Integrand size = 251, antiderivative size = 31 \[ \int \frac {24 x+16 x^3+e^x \left (12+8 x^2\right )+\left (6 x-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-8 x^3-4 e^x x^3\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )\right )}{\left (-12 x^3+80 x^5+e^{2 x} \left (-3 x+20 x^3\right )+e^x \left (-12 x^2+80 x^4\right )+\left (4 e^{2 x} x^3+16 e^x x^4+16 x^5\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )} \, dx=\frac {\log \left (\log ^2\left (\log \left (5-\frac {3}{4 x^2}+\log (x)\right )\right )\right )}{\left (2+\frac {e^x}{x}\right ) x} \]
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {24 x+16 x^3+e^x \left (12+8 x^2\right )+\left (6 x-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-8 x^3-4 e^x x^3\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )\right )}{\left (-12 x^3+80 x^5+e^{2 x} \left (-3 x+20 x^3\right )+e^x \left (-12 x^2+80 x^4\right )+\left (4 e^{2 x} x^3+16 e^x x^4+16 x^5\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )} \, dx=\frac {\log \left (\log ^2\left (\log \left (5-\frac {3}{4 x^2}+\log (x)\right )\right )\right )}{e^x+2 x} \]
Integrate[(24*x + 16*x^3 + E^x*(12 + 8*x^2) + (6*x - 40*x^3 + E^x*(3*x - 2 0*x^3) + (-8*x^3 - 4*E^x*x^3)*Log[x])*Log[(-3 + 20*x^2 + 4*x^2*Log[x])/(4* x^2)]*Log[Log[(-3 + 20*x^2 + 4*x^2*Log[x])/(4*x^2)]]*Log[Log[Log[(-3 + 20* x^2 + 4*x^2*Log[x])/(4*x^2)]]^2])/((-12*x^3 + 80*x^5 + E^(2*x)*(-3*x + 20* x^3) + E^x*(-12*x^2 + 80*x^4) + (4*E^(2*x)*x^3 + 16*E^x*x^4 + 16*x^5)*Log[ x])*Log[(-3 + 20*x^2 + 4*x^2*Log[x])/(4*x^2)]*Log[Log[(-3 + 20*x^2 + 4*x^2 *Log[x])/(4*x^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.
| \(\displaystyle \int \frac {16 x^3+e^x \left (8 x^2+12\right )+\left (-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-4 e^x x^3-8 x^3\right ) \log (x)+6 x\right ) \log \left (\frac {20 x^2+4 x^2 \log (x)-3}{4 x^2}\right ) \log \left (\log \left (\frac {20 x^2+4 x^2 \log (x)-3}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {20 x^2+4 x^2 \log (x)-3}{4 x^2}\right )\right )\right )+24 x}{\left (80 x^5-12 x^3+e^{2 x} \left (20 x^3-3 x\right )+e^x \left (80 x^4-12 x^2\right )+\left (16 x^5+16 e^x x^4+4 e^{2 x} x^3\right ) \log (x)\right ) \log \left (\frac {20 x^2+4 x^2 \log (x)-3}{4 x^2}\right ) \log \left (\log \left (\frac {20 x^2+4 x^2 \log (x)-3}{4 x^2}\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-16 x^3-4 e^x \left (2 x^2+3\right )+\left (e^x+2\right ) x \left (20 x^2+4 x^2 \log (x)-3\right ) \log \left (-\frac {3}{4 x^2}+\log (x)+5\right ) \log \left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right ) \log \left (\log ^2\left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right )\right )-24 x}{x \left (2 x+e^x\right )^2 \left (-20 x^2-4 x^2 \log (x)+3\right ) \log \left (-\frac {3}{4 x^2}+\log (x)+5\right ) \log \left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 (x-1) \log \left (\log ^2\left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right )\right )}{\left (2 x+e^x\right )^2}-\frac {-8 x^2-3 x \log \left (-\frac {3}{4 x^2}+\log (x)+5\right ) \log \left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right ) \log \left (\log ^2\left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right )\right )+4 x^3 \log (x) \log \left (-\frac {3}{4 x^2}+\log (x)+5\right ) \log \left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right ) \log \left (\log ^2\left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right )\right )+20 x^3 \log \left (-\frac {3}{4 x^2}+\log (x)+5\right ) \log \left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right ) \log \left (\log ^2\left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right )\right )-12}{x \left (2 x+e^x\right ) \left (20 x^2+4 x^2 \log (x)-3\right ) \log \left (-\frac {3}{4 x^2}+\log (x)+5\right ) \log \left (\log \left (-\frac {3}{4 x^2}+\log (x)+5\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \int \frac {\log \left (\log ^2\left (\log \left (\log (x)-\frac {3}{4 x^2}+5\right )\right )\right )}{\left (2 x+e^x\right )^2}dx+2 \int \frac {x \log \left (\log ^2\left (\log \left (\log (x)-\frac {3}{4 x^2}+5\right )\right )\right )}{\left (2 x+e^x\right )^2}dx-\int \frac {\log \left (\log ^2\left (\log \left (\log (x)-\frac {3}{4 x^2}+5\right )\right )\right )}{2 x+e^x}dx+12 \int \frac {1}{x \left (2 x+e^x\right ) \left (4 \log (x) x^2+20 x^2-3\right ) \log \left (\log (x)-\frac {3}{4 x^2}+5\right ) \log \left (\log \left (\log (x)-\frac {3}{4 x^2}+5\right )\right )}dx+8 \int \frac {x}{\left (2 x+e^x\right ) \left (4 \log (x) x^2+20 x^2-3\right ) \log \left (\log (x)-\frac {3}{4 x^2}+5\right ) \log \left (\log \left (\log (x)-\frac {3}{4 x^2}+5\right )\right )}dx\) |
Int[(24*x + 16*x^3 + E^x*(12 + 8*x^2) + (6*x - 40*x^3 + E^x*(3*x - 20*x^3) + (-8*x^3 - 4*E^x*x^3)*Log[x])*Log[(-3 + 20*x^2 + 4*x^2*Log[x])/(4*x^2)]* Log[Log[(-3 + 20*x^2 + 4*x^2*Log[x])/(4*x^2)]]*Log[Log[Log[(-3 + 20*x^2 + 4*x^2*Log[x])/(4*x^2)]]^2])/((-12*x^3 + 80*x^5 + E^(2*x)*(-3*x + 20*x^3) + E^x*(-12*x^2 + 80*x^4) + (4*E^(2*x)*x^3 + 16*E^x*x^4 + 16*x^5)*Log[x])*Lo g[(-3 + 20*x^2 + 4*x^2*Log[x])/(4*x^2)]*Log[Log[(-3 + 20*x^2 + 4*x^2*Log[x ])/(4*x^2)]]),x]
3.9.2.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 8.56 (sec) , antiderivative size = 820, normalized size of antiderivative = 26.45
\[\text {Expression too large to display}\]
int((((-4*exp(x)*x^3-8*x^3)*ln(x)+(-20*x^3+3*x)*exp(x)-40*x^3+6*x)*ln(1/4* (4*x^2*ln(x)+20*x^2-3)/x^2)*ln(ln(1/4*(4*x^2*ln(x)+20*x^2-3)/x^2))*ln(ln(l n(1/4*(4*x^2*ln(x)+20*x^2-3)/x^2))^2)+(8*x^2+12)*exp(x)+16*x^3+24*x)/((4*e xp(x)^2*x^3+16*exp(x)*x^4+16*x^5)*ln(x)+(20*x^3-3*x)*exp(x)^2+(80*x^4-12*x ^2)*exp(x)+80*x^5-12*x^3)/ln(1/4*(4*x^2*ln(x)+20*x^2-3)/x^2)/ln(ln(1/4*(4* x^2*ln(x)+20*x^2-3)/x^2)),x)
2/(exp(x)+2*x)*ln(ln(-2*ln(x)+ln(-3/4+(5+ln(x))*x^2)+1/2*I*Pi*csgn(I*x^2)* (-csgn(I*x^2)+csgn(I*x))^2-1/2*I*Pi*csgn(I/x^2*(-3/4+(5+ln(x))*x^2))*(-csg n(I/x^2*(-3/4+(5+ln(x))*x^2))+csgn(I/x^2))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^ 2))+csgn(I*(-3/4+(5+ln(x))*x^2)))))-1/2*I*Pi*csgn(I*ln(-2*ln(x)+ln(-3/4+(5 +ln(x))*x^2)+1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2-1/2*I*Pi*csgn (I/x^2*(-3/4+(5+ln(x))*x^2))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^2))+csgn(I/x^2 ))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^2))+csgn(I*(-3/4+(5+ln(x))*x^2))))^2)*(c sgn(I*ln(-2*ln(x)+ln(-3/4+(5+ln(x))*x^2)+1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2 )+csgn(I*x))^2-1/2*I*Pi*csgn(I/x^2*(-3/4+(5+ln(x))*x^2))*(-csgn(I/x^2*(-3/ 4+(5+ln(x))*x^2))+csgn(I/x^2))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^2))+csgn(I*( -3/4+(5+ln(x))*x^2)))))^2-2*csgn(I*ln(-2*ln(x)+ln(-3/4+(5+ln(x))*x^2)+1/2* I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2-1/2*I*Pi*csgn(I/x^2*(-3/4+(5+l n(x))*x^2))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^2))+csgn(I/x^2))*(-csgn(I/x^2*( -3/4+(5+ln(x))*x^2))+csgn(I*(-3/4+(5+ln(x))*x^2))))^2)*csgn(I*ln(-2*ln(x)+ ln(-3/4+(5+ln(x))*x^2)+1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2-1/2 *I*Pi*csgn(I/x^2*(-3/4+(5+ln(x))*x^2))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^2))+ csgn(I/x^2))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^2))+csgn(I*(-3/4+(5+ln(x))*x^2 )))))+csgn(I*ln(-2*ln(x)+ln(-3/4+(5+ln(x))*x^2)+1/2*I*Pi*csgn(I*x^2)*(-csg n(I*x^2)+csgn(I*x))^2-1/2*I*Pi*csgn(I/x^2*(-3/4+(5+ln(x))*x^2))*(-csgn(I/x ^2*(-3/4+(5+ln(x))*x^2))+csgn(I/x^2))*(-csgn(I/x^2*(-3/4+(5+ln(x))*x^2)...
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {24 x+16 x^3+e^x \left (12+8 x^2\right )+\left (6 x-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-8 x^3-4 e^x x^3\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )\right )}{\left (-12 x^3+80 x^5+e^{2 x} \left (-3 x+20 x^3\right )+e^x \left (-12 x^2+80 x^4\right )+\left (4 e^{2 x} x^3+16 e^x x^4+16 x^5\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )} \, dx=\frac {\log \left (\log \left (\log \left (\frac {4 \, x^{2} \log \left (x\right ) + 20 \, x^{2} - 3}{4 \, x^{2}}\right )\right )^{2}\right )}{2 \, x + e^{x}} \]
integrate((((-4*exp(x)*x^3-8*x^3)*log(x)+(-20*x^3+3*x)*exp(x)-40*x^3+6*x)* log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2)*log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x ^2))*log(log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2))^2)+(8*x^2+12)*exp(x)+16 *x^3+24*x)/((4*exp(x)^2*x^3+16*exp(x)*x^4+16*x^5)*log(x)+(20*x^3-3*x)*exp( x)^2+(80*x^4-12*x^2)*exp(x)+80*x^5-12*x^3)/log(1/4*(4*x^2*log(x)+20*x^2-3) /x^2)/log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2)),x, algorithm=\
Timed out. \[ \int \frac {24 x+16 x^3+e^x \left (12+8 x^2\right )+\left (6 x-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-8 x^3-4 e^x x^3\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )\right )}{\left (-12 x^3+80 x^5+e^{2 x} \left (-3 x+20 x^3\right )+e^x \left (-12 x^2+80 x^4\right )+\left (4 e^{2 x} x^3+16 e^x x^4+16 x^5\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )} \, dx=\text {Timed out} \]
integrate((((-4*exp(x)*x**3-8*x**3)*ln(x)+(-20*x**3+3*x)*exp(x)-40*x**3+6* x)*ln(1/4*(4*x**2*ln(x)+20*x**2-3)/x**2)*ln(ln(1/4*(4*x**2*ln(x)+20*x**2-3 )/x**2))*ln(ln(ln(1/4*(4*x**2*ln(x)+20*x**2-3)/x**2))**2)+(8*x**2+12)*exp( x)+16*x**3+24*x)/((4*exp(x)**2*x**3+16*exp(x)*x**4+16*x**5)*ln(x)+(20*x**3 -3*x)*exp(x)**2+(80*x**4-12*x**2)*exp(x)+80*x**5-12*x**3)/ln(1/4*(4*x**2*l n(x)+20*x**2-3)/x**2)/ln(ln(1/4*(4*x**2*ln(x)+20*x**2-3)/x**2)),x)
Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {24 x+16 x^3+e^x \left (12+8 x^2\right )+\left (6 x-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-8 x^3-4 e^x x^3\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )\right )}{\left (-12 x^3+80 x^5+e^{2 x} \left (-3 x+20 x^3\right )+e^x \left (-12 x^2+80 x^4\right )+\left (4 e^{2 x} x^3+16 e^x x^4+16 x^5\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )} \, dx=\frac {2 \, \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (4 \, x^{2} \log \left (x\right ) + 20 \, x^{2} - 3\right ) - 2 \, \log \left (x\right )\right )\right )}{2 \, x + e^{x}} \]
integrate((((-4*exp(x)*x^3-8*x^3)*log(x)+(-20*x^3+3*x)*exp(x)-40*x^3+6*x)* log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2)*log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x ^2))*log(log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2))^2)+(8*x^2+12)*exp(x)+16 *x^3+24*x)/((4*exp(x)^2*x^3+16*exp(x)*x^4+16*x^5)*log(x)+(20*x^3-3*x)*exp( x)^2+(80*x^4-12*x^2)*exp(x)+80*x^5-12*x^3)/log(1/4*(4*x^2*log(x)+20*x^2-3) /x^2)/log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2)),x, algorithm=\
Time = 2.53 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {24 x+16 x^3+e^x \left (12+8 x^2\right )+\left (6 x-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-8 x^3-4 e^x x^3\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )\right )}{\left (-12 x^3+80 x^5+e^{2 x} \left (-3 x+20 x^3\right )+e^x \left (-12 x^2+80 x^4\right )+\left (4 e^{2 x} x^3+16 e^x x^4+16 x^5\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )} \, dx=\frac {\log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (4 \, x^{2} \log \left (x\right ) + 20 \, x^{2} - 3\right ) - 2 \, \log \left (x\right )\right )^{2}\right )}{2 \, x + e^{x}} \]
integrate((((-4*exp(x)*x^3-8*x^3)*log(x)+(-20*x^3+3*x)*exp(x)-40*x^3+6*x)* log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2)*log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x ^2))*log(log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2))^2)+(8*x^2+12)*exp(x)+16 *x^3+24*x)/((4*exp(x)^2*x^3+16*exp(x)*x^4+16*x^5)*log(x)+(20*x^3-3*x)*exp( x)^2+(80*x^4-12*x^2)*exp(x)+80*x^5-12*x^3)/log(1/4*(4*x^2*log(x)+20*x^2-3) /x^2)/log(log(1/4*(4*x^2*log(x)+20*x^2-3)/x^2)),x, algorithm=\
Timed out. \[ \int \frac {24 x+16 x^3+e^x \left (12+8 x^2\right )+\left (6 x-40 x^3+e^x \left (3 x-20 x^3\right )+\left (-8 x^3-4 e^x x^3\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right ) \log \left (\log ^2\left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )\right )}{\left (-12 x^3+80 x^5+e^{2 x} \left (-3 x+20 x^3\right )+e^x \left (-12 x^2+80 x^4\right )+\left (4 e^{2 x} x^3+16 e^x x^4+16 x^5\right ) \log (x)\right ) \log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right ) \log \left (\log \left (\frac {-3+20 x^2+4 x^2 \log (x)}{4 x^2}\right )\right )} \, dx=-\int \frac {24\,x+{\mathrm {e}}^x\,\left (8\,x^2+12\right )+16\,x^3+\ln \left (\frac {x^2\,\ln \left (x\right )+5\,x^2-\frac {3}{4}}{x^2}\right )\,\ln \left ({\ln \left (\ln \left (\frac {x^2\,\ln \left (x\right )+5\,x^2-\frac {3}{4}}{x^2}\right )\right )}^2\right )\,\ln \left (\ln \left (\frac {x^2\,\ln \left (x\right )+5\,x^2-\frac {3}{4}}{x^2}\right )\right )\,\left (6\,x+{\mathrm {e}}^x\,\left (3\,x-20\,x^3\right )-40\,x^3-\ln \left (x\right )\,\left (4\,x^3\,{\mathrm {e}}^x+8\,x^3\right )\right )}{\ln \left (\frac {x^2\,\ln \left (x\right )+5\,x^2-\frac {3}{4}}{x^2}\right )\,\ln \left (\ln \left (\frac {x^2\,\ln \left (x\right )+5\,x^2-\frac {3}{4}}{x^2}\right )\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (3\,x-20\,x^3\right )+{\mathrm {e}}^x\,\left (12\,x^2-80\,x^4\right )-\ln \left (x\right )\,\left (16\,x^4\,{\mathrm {e}}^x+4\,x^3\,{\mathrm {e}}^{2\,x}+16\,x^5\right )+12\,x^3-80\,x^5\right )} \,d x \]
int(-(24*x + exp(x)*(8*x^2 + 12) + 16*x^3 + log((x^2*log(x) + 5*x^2 - 3/4) /x^2)*log(log(log((x^2*log(x) + 5*x^2 - 3/4)/x^2))^2)*log(log((x^2*log(x) + 5*x^2 - 3/4)/x^2))*(6*x + exp(x)*(3*x - 20*x^3) - 40*x^3 - log(x)*(4*x^3 *exp(x) + 8*x^3)))/(log((x^2*log(x) + 5*x^2 - 3/4)/x^2)*log(log((x^2*log(x ) + 5*x^2 - 3/4)/x^2))*(exp(2*x)*(3*x - 20*x^3) + exp(x)*(12*x^2 - 80*x^4) - log(x)*(16*x^4*exp(x) + 4*x^3*exp(2*x) + 16*x^5) + 12*x^3 - 80*x^5)),x)
-int((24*x + exp(x)*(8*x^2 + 12) + 16*x^3 + log((x^2*log(x) + 5*x^2 - 3/4) /x^2)*log(log(log((x^2*log(x) + 5*x^2 - 3/4)/x^2))^2)*log(log((x^2*log(x) + 5*x^2 - 3/4)/x^2))*(6*x + exp(x)*(3*x - 20*x^3) - 40*x^3 - log(x)*(4*x^3 *exp(x) + 8*x^3)))/(log((x^2*log(x) + 5*x^2 - 3/4)/x^2)*log(log((x^2*log(x ) + 5*x^2 - 3/4)/x^2))*(exp(2*x)*(3*x - 20*x^3) + exp(x)*(12*x^2 - 80*x^4) - log(x)*(16*x^4*exp(x) + 4*x^3*exp(2*x) + 16*x^5) + 12*x^3 - 80*x^5)), x )