\(\int \frac {(-x-16 e x^2-64 x^4) \log (x)+(-e^2-x-8 e x^2-16 x^4+(e^2+x+8 e x^2+16 x^4) \log (x)) \log (e^2+x+8 e x^2+16 x^4)+(e^2+x+8 e x^2+16 x^4+(-e^2-x-8 e x^2-16 x^4) \log (x)+(e^2+x+8 e x^2+16 x^4) \log ^2(x)) \log ^2(e^2+x+8 e x^2+16 x^4)}{(e^2+x+8 e x^2+16 x^4) \log ^2(x) \log ^2(e^2+x+8 e x^2+16 x^4)} \, dx\) [2173]

Optimal result

Integrand size = 198, antiderivative size = 27 \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=x+\frac {-x+\frac {x}{\log \left (x+\left (e+4 x^2\right )^2\right )}}{\log (x)} \] Output:

x+(x/ln(x+(exp(1)+4*x^2)^2)-x)/ln(x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=x-\frac {x}{\log (x)}+\frac {x}{\log (x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \] Input:

Integrate[((-x - 16*E*x^2 - 64*x^4)*Log[x] + (-E^2 - x - 8*E*x^2 - 16*x^4 
+ (E^2 + x + 8*E*x^2 + 16*x^4)*Log[x])*Log[E^2 + x + 8*E*x^2 + 16*x^4] + ( 
E^2 + x + 8*E*x^2 + 16*x^4 + (-E^2 - x - 8*E*x^2 - 16*x^4)*Log[x] + (E^2 + 
 x + 8*E*x^2 + 16*x^4)*Log[x]^2)*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2)/((E^2 
+ x + 8*E*x^2 + 16*x^4)*Log[x]^2*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2),x]
 

Output:

x - x/Log[x] + x/(Log[x]*Log[E^2 + x + 8*E*x^2 + 16*x^4])
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (16 x^4+8 e x^2+\left (16 x^4+8 e x^2+x+e^2\right ) \log ^2(x)+\left (-16 x^4-8 e x^2-x-e^2\right ) \log (x)+x+e^2\right ) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )+\left (-16 x^4-8 e x^2+\left (16 x^4+8 e x^2+x+e^2\right ) \log (x)-x-e^2\right ) \log \left (16 x^4+8 e x^2+x+e^2\right )+\left (-64 x^4-16 e x^2-x\right ) \log (x)}{\left (16 x^4+8 e x^2+x+e^2\right ) \log ^2(x) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\frac {\log (x)-1}{\log \left (16 x^4+8 e x^2+x+e^2\right )}-\frac {x \left (64 x^3+16 e x+1\right ) \log (x)}{\left (16 x^4+8 e x^2+x+e^2\right ) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )}+\log ^2(x)-\log (x)+1}{\log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\log (x)-1}{\log ^2(x) \log \left (16 x^4+8 e x^2+x+e^2\right )}-\frac {x \left (64 x^3+16 e x+1\right )}{\left (16 x^4+8 e x^2+x+e^2\right ) \log (x) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )}+\frac {\log ^2(x)-\log (x)+1}{\log ^2(x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \int \frac {1}{\log (x) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )}dx+4 e^2 \int \frac {1}{\left (16 x^4+8 e x^2+x+e^2\right ) \log (x) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )}dx+3 \int \frac {x}{\left (16 x^4+8 e x^2+x+e^2\right ) \log (x) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )}dx+16 e \int \frac {x^2}{\left (16 x^4+8 e x^2+x+e^2\right ) \log (x) \log ^2\left (16 x^4+8 e x^2+x+e^2\right )}dx-\int \frac {1}{\log ^2(x) \log \left (16 x^4+8 e x^2+x+e^2\right )}dx+\int \frac {1}{\log (x) \log \left (16 x^4+8 e x^2+x+e^2\right )}dx+x-\frac {x}{\log (x)}\)

Input:

Int[((-x - 16*E*x^2 - 64*x^4)*Log[x] + (-E^2 - x - 8*E*x^2 - 16*x^4 + (E^2 
 + x + 8*E*x^2 + 16*x^4)*Log[x])*Log[E^2 + x + 8*E*x^2 + 16*x^4] + (E^2 + 
x + 8*E*x^2 + 16*x^4 + (-E^2 - x - 8*E*x^2 - 16*x^4)*Log[x] + (E^2 + x + 8 
*E*x^2 + 16*x^4)*Log[x]^2)*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2)/((E^2 + x + 
8*E*x^2 + 16*x^4)*Log[x]^2*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37

\[x -\frac {x}{\ln \left (x \right )}+\frac {x}{\ln \left ({\mathrm e}^{2}+8 x^{2} {\mathrm e}+16 x^{4}+x \right ) \ln \left (x \right )}\]

Input:

int((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*ln(x)^2+(-exp(1)^2-8*x^2*exp(1)-16* 
x^4-x)*ln(x)+exp(1)^2+8*x^2*exp(1)+16*x^4+x)*ln(exp(1)^2+8*x^2*exp(1)+16*x 
^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*ln(x)-exp(1)^2-8*x^2*exp(1)-16*x 
^4-x)*ln(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4-x)*ln(x))/ 
(exp(1)^2+8*x^2*exp(1)+16*x^4+x)/ln(x)^2/ln(exp(1)^2+8*x^2*exp(1)+16*x^4+x 
)^2,x)
 

Output:

x-x/ln(x)+x/ln(exp(1)^2+8*x^2*exp(1)+16*x^4+x)/ln(x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=\frac {{\left (x \log \left (x\right ) - x\right )} \log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) + x}{\log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) \log \left (x\right )} \] Input:

integrate((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)^2+(-exp(1)^2-8*x^2*exp 
(1)-16*x^4-x)*log(x)+exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(exp(1)^2+8*x^2*ex 
p(1)+16*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)-exp(1)^2-8*x^2*e 
xp(1)-16*x^4-x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4 
-x)*log(x))/(exp(1)^2+8*x^2*exp(1)+16*x^4+x)/log(x)^2/log(exp(1)^2+8*x^2*e 
xp(1)+16*x^4+x)^2,x, algorithm="fricas")
 

Output:

((x*log(x) - x)*log(16*x^4 + 8*x^2*e + x + e^2) + x)/(log(16*x^4 + 8*x^2*e 
 + x + e^2)*log(x))
 

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=x - \frac {x}{\log {\left (x \right )}} + \frac {x}{\log {\left (x \right )} \log {\left (16 x^{4} + 8 e x^{2} + x + e^{2} \right )}} \] Input:

integrate((((exp(1)**2+8*x**2*exp(1)+16*x**4+x)*ln(x)**2+(-exp(1)**2-8*x** 
2*exp(1)-16*x**4-x)*ln(x)+exp(1)**2+8*x**2*exp(1)+16*x**4+x)*ln(exp(1)**2+ 
8*x**2*exp(1)+16*x**4+x)**2+((exp(1)**2+8*x**2*exp(1)+16*x**4+x)*ln(x)-exp 
(1)**2-8*x**2*exp(1)-16*x**4-x)*ln(exp(1)**2+8*x**2*exp(1)+16*x**4+x)+(-16 
*x**2*exp(1)-64*x**4-x)*ln(x))/(exp(1)**2+8*x**2*exp(1)+16*x**4+x)/ln(x)** 
2/ln(exp(1)**2+8*x**2*exp(1)+16*x**4+x)**2,x)
 

Output:

x - x/log(x) + x/(log(x)*log(16*x**4 + 8*E*x**2 + x + exp(2)))
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=\frac {{\left (x \log \left (x\right ) - x\right )} \log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) + x}{\log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) \log \left (x\right )} \] Input:

integrate((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)^2+(-exp(1)^2-8*x^2*exp 
(1)-16*x^4-x)*log(x)+exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(exp(1)^2+8*x^2*ex 
p(1)+16*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)-exp(1)^2-8*x^2*e 
xp(1)-16*x^4-x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4 
-x)*log(x))/(exp(1)^2+8*x^2*exp(1)+16*x^4+x)/log(x)^2/log(exp(1)^2+8*x^2*e 
xp(1)+16*x^4+x)^2,x, algorithm="maxima")
 

Output:

((x*log(x) - x)*log(16*x^4 + 8*x^2*e + x + e^2) + x)/(log(16*x^4 + 8*x^2*e 
 + x + e^2)*log(x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).

Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.48 \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=\frac {x \log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) \log \left (x\right ) - x \log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) + x}{\log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) \log \left (x\right )} \] Input:

integrate((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)^2+(-exp(1)^2-8*x^2*exp 
(1)-16*x^4-x)*log(x)+exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(exp(1)^2+8*x^2*ex 
p(1)+16*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)-exp(1)^2-8*x^2*e 
xp(1)-16*x^4-x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4 
-x)*log(x))/(exp(1)^2+8*x^2*exp(1)+16*x^4+x)/log(x)^2/log(exp(1)^2+8*x^2*e 
xp(1)+16*x^4+x)^2,x, algorithm="giac")
 

Output:

(x*log(16*x^4 + 8*x^2*e + x + e^2)*log(x) - x*log(16*x^4 + 8*x^2*e + x + e 
^2) + x)/(log(16*x^4 + 8*x^2*e + x + e^2)*log(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=-\int \frac {\left (\ln \left (x\right )\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )-{\mathrm {e}}^2-x-8\,x^2\,\mathrm {e}-{\ln \left (x\right )}^2\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )-16\,x^4\right )\,{\ln \left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )}^2+\left (x+{\mathrm {e}}^2-\ln \left (x\right )\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )+8\,x^2\,\mathrm {e}+16\,x^4\right )\,\ln \left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )+\ln \left (x\right )\,\left (64\,x^4+16\,\mathrm {e}\,x^2+x\right )}{{\ln \left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )}^2\,{\ln \left (x\right )}^2\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )} \,d x \] Input:

int(-(log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)*(x + exp(2) - log(x)*(x + ex 
p(2) + 8*x^2*exp(1) + 16*x^4) + 8*x^2*exp(1) + 16*x^4) - log(x + exp(2) + 
8*x^2*exp(1) + 16*x^4)^2*(x + exp(2) - log(x)*(x + exp(2) + 8*x^2*exp(1) + 
 16*x^4) + 8*x^2*exp(1) + log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4) + 
16*x^4) + log(x)*(x + 16*x^2*exp(1) + 64*x^4))/(log(x + exp(2) + 8*x^2*exp 
(1) + 16*x^4)^2*log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4)),x)
                                                                                    
                                                                                    
 

Output:

-int((log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)*(x + exp(2) - log(x)*(x + ex 
p(2) + 8*x^2*exp(1) + 16*x^4) + 8*x^2*exp(1) + 16*x^4) - log(x + exp(2) + 
8*x^2*exp(1) + 16*x^4)^2*(x + exp(2) - log(x)*(x + exp(2) + 8*x^2*exp(1) + 
 16*x^4) + 8*x^2*exp(1) + log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4) + 
16*x^4) + log(x)*(x + 16*x^2*exp(1) + 64*x^4))/(log(x + exp(2) + 8*x^2*exp 
(1) + 16*x^4)^2*log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx=\frac {x \left (\mathrm {log}\left (16 x^{4}+8 e \,x^{2}+e^{2}+x \right ) \mathrm {log}\left (x \right )-\mathrm {log}\left (16 x^{4}+8 e \,x^{2}+e^{2}+x \right )+1\right )}{\mathrm {log}\left (16 x^{4}+8 e \,x^{2}+e^{2}+x \right ) \mathrm {log}\left (x \right )} \] Input:

int((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)^2+(-exp(1)^2-8*x^2*exp(1)-16 
*x^4-x)*log(x)+exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(exp(1)^2+8*x^2*exp(1)+1 
6*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)-exp(1)^2-8*x^2*exp(1)- 
16*x^4-x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4-x)*lo 
g(x))/(exp(1)^2+8*x^2*exp(1)+16*x^4+x)/log(x)^2/log(exp(1)^2+8*x^2*exp(1)+ 
16*x^4+x)^2,x)
 

Output:

(x*(log(e**2 + 8*e*x**2 + 16*x**4 + x)*log(x) - log(e**2 + 8*e*x**2 + 16*x 
**4 + x) + 1))/(log(e**2 + 8*e*x**2 + 16*x**4 + x)*log(x))