\right )+e^x \log (4)-\log (4)\right ) \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )}-\frac {\log (2) \left (x^3+x^2-2 x \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )-4 \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )-6 x\right )}{(x-2)^2 (x \log (2)+\log (4)) \log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x^2+2 x+e^x (x+2) \log (2)-(x+2) \log (2)}\right )\right )}\right )dx\) |
Int[(4*x + 6*x^2 - 4*x^3 + (-2*x + x^2)*Log[2] + E^x*(6*x - x^2 - x^3)*Log [2] + (4*x + 4*x^2 + (-4 - 2*x)*Log[2] + E^x*(4 + 2*x)*Log[2])*Log[4/(2*x + 2*x^2 + (-2 - x)*Log[2] + E^x*(2 + x)*Log[2])]*Log[Log[4/(2*x + 2*x^2 + (-2 - x)*Log[2] + E^x*(2 + x)*Log[2])]])/((8*x - 6*x^3 + 2*x^4 + (-8 + 4*x + 2*x^2 - x^3)*Log[2] + E^x*(8 - 4*x - 2*x^2 + x^3)*Log[2])*Log[4/(2*x + 2*x^2 + (-2 - x)*Log[2] + E^x*(2 + x)*Log[2])]*Log[Log[4/(2*x + 2*x^2 + (- 2 - x)*Log[2] + E^x*(2 + x)*Log[2])]]^2),x]
3.1.83.3.1 Defintions of rubi rules used
Time = 95.36 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16
| method | result | size |
| risch | \(-\frac {x}{\left (-2+x \right ) \ln \left (2 \ln \left (2\right )-\ln \left (\left (\left ({\mathrm e}^{x}-1\right ) x +2 \,{\mathrm e}^{x}-2\right ) \ln \left (2\right )+2 x^{2}+2 x \right )\right )}\) | \(44\) |
| parallelrisch | \(-\frac {x}{\left (-2+x \right ) \ln \left (\ln \left (\frac {4}{x \ln \left (2\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (2\right )-x \ln \left (2\right )+2 x^{2}-2 \ln \left (2\right )+2 x}\right )\right )}\) | \(47\) |
int((((4+2*x)*ln(2)*exp(x)+(-2*x-4)*ln(2)+4*x^2+4*x)*ln(4/((2+x)*ln(2)*exp (x)+(-2-x)*ln(2)+2*x^2+2*x))*ln(ln(4/((2+x)*ln(2)*exp(x)+(-2-x)*ln(2)+2*x^ 2+2*x)))+(-x^3-x^2+6*x)*ln(2)*exp(x)+(x^2-2*x)*ln(2)-4*x^3+6*x^2+4*x)/((x^ 3-2*x^2-4*x+8)*ln(2)*exp(x)+(-x^3+2*x^2+4*x-8)*ln(2)+2*x^4-6*x^3+8*x)/ln(4 /((2+x)*ln(2)*exp(x)+(-2-x)*ln(2)+2*x^2+2*x))/ln(ln(4/((2+x)*ln(2)*exp(x)+ (-2-x)*ln(2)+2*x^2+2*x)))^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {4 x+6 x^2-4 x^3+\left (-2 x+x^2\right ) \log (2)+e^x \left (6 x-x^2-x^3\right ) \log (2)+\left (4 x+4 x^2+(-4-2 x) \log (2)+e^x (4+2 x) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )}{\left (8 x-6 x^3+2 x^4+\left (-8+4 x+2 x^2-x^3\right ) \log (2)+e^x \left (8-4 x-2 x^2+x^3\right ) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \, dx=-\frac {x}{{\left (x - 2\right )} \log \left (\log \left (\frac {4}{{\left (x + 2\right )} e^{x} \log \left (2\right ) + 2 \, x^{2} - {\left (x + 2\right )} \log \left (2\right ) + 2 \, x}\right )\right )} \]
integrate((((4+2*x)*log(2)*exp(x)+(-2*x-4)*log(2)+4*x^2+4*x)*log(4/((2+x)* log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x))*log(log(4/((2+x)*log(2)*exp(x)+(-2 -x)*log(2)+2*x^2+2*x)))+(-x^3-x^2+6*x)*log(2)*exp(x)+(x^2-2*x)*log(2)-4*x^ 3+6*x^2+4*x)/((x^3-2*x^2-4*x+8)*log(2)*exp(x)+(-x^3+2*x^2+4*x-8)*log(2)+2* x^4-6*x^3+8*x)/log(4/((2+x)*log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x))/log(lo g(4/((2+x)*log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x)))^2,x, algorithm=\
Time = 3.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {4 x+6 x^2-4 x^3+\left (-2 x+x^2\right ) \log (2)+e^x \left (6 x-x^2-x^3\right ) \log (2)+\left (4 x+4 x^2+(-4-2 x) \log (2)+e^x (4+2 x) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )}{\left (8 x-6 x^3+2 x^4+\left (-8+4 x+2 x^2-x^3\right ) \log (2)+e^x \left (8-4 x-2 x^2+x^3\right ) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \, dx=- \frac {x}{\left (x - 2\right ) \log {\left (\log {\left (\frac {4}{2 x^{2} + 2 x + \left (- x - 2\right ) \log {\left (2 \right )} + \left (x + 2\right ) e^{x} \log {\left (2 \right )}} \right )} \right )}} \]
integrate((((4+2*x)*ln(2)*exp(x)+(-2*x-4)*ln(2)+4*x**2+4*x)*ln(4/((2+x)*ln (2)*exp(x)+(-2-x)*ln(2)+2*x**2+2*x))*ln(ln(4/((2+x)*ln(2)*exp(x)+(-2-x)*ln (2)+2*x**2+2*x)))+(-x**3-x**2+6*x)*ln(2)*exp(x)+(x**2-2*x)*ln(2)-4*x**3+6* x**2+4*x)/((x**3-2*x**2-4*x+8)*ln(2)*exp(x)+(-x**3+2*x**2+4*x-8)*ln(2)+2*x **4-6*x**3+8*x)/ln(4/((2+x)*ln(2)*exp(x)+(-2-x)*ln(2)+2*x**2+2*x))/ln(ln(4 /((2+x)*ln(2)*exp(x)+(-2-x)*ln(2)+2*x**2+2*x)))**2,x)
Time = 1.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \frac {4 x+6 x^2-4 x^3+\left (-2 x+x^2\right ) \log (2)+e^x \left (6 x-x^2-x^3\right ) \log (2)+\left (4 x+4 x^2+(-4-2 x) \log (2)+e^x (4+2 x) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )}{\left (8 x-6 x^3+2 x^4+\left (-8+4 x+2 x^2-x^3\right ) \log (2)+e^x \left (8-4 x-2 x^2+x^3\right ) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \, dx=-\frac {x}{{\left (x - 2\right )} \log \left (2 \, \log \left (2\right ) - \log \left (2 \, x^{2} - x {\left (\log \left (2\right ) - 2\right )} + {\left (x \log \left (2\right ) + 2 \, \log \left (2\right )\right )} e^{x} - 2 \, \log \left (2\right )\right )\right )} \]
integrate((((4+2*x)*log(2)*exp(x)+(-2*x-4)*log(2)+4*x^2+4*x)*log(4/((2+x)* log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x))*log(log(4/((2+x)*log(2)*exp(x)+(-2 -x)*log(2)+2*x^2+2*x)))+(-x^3-x^2+6*x)*log(2)*exp(x)+(x^2-2*x)*log(2)-4*x^ 3+6*x^2+4*x)/((x^3-2*x^2-4*x+8)*log(2)*exp(x)+(-x^3+2*x^2+4*x-8)*log(2)+2* x^4-6*x^3+8*x)/log(4/((2+x)*log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x))/log(lo g(4/((2+x)*log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x)))^2,x, algorithm=\
-x/((x - 2)*log(2*log(2) - log(2*x^2 - x*(log(2) - 2) + (x*log(2) + 2*log( 2))*e^x - 2*log(2))))
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (33) = 66\).
Time = 1.41 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.32 \[ \int \frac {4 x+6 x^2-4 x^3+\left (-2 x+x^2\right ) \log (2)+e^x \left (6 x-x^2-x^3\right ) \log (2)+\left (4 x+4 x^2+(-4-2 x) \log (2)+e^x (4+2 x) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )}{\left (8 x-6 x^3+2 x^4+\left (-8+4 x+2 x^2-x^3\right ) \log (2)+e^x \left (8-4 x-2 x^2+x^3\right ) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \, dx=-\frac {x}{x \log \left (2 \, \log \left (2\right ) - \log \left (x e^{x} \log \left (2\right ) + 2 \, x^{2} - x \log \left (2\right ) + 2 \, e^{x} \log \left (2\right ) + 2 \, x - 2 \, \log \left (2\right )\right )\right ) - 2 \, \log \left (2 \, \log \left (2\right ) - \log \left (x e^{x} \log \left (2\right ) + 2 \, x^{2} - x \log \left (2\right ) + 2 \, e^{x} \log \left (2\right ) + 2 \, x - 2 \, \log \left (2\right )\right )\right )} \]
integrate((((4+2*x)*log(2)*exp(x)+(-2*x-4)*log(2)+4*x^2+4*x)*log(4/((2+x)* log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x))*log(log(4/((2+x)*log(2)*exp(x)+(-2 -x)*log(2)+2*x^2+2*x)))+(-x^3-x^2+6*x)*log(2)*exp(x)+(x^2-2*x)*log(2)-4*x^ 3+6*x^2+4*x)/((x^3-2*x^2-4*x+8)*log(2)*exp(x)+(-x^3+2*x^2+4*x-8)*log(2)+2* x^4-6*x^3+8*x)/log(4/((2+x)*log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x))/log(lo g(4/((2+x)*log(2)*exp(x)+(-2-x)*log(2)+2*x^2+2*x)))^2,x, algorithm=\
-x/(x*log(2*log(2) - log(x*e^x*log(2) + 2*x^2 - x*log(2) + 2*e^x*log(2) + 2*x - 2*log(2))) - 2*log(2*log(2) - log(x*e^x*log(2) + 2*x^2 - x*log(2) + 2*e^x*log(2) + 2*x - 2*log(2))))
Timed out. \[ \int \frac {4 x+6 x^2-4 x^3+\left (-2 x+x^2\right ) \log (2)+e^x \left (6 x-x^2-x^3\right ) \log (2)+\left (4 x+4 x^2+(-4-2 x) \log (2)+e^x (4+2 x) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log \left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )}{\left (8 x-6 x^3+2 x^4+\left (-8+4 x+2 x^2-x^3\right ) \log (2)+e^x \left (8-4 x-2 x^2+x^3\right ) \log (2)\right ) \log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right ) \log ^2\left (\log \left (\frac {4}{2 x+2 x^2+(-2-x) \log (2)+e^x (2+x) \log (2)}\right )\right )} \, dx=\int \frac {4\,x-\ln \left (2\right )\,\left (2\,x-x^2\right )+6\,x^2-4\,x^3-{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x^3+x^2-6\,x\right )+\ln \left (\ln \left (\frac {4}{2\,x-\ln \left (2\right )\,\left (x+2\right )+2\,x^2+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x+2\right )}\right )\right )\,\ln \left (\frac {4}{2\,x-\ln \left (2\right )\,\left (x+2\right )+2\,x^2+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x+2\right )}\right )\,\left (4\,x-\ln \left (2\right )\,\left (2\,x+4\right )+4\,x^2+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (2\,x+4\right )\right )}{{\ln \left (\ln \left (\frac {4}{2\,x-\ln \left (2\right )\,\left (x+2\right )+2\,x^2+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x+2\right )}\right )\right )}^2\,\ln \left (\frac {4}{2\,x-\ln \left (2\right )\,\left (x+2\right )+2\,x^2+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x+2\right )}\right )\,\left (8\,x+\ln \left (2\right )\,\left (-x^3+2\,x^2+4\,x-8\right )-6\,x^3+2\,x^4-{\mathrm {e}}^x\,\ln \left (2\right )\,\left (-x^3+2\,x^2+4\,x-8\right )\right )} \,d x \]
int((4*x - log(2)*(2*x - x^2) + 6*x^2 - 4*x^3 - exp(x)*log(2)*(x^2 - 6*x + x^3) + log(log(4/(2*x - log(2)*(x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2)))) *log(4/(2*x - log(2)*(x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2)))*(4*x - log( 2)*(2*x + 4) + 4*x^2 + exp(x)*log(2)*(2*x + 4)))/(log(log(4/(2*x - log(2)* (x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2))))^2*log(4/(2*x - log(2)*(x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2)))*(8*x + log(2)*(4*x + 2*x^2 - x^3 - 8) - 6 *x^3 + 2*x^4 - exp(x)*log(2)*(4*x + 2*x^2 - x^3 - 8))),x)
int((4*x - log(2)*(2*x - x^2) + 6*x^2 - 4*x^3 - exp(x)*log(2)*(x^2 - 6*x + x^3) + log(log(4/(2*x - log(2)*(x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2)))) *log(4/(2*x - log(2)*(x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2)))*(4*x - log( 2)*(2*x + 4) + 4*x^2 + exp(x)*log(2)*(2*x + 4)))/(log(log(4/(2*x - log(2)* (x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2))))^2*log(4/(2*x - log(2)*(x + 2) + 2*x^2 + exp(x)*log(2)*(x + 2)))*(8*x + log(2)*(4*x + 2*x^2 - x^3 - 8) - 6 *x^3 + 2*x^4 - exp(x)*log(2)*(4*x + 2*x^2 - x^3 - 8))), x)