Integrand size = 290, antiderivative size = 33 \[ \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx=x^2 \log ^2\left (\frac {5}{x \log \left ((4-x)^2 x \left (-e^2+2 x\right )\right )}\right ) \]
\[ \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx=\int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx \]
Integrate[((-32*x^2 + 16*x^3 + E^2*(8*x - 6*x^2) + (-16*x^2 + 4*x^3 + E^2* (8*x - 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])*L og[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])] + (16*x ^2 - 4*x^3 + E^2*(-8*x + 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)]*Log[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])]^2)/((E^2*(-4 + x) + 8*x - 2*x^2)*Log[32*x^2 - 16*x^3 + 2*x^4 + E ^2*(-16*x + 8*x^2 - x^3)]),x]
Integrate[((-32*x^2 + 16*x^3 + E^2*(8*x - 6*x^2) + (-16*x^2 + 4*x^3 + E^2* (8*x - 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])*L og[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])] + (16*x ^2 - 4*x^3 + E^2*(-8*x + 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)]*Log[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])]^2)/((E^2*(-4 + x) + 8*x - 2*x^2)*Log[32*x^2 - 16*x^3 + 2*x^4 + E ^2*(-16*x + 8*x^2 - x^3)]), 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.
\(\displaystyle \int \frac {\left (-4 x^3+16 x^2+e^2 \left (2 x^2-8 x\right )\right ) \log \left (2 x^4-16 x^3+32 x^2+e^2 \left (-x^3+8 x^2-16 x\right )\right ) \log ^2\left (\frac {5}{x \log \left (2 x^4-16 x^3+32 x^2+e^2 \left (-x^3+8 x^2-16 x\right )\right )}\right )+\left (16 x^3-32 x^2+e^2 \left (8 x-6 x^2\right )+\left (4 x^3-16 x^2+e^2 \left (8 x-2 x^2\right )\right ) \log \left (2 x^4-16 x^3+32 x^2+e^2 \left (-x^3+8 x^2-16 x\right )\right )\right ) \log \left (\frac {5}{x \log \left (2 x^4-16 x^3+32 x^2+e^2 \left (-x^3+8 x^2-16 x\right )\right )}\right )}{\left (-2 x^2+8 x+e^2 (x-4)\right ) \log \left (2 x^4-16 x^3+32 x^2+e^2 \left (-x^3+8 x^2-16 x\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}\right ) \left (\frac {2 x \left (-8 (x-2) x+e^2 (3 x-4)+\left (e^2-2 x\right ) (x-4) \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )\right )}{\log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}+2 (x-4) x \left (2 x-e^2\right ) \log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}\right )\right )}{\left (e^2-2 x\right ) (4-x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x \left (-8 x^2-2 x^2 \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )+16 \left (1+\frac {3 e^2}{16}\right ) x+8 \left (1+\frac {e^2}{8}\right ) x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )-4 e^2 \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )-4 e^2\right ) \log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}\right )}{\left (e^2-2 x\right ) (4-x) \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}+2 x \log ^2\left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}\right )\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {2 x \left (-8 x^2-2 x^2 \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )+16 \left (1+\frac {3 e^2}{16}\right ) x+8 \left (1+\frac {e^2}{8}\right ) x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )-4 e^2 \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )-4 e^2\right ) \log \left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}\right )}{\left (e^2-2 x\right ) (4-x) \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}+2 x \log ^2\left (\frac {5}{x \log \left (-\left (\left (e^2-2 x\right ) (x-4)^2 x\right )\right )}\right )\right )dx\) |
Int[((-32*x^2 + 16*x^3 + E^2*(8*x - 6*x^2) + (-16*x^2 + 4*x^3 + E^2*(8*x - 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])*Log[5/( x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3)])] + (16*x^2 - 4 *x^3 + E^2*(-8*x + 2*x^2))*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^ 2 - x^3)]*Log[5/(x*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-16*x + 8*x^2 - x^3) ])]^2)/((E^2*(-4 + x) + 8*x - 2*x^2)*Log[32*x^2 - 16*x^3 + 2*x^4 + E^2*(-1 6*x + 8*x^2 - x^3)]),x]
3.7.97.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 806.56 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48
method | result | size |
parallelrisch | \(x^{2} {\ln \left (\frac {5}{x \ln \left (\left (-x^{3}+8 x^{2}-16 x \right ) {\mathrm e}^{2}+2 x^{4}-16 x^{3}+32 x^{2}\right )}\right )}^{2}\) | \(49\) |
int((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-1 6*x^3+32*x^2)*ln(5/x/ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2))^2+( ((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*ln((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^ 3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*ln(5/x/ln((-x^3+8*x^2-16*x)*e xp(2)+2*x^4-16*x^3+32*x^2)))/((x-4)*exp(2)-2*x^2+8*x)/ln((-x^3+8*x^2-16*x) *exp(2)+2*x^4-16*x^3+32*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx=x^{2} \log \left (\frac {5}{x \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )}\right )^{2} \]
integrate((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+ 2*x^4-16*x^3+32*x^2)*log(5/x/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32* x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+2 *x^4-16*x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*log(5/x/log((-x^3+8 *x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)))/((x-4)*exp(2)-2*x^2+8*x)/log((-x^ 3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x, algorithm=\
Time = 1.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx=x^{2} \log {\left (\frac {5}{x \log {\left (2 x^{4} - 16 x^{3} + 32 x^{2} + \left (- x^{3} + 8 x^{2} - 16 x\right ) e^{2} \right )}} \right )}^{2} \]
integrate((((2*x**2-8*x)*exp(2)-4*x**3+16*x**2)*ln((-x**3+8*x**2-16*x)*exp (2)+2*x**4-16*x**3+32*x**2)*ln(5/x/ln((-x**3+8*x**2-16*x)*exp(2)+2*x**4-16 *x**3+32*x**2))**2+(((-2*x**2+8*x)*exp(2)+4*x**3-16*x**2)*ln((-x**3+8*x**2 -16*x)*exp(2)+2*x**4-16*x**3+32*x**2)+(-6*x**2+8*x)*exp(2)+16*x**3-32*x**2 )*ln(5/x/ln((-x**3+8*x**2-16*x)*exp(2)+2*x**4-16*x**3+32*x**2)))/((x-4)*ex p(2)-2*x**2+8*x)/ln((-x**3+8*x**2-16*x)*exp(2)+2*x**4-16*x**3+32*x**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (30) = 60\).
Time = 0.36 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.61 \[ \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx=x^{2} \log \left (5\right )^{2} - 2 \, x^{2} \log \left (5\right ) \log \left (x\right ) + x^{2} \log \left (x\right )^{2} + x^{2} \log \left (\log \left (2 \, x - e^{2}\right ) + 2 \, \log \left (x - 4\right ) + \log \left (x\right )\right )^{2} - 2 \, {\left (x^{2} \log \left (5\right ) - x^{2} \log \left (x\right )\right )} \log \left (\log \left (2 \, x - e^{2}\right ) + 2 \, \log \left (x - 4\right ) + \log \left (x\right )\right ) \]
integrate((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+ 2*x^4-16*x^3+32*x^2)*log(5/x/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32* x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+2 *x^4-16*x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*log(5/x/log((-x^3+8 *x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)))/((x-4)*exp(2)-2*x^2+8*x)/log((-x^ 3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x, algorithm=\
x^2*log(5)^2 - 2*x^2*log(5)*log(x) + x^2*log(x)^2 + x^2*log(log(2*x - e^2) + 2*log(x - 4) + log(x))^2 - 2*(x^2*log(5) - x^2*log(x))*log(log(2*x - e^ 2) + 2*log(x - 4) + log(x))
\[ \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx=\int { \frac {2 \, {\left ({\left (2 \, x^{3} - 8 \, x^{2} - {\left (x^{2} - 4 \, x\right )} e^{2}\right )} \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right ) \log \left (\frac {5}{x \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )}\right )^{2} - {\left (8 \, x^{3} - 16 \, x^{2} - {\left (3 \, x^{2} - 4 \, x\right )} e^{2} + {\left (2 \, x^{3} - 8 \, x^{2} - {\left (x^{2} - 4 \, x\right )} e^{2}\right )} \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )\right )} \log \left (\frac {5}{x \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )}\right )\right )}}{{\left (2 \, x^{2} - {\left (x - 4\right )} e^{2} - 8 \, x\right )} \log \left (2 \, x^{4} - 16 \, x^{3} + 32 \, x^{2} - {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{2}\right )} \,d x } \]
integrate((((2*x^2-8*x)*exp(2)-4*x^3+16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+ 2*x^4-16*x^3+32*x^2)*log(5/x/log((-x^3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32* x^2))^2+(((-2*x^2+8*x)*exp(2)+4*x^3-16*x^2)*log((-x^3+8*x^2-16*x)*exp(2)+2 *x^4-16*x^3+32*x^2)+(-6*x^2+8*x)*exp(2)+16*x^3-32*x^2)*log(5/x/log((-x^3+8 *x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2)))/((x-4)*exp(2)-2*x^2+8*x)/log((-x^ 3+8*x^2-16*x)*exp(2)+2*x^4-16*x^3+32*x^2),x, algorithm=\
integrate(2*((2*x^3 - 8*x^2 - (x^2 - 4*x)*e^2)*log(2*x^4 - 16*x^3 + 32*x^2 - (x^3 - 8*x^2 + 16*x)*e^2)*log(5/(x*log(2*x^4 - 16*x^3 + 32*x^2 - (x^3 - 8*x^2 + 16*x)*e^2)))^2 - (8*x^3 - 16*x^2 - (3*x^2 - 4*x)*e^2 + (2*x^3 - 8 *x^2 - (x^2 - 4*x)*e^2)*log(2*x^4 - 16*x^3 + 32*x^2 - (x^3 - 8*x^2 + 16*x) *e^2))*log(5/(x*log(2*x^4 - 16*x^3 + 32*x^2 - (x^3 - 8*x^2 + 16*x)*e^2)))) /((2*x^2 - (x - 4)*e^2 - 8*x)*log(2*x^4 - 16*x^3 + 32*x^2 - (x^3 - 8*x^2 + 16*x)*e^2)), x)
Time = 10.41 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-32 x^2+16 x^3+e^2 \left (8 x-6 x^2\right )+\left (-16 x^2+4 x^3+e^2 \left (8 x-2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )\right ) \log \left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )+\left (16 x^2-4 x^3+e^2 \left (-8 x+2 x^2\right )\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right ) \log ^2\left (\frac {5}{x \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )}\right )}{\left (e^2 (-4+x)+8 x-2 x^2\right ) \log \left (32 x^2-16 x^3+2 x^4+e^2 \left (-16 x+8 x^2-x^3\right )\right )} \, dx=x^2\,{\ln \left (\frac {5}{x\,\ln \left (32\,x^2-{\mathrm {e}}^2\,\left (x^3-8\,x^2+16\,x\right )-16\,x^3+2\,x^4\right )}\right )}^2 \]
int((log(5/(x*log(32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^3 + 2*x^4))) *(exp(2)*(8*x - 6*x^2) - 32*x^2 + 16*x^3 + log(32*x^2 - exp(2)*(16*x - 8*x ^2 + x^3) - 16*x^3 + 2*x^4)*(exp(2)*(8*x - 2*x^2) - 16*x^2 + 4*x^3)) - log (32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^3 + 2*x^4)*log(5/(x*log(32*x^ 2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^3 + 2*x^4)))^2*(exp(2)*(8*x - 2*x^2 ) - 16*x^2 + 4*x^3))/(log(32*x^2 - exp(2)*(16*x - 8*x^2 + x^3) - 16*x^3 + 2*x^4)*(8*x + exp(2)*(x - 4) - 2*x^2)),x)