Integrand size = 265, antiderivative size = 28 \[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=x+\frac {\log \left (2+3 \left (x+x^2\right )\right )}{4-e^{4 x}-\log (x)} \] Output:
ln(3*x^2+3*x+2)/(4-exp(2*x)^2-ln(x))+x
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=x-\frac {\log \left (2+3 x+3 x^2\right )}{-4+e^{4 x}+\log (x)} \] Input:
Integrate[(44*x + 72*x^2 + 48*x^3 + E^(4*x)*(-19*x - 30*x^2 - 24*x^3) + E^ (8*x)*(2*x + 3*x^2 + 3*x^3) + (-19*x - 30*x^2 - 24*x^3 + E^(4*x)*(4*x + 6* x^2 + 6*x^3))*Log[x] + (2*x + 3*x^2 + 3*x^3)*Log[x]^2 + (2 + 3*x + 3*x^2 + E^(4*x)*(8*x + 12*x^2 + 12*x^3))*Log[2 + 3*x + 3*x^2])/(32*x + 48*x^2 + 4 8*x^3 + E^(4*x)*(-16*x - 24*x^2 - 24*x^3) + E^(8*x)*(2*x + 3*x^2 + 3*x^3) + (-16*x - 24*x^2 - 24*x^3 + E^(4*x)*(4*x + 6*x^2 + 6*x^3))*Log[x] + (2*x + 3*x^2 + 3*x^3)*Log[x]^2),x]
Output:
x - Log[2 + 3*x + 3*x^2]/(-4 + E^(4*x) + Log[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.
\(\displaystyle \int \frac {48 x^3+72 x^2+e^{4 x} \left (-24 x^3-30 x^2-19 x\right )+e^{8 x} \left (3 x^3+3 x^2+2 x\right )+\left (3 x^3+3 x^2+2 x\right ) \log ^2(x)+\left (-24 x^3-30 x^2+e^{4 x} \left (6 x^3+6 x^2+4 x\right )-19 x\right ) \log (x)+\left (3 x^2+e^{4 x} \left (12 x^3+12 x^2+8 x\right )+3 x+2\right ) \log \left (3 x^2+3 x+2\right )+44 x}{48 x^3+48 x^2+e^{4 x} \left (-24 x^3-24 x^2-16 x\right )+e^{8 x} \left (3 x^3+3 x^2+2 x\right )+\left (3 x^3+3 x^2+2 x\right ) \log ^2(x)+\left (-24 x^3-24 x^2+e^{4 x} \left (6 x^3+6 x^2+4 x\right )-16 x\right ) \log (x)+32 x} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (e^{4 x}-4\right ) x \left (-12 x^2+e^{4 x} \left (3 x^2+3 x+2\right )-18 x-11\right )+x \left (3 x^2+3 x+2\right ) \log ^2(x)+x \left (-24 x^2+e^{4 x} \left (6 x^2+6 x+4\right )-30 x-19\right ) \log (x)+\left (4 e^{4 x} x+1\right ) \left (3 x^2+3 x+2\right ) \log \left (3 x^2+3 x+2\right )}{x \left (3 x^2+3 x+2\right ) \left (-e^{4 x}-\log (x)+4\right )^2}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (-\frac {(-16 x+4 x \log (x)-1) \log \left (3 x^2+3 x+2\right )}{x \left (e^{4 x}+\log (x)-4\right )^2}+\frac {12 x^2 \log \left (3 x^2+3 x+2\right )+12 x \log \left (3 x^2+3 x+2\right )+8 \log \left (3 x^2+3 x+2\right )-6 x-3}{\left (3 x^2+3 x+2\right ) \left (e^{4 x}+\log (x)-4\right )}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 16 \int \frac {\log \left (3 x^2+3 x+2\right )}{\left (\log (x)+e^{4 x}-4\right )^2}dx+\int \frac {\log \left (3 x^2+3 x+2\right )}{x \left (\log (x)+e^{4 x}-4\right )^2}dx-4 \int \frac {\log (x) \log \left (3 x^2+3 x+2\right )}{\left (\log (x)+e^{4 x}-4\right )^2}dx+4 \int \frac {\log \left (3 x^2+3 x+2\right )}{\log (x)+e^{4 x}-4}dx-6 i \sqrt {\frac {3}{5}} \int \frac {1}{\left (-6 x+i \sqrt {15}-3\right ) \left (\log (x)+e^{4 x}-4\right )}dx-\frac {6}{5} \left (5+i \sqrt {15}\right ) \int \frac {1}{\left (6 x-i \sqrt {15}+3\right ) \left (\log (x)+e^{4 x}-4\right )}dx-\frac {6}{5} \left (5-i \sqrt {15}\right ) \int \frac {1}{\left (6 x+i \sqrt {15}+3\right ) \left (\log (x)+e^{4 x}-4\right )}dx-6 i \sqrt {\frac {3}{5}} \int \frac {1}{\left (6 x+i \sqrt {15}+3\right ) \left (\log (x)+e^{4 x}-4\right )}dx+x\) |
Input:
Int[(44*x + 72*x^2 + 48*x^3 + E^(4*x)*(-19*x - 30*x^2 - 24*x^3) + E^(8*x)* (2*x + 3*x^2 + 3*x^3) + (-19*x - 30*x^2 - 24*x^3 + E^(4*x)*(4*x + 6*x^2 + 6*x^3))*Log[x] + (2*x + 3*x^2 + 3*x^3)*Log[x]^2 + (2 + 3*x + 3*x^2 + E^(4* x)*(8*x + 12*x^2 + 12*x^3))*Log[2 + 3*x + 3*x^2])/(32*x + 48*x^2 + 48*x^3 + E^(4*x)*(-16*x - 24*x^2 - 24*x^3) + E^(8*x)*(2*x + 3*x^2 + 3*x^3) + (-16 *x - 24*x^2 - 24*x^3 + E^(4*x)*(4*x + 6*x^2 + 6*x^3))*Log[x] + (2*x + 3*x^ 2 + 3*x^3)*Log[x]^2),x]
Output:
$Aborted
Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
\[-\frac {\ln \left (3 x^{2}+3 x +2\right )}{{\mathrm e}^{4 x}+\ln \left (x \right )-4}+x\]
Input:
int((((12*x^3+12*x^2+8*x)*exp(2*x)^2+3*x^2+3*x+2)*ln(3*x^2+3*x+2)+(3*x^3+3 *x^2+2*x)*ln(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2-24*x^3-30*x^2-19*x)*ln(x)+ (3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-30*x^2-19*x)*exp(2*x)^2+48*x^3+72*x^ 2+44*x)/((3*x^3+3*x^2+2*x)*ln(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2-24*x^3-24 *x^2-16*x)*ln(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-24*x^2-16*x)*exp(2* x)^2+48*x^3+48*x^2+32*x),x)
Output:
-1/(exp(4*x)+ln(x)-4)*ln(3*x^2+3*x+2)+x
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=\frac {x e^{\left (4 \, x\right )} + x \log \left (x\right ) - 4 \, x - \log \left (3 \, x^{2} + 3 \, x + 2\right )}{e^{\left (4 \, x\right )} + \log \left (x\right ) - 4} \] Input:
integrate((((12*x^3+12*x^2+8*x)*exp(2*x)^2+3*x^2+3*x+2)*log(3*x^2+3*x+2)+( 3*x^3+3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2-24*x^3-30*x^2-19*x )*log(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-30*x^2-19*x)*exp(2*x)^2+48* x^3+72*x^2+44*x)/((3*x^3+3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2 -24*x^3-24*x^2-16*x)*log(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-24*x^2-1 6*x)*exp(2*x)^2+48*x^3+48*x^2+32*x),x, algorithm="fricas")
Output:
(x*e^(4*x) + x*log(x) - 4*x - log(3*x^2 + 3*x + 2))/(e^(4*x) + log(x) - 4)
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=x - \frac {\log {\left (3 x^{2} + 3 x + 2 \right )}}{e^{4 x} + \log {\left (x \right )} - 4} \] Input:
integrate((((12*x**3+12*x**2+8*x)*exp(2*x)**2+3*x**2+3*x+2)*ln(3*x**2+3*x+ 2)+(3*x**3+3*x**2+2*x)*ln(x)**2+((6*x**3+6*x**2+4*x)*exp(2*x)**2-24*x**3-3 0*x**2-19*x)*ln(x)+(3*x**3+3*x**2+2*x)*exp(2*x)**4+(-24*x**3-30*x**2-19*x) *exp(2*x)**2+48*x**3+72*x**2+44*x)/((3*x**3+3*x**2+2*x)*ln(x)**2+((6*x**3+ 6*x**2+4*x)*exp(2*x)**2-24*x**3-24*x**2-16*x)*ln(x)+(3*x**3+3*x**2+2*x)*ex p(2*x)**4+(-24*x**3-24*x**2-16*x)*exp(2*x)**2+48*x**3+48*x**2+32*x),x)
Output:
x - log(3*x**2 + 3*x + 2)/(exp(4*x) + log(x) - 4)
Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=\frac {x e^{\left (4 \, x\right )} + x \log \left (x\right ) - 4 \, x - \log \left (3 \, x^{2} + 3 \, x + 2\right )}{e^{\left (4 \, x\right )} + \log \left (x\right ) - 4} \] Input:
integrate((((12*x^3+12*x^2+8*x)*exp(2*x)^2+3*x^2+3*x+2)*log(3*x^2+3*x+2)+( 3*x^3+3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2-24*x^3-30*x^2-19*x )*log(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-30*x^2-19*x)*exp(2*x)^2+48* x^3+72*x^2+44*x)/((3*x^3+3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2 -24*x^3-24*x^2-16*x)*log(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-24*x^2-1 6*x)*exp(2*x)^2+48*x^3+48*x^2+32*x),x, algorithm="maxima")
Output:
(x*e^(4*x) + x*log(x) - 4*x - log(3*x^2 + 3*x + 2))/(e^(4*x) + log(x) - 4)
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=\frac {x e^{\left (4 \, x\right )} + x \log \left (x\right ) - 4 \, x - \log \left (3 \, x^{2} + 3 \, x + 2\right )}{e^{\left (4 \, x\right )} + \log \left (x\right ) - 4} \] Input:
integrate((((12*x^3+12*x^2+8*x)*exp(2*x)^2+3*x^2+3*x+2)*log(3*x^2+3*x+2)+( 3*x^3+3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2-24*x^3-30*x^2-19*x )*log(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-30*x^2-19*x)*exp(2*x)^2+48* x^3+72*x^2+44*x)/((3*x^3+3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2 -24*x^3-24*x^2-16*x)*log(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-24*x^2-1 6*x)*exp(2*x)^2+48*x^3+48*x^2+32*x),x, algorithm="giac")
Output:
(x*e^(4*x) + x*log(x) - 4*x - log(3*x^2 + 3*x + 2))/(e^(4*x) + log(x) - 4)
Time = 3.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=x-\frac {\ln \left (3\,x^2+3\,x+2\right )}{{\mathrm {e}}^{4\,x}+\ln \left (x\right )-4} \] Input:
int((44*x - log(x)*(19*x - exp(4*x)*(4*x + 6*x^2 + 6*x^3) + 30*x^2 + 24*x^ 3) + exp(8*x)*(2*x + 3*x^2 + 3*x^3) - exp(4*x)*(19*x + 30*x^2 + 24*x^3) + log(x)^2*(2*x + 3*x^2 + 3*x^3) + 72*x^2 + 48*x^3 + log(3*x + 3*x^2 + 2)*(3 *x + exp(4*x)*(8*x + 12*x^2 + 12*x^3) + 3*x^2 + 2))/(32*x - log(x)*(16*x - exp(4*x)*(4*x + 6*x^2 + 6*x^3) + 24*x^2 + 24*x^3) + exp(8*x)*(2*x + 3*x^2 + 3*x^3) - exp(4*x)*(16*x + 24*x^2 + 24*x^3) + log(x)^2*(2*x + 3*x^2 + 3* x^3) + 48*x^2 + 48*x^3),x)
Output:
x - log(3*x + 3*x^2 + 2)/(exp(4*x) + log(x) - 4)
\[ \int \frac {44 x+72 x^2+48 x^3+e^{4 x} \left (-19 x-30 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-19 x-30 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)+\left (2+3 x+3 x^2+e^{4 x} \left (8 x+12 x^2+12 x^3\right )\right ) \log \left (2+3 x+3 x^2\right )}{32 x+48 x^2+48 x^3+e^{4 x} \left (-16 x-24 x^2-24 x^3\right )+e^{8 x} \left (2 x+3 x^2+3 x^3\right )+\left (-16 x-24 x^2-24 x^3+e^{4 x} \left (4 x+6 x^2+6 x^3\right )\right ) \log (x)+\left (2 x+3 x^2+3 x^3\right ) \log ^2(x)} \, dx=\text {too large to display} \] Input:
int((((12*x^3+12*x^2+8*x)*exp(2*x)^2+3*x^2+3*x+2)*log(3*x^2+3*x+2)+(3*x^3+ 3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2-24*x^3-30*x^2-19*x)*log( x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-30*x^2-19*x)*exp(2*x)^2+48*x^3+72 *x^2+44*x)/((3*x^3+3*x^2+2*x)*log(x)^2+((6*x^3+6*x^2+4*x)*exp(2*x)^2-24*x^ 3-24*x^2-16*x)*log(x)+(3*x^3+3*x^2+2*x)*exp(2*x)^4+(-24*x^3-24*x^2-16*x)*e xp(2*x)^2+48*x^3+48*x^2+32*x),x)
Output:
2*int(log(x)**2/(3*e**(8*x)*x**2 + 3*e**(8*x)*x + 2*e**(8*x) + 6*e**(4*x)* log(x)*x**2 + 6*e**(4*x)*log(x)*x + 4*e**(4*x)*log(x) - 24*e**(4*x)*x**2 - 24*e**(4*x)*x - 16*e**(4*x) + 3*log(x)**2*x**2 + 3*log(x)**2*x + 2*log(x) **2 - 24*log(x)*x**2 - 24*log(x)*x - 16*log(x) + 48*x**2 + 48*x + 32),x) + 2*int(e**(8*x)/(3*e**(8*x)*x**2 + 3*e**(8*x)*x + 2*e**(8*x) + 6*e**(4*x)* log(x)*x**2 + 6*e**(4*x)*log(x)*x + 4*e**(4*x)*log(x) - 24*e**(4*x)*x**2 - 24*e**(4*x)*x - 16*e**(4*x) + 3*log(x)**2*x**2 + 3*log(x)**2*x + 2*log(x) **2 - 24*log(x)*x**2 - 24*log(x)*x - 16*log(x) + 48*x**2 + 48*x + 32),x) - 19*int(e**(4*x)/(3*e**(8*x)*x**2 + 3*e**(8*x)*x + 2*e**(8*x) + 6*e**(4*x) *log(x)*x**2 + 6*e**(4*x)*log(x)*x + 4*e**(4*x)*log(x) - 24*e**(4*x)*x**2 - 24*e**(4*x)*x - 16*e**(4*x) + 3*log(x)**2*x**2 + 3*log(x)**2*x + 2*log(x )**2 - 24*log(x)*x**2 - 24*log(x)*x - 16*log(x) + 48*x**2 + 48*x + 32),x) + 48*int(x**2/(3*e**(8*x)*x**2 + 3*e**(8*x)*x + 2*e**(8*x) + 6*e**(4*x)*lo g(x)*x**2 + 6*e**(4*x)*log(x)*x + 4*e**(4*x)*log(x) - 24*e**(4*x)*x**2 - 2 4*e**(4*x)*x - 16*e**(4*x) + 3*log(x)**2*x**2 + 3*log(x)**2*x + 2*log(x)** 2 - 24*log(x)*x**2 - 24*log(x)*x - 16*log(x) + 48*x**2 + 48*x + 32),x) + 2 *int(log(3*x**2 + 3*x + 2)/(3*e**(8*x)*x**3 + 3*e**(8*x)*x**2 + 2*e**(8*x) *x + 6*e**(4*x)*log(x)*x**3 + 6*e**(4*x)*log(x)*x**2 + 4*e**(4*x)*log(x)*x - 24*e**(4*x)*x**3 - 24*e**(4*x)*x**2 - 16*e**(4*x)*x + 3*log(x)**2*x**3 + 3*log(x)**2*x**2 + 2*log(x)**2*x - 24*log(x)*x**3 - 24*log(x)*x**2 - ...