\(\int \frac {1+e^{2 x} (x+x^2+3 x^3-x^4)+(-x+x^2+3 x^3-x^4) \log (x)+(-e^{2 x} x-x \log (x)) \log (e^{2 x}+\log (x))}{e^{2 x} (x^3-x^5)+(x^3-x^5) \log (x)+(-e^{2 x} x^2-x^2 \log (x)) \log (e^{2 x}+\log (x))+(e^{2 x} (-x^2+x^4)+(-x^2+x^4) \log (x)+(e^{2 x} x+x \log (x)) \log (e^{2 x}+\log (x))) \log (\frac {1}{2} (-x+x^3+\log (e^{2 x}+\log (x))))} \, dx\) [2940]

Optimal result

Integrand size = 198, antiderivative size = 26 \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\log \left (x-\log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )\right ) \] Output:

ln(x-ln(1/2*ln(ln(x)+exp(x)^2)+1/2*x^3-1/2*x))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\log \left (x+\log (2)-\log \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right ) \] Input:

Integrate[(1 + E^(2*x)*(x + x^2 + 3*x^3 - x^4) + (-x + x^2 + 3*x^3 - x^4)* 
Log[x] + (-(E^(2*x)*x) - x*Log[x])*Log[E^(2*x) + Log[x]])/(E^(2*x)*(x^3 - 
x^5) + (x^3 - x^5)*Log[x] + (-(E^(2*x)*x^2) - x^2*Log[x])*Log[E^(2*x) + Lo 
g[x]] + (E^(2*x)*(-x^2 + x^4) + (-x^2 + x^4)*Log[x] + (E^(2*x)*x + x*Log[x 
])*Log[E^(2*x) + Log[x]])*Log[(-x + x^3 + Log[E^(2*x) + Log[x]])/2]),x]
 

Output:

Log[x + Log[2] - Log[-x + x^3 + Log[E^(2*x) + Log[x]]]]
 

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {7292, 7235}

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.

\(\displaystyle \int \frac {e^{2 x} \left (-x^4+3 x^3+x^2+x\right )+\left (-x^4+3 x^3+x^2-x\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+1}{\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (e^{2 x} \left (x^4-x^2\right )+\left (x^4-x^2\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (x^3-x+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx\)

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7235

\(\displaystyle \log \left (-\log \left (x^3-x+\log \left (e^{2 x}+\log (x)\right )\right )+x+\log (2)\right )\)

Input:

Int[(1 + E^(2*x)*(x + x^2 + 3*x^3 - x^4) + (-x + x^2 + 3*x^3 - x^4)*Log[x] 
 + (-(E^(2*x)*x) - x*Log[x])*Log[E^(2*x) + Log[x]])/(E^(2*x)*(x^3 - x^5) + 
 (x^3 - x^5)*Log[x] + (-(E^(2*x)*x^2) - x^2*Log[x])*Log[E^(2*x) + Log[x]] 
+ (E^(2*x)*(-x^2 + x^4) + (-x^2 + x^4)*Log[x] + (E^(2*x)*x + x*Log[x])*Log 
[E^(2*x) + Log[x]])*Log[(-x + x^3 + Log[E^(2*x) + Log[x]])/2]),x]
 

Output:

Log[x + Log[2] - Log[-x + x^3 + Log[E^(2*x) + Log[x]]]]
 

Defintions of rubi rules used

rule 7235

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L 
og[RemoveContent[y, x]], x] /;  !FalseQ[q]]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00

\[\ln \left (\ln \left (\frac {\ln \left (\ln \left (x \right )+{\mathrm e}^{2 x}\right )}{2}+\frac {x^{3}}{2}-\frac {x}{2}\right )-x \right )\]

Input:

int(((-x*ln(x)-x*exp(x)^2)*ln(ln(x)+exp(x)^2)+(-x^4+3*x^3+x^2-x)*ln(x)+(-x 
^4+3*x^3+x^2+x)*exp(x)^2+1)/(((x*ln(x)+x*exp(x)^2)*ln(ln(x)+exp(x)^2)+(x^4 
-x^2)*ln(x)+(x^4-x^2)*exp(x)^2)*ln(1/2*ln(ln(x)+exp(x)^2)+1/2*x^3-1/2*x)+( 
-x^2*ln(x)-exp(x)^2*x^2)*ln(ln(x)+exp(x)^2)+(-x^5+x^3)*ln(x)+(-x^5+x^3)*ex 
p(x)^2),x)
 

Output:

ln(ln(1/2*ln(ln(x)+exp(2*x))+1/2*x^3-1/2*x)-x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\log \left (-x + \log \left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x + \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + \log \left (x\right )\right )\right )\right ) \] Input:

integrate(((-x*log(x)-x*exp(x)^2)*log(log(x)+exp(x)^2)+(-x^4+3*x^3+x^2-x)* 
log(x)+(-x^4+3*x^3+x^2+x)*exp(x)^2+1)/(((x*log(x)+x*exp(x)^2)*log(log(x)+e 
xp(x)^2)+(x^4-x^2)*log(x)+(x^4-x^2)*exp(x)^2)*log(1/2*log(log(x)+exp(x)^2) 
+1/2*x^3-1/2*x)+(-x^2*log(x)-exp(x)^2*x^2)*log(log(x)+exp(x)^2)+(-x^5+x^3) 
*log(x)+(-x^5+x^3)*exp(x)^2),x, algorithm="fricas")
 

Output:

log(-x + log(1/2*x^3 - 1/2*x + 1/2*log(e^(2*x) + log(x))))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\text {Timed out} \] Input:

integrate(((-x*ln(x)-x*exp(x)**2)*ln(ln(x)+exp(x)**2)+(-x**4+3*x**3+x**2-x 
)*ln(x)+(-x**4+3*x**3+x**2+x)*exp(x)**2+1)/(((x*ln(x)+x*exp(x)**2)*ln(ln(x 
)+exp(x)**2)+(x**4-x**2)*ln(x)+(x**4-x**2)*exp(x)**2)*ln(1/2*ln(ln(x)+exp( 
x)**2)+1/2*x**3-1/2*x)+(-x**2*ln(x)-exp(x)**2*x**2)*ln(ln(x)+exp(x)**2)+(- 
x**5+x**3)*ln(x)+(-x**5+x**3)*exp(x)**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\log \left (-x - \log \left (2\right ) + \log \left (x^{3} - x + \log \left (e^{\left (2 \, x\right )} + \log \left (x\right )\right )\right )\right ) \] Input:

integrate(((-x*log(x)-x*exp(x)^2)*log(log(x)+exp(x)^2)+(-x^4+3*x^3+x^2-x)* 
log(x)+(-x^4+3*x^3+x^2+x)*exp(x)^2+1)/(((x*log(x)+x*exp(x)^2)*log(log(x)+e 
xp(x)^2)+(x^4-x^2)*log(x)+(x^4-x^2)*exp(x)^2)*log(1/2*log(log(x)+exp(x)^2) 
+1/2*x^3-1/2*x)+(-x^2*log(x)-exp(x)^2*x^2)*log(log(x)+exp(x)^2)+(-x^5+x^3) 
*log(x)+(-x^5+x^3)*exp(x)^2),x, algorithm="maxima")
 

Output:

log(-x - log(2) + log(x^3 - x + log(e^(2*x) + log(x))))
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\log \left (x + \log \left (2\right ) - \log \left (x^{3} - x + \log \left (e^{\left (2 \, x\right )} + \log \left (x\right )\right )\right )\right ) \] Input:

integrate(((-x*log(x)-x*exp(x)^2)*log(log(x)+exp(x)^2)+(-x^4+3*x^3+x^2-x)* 
log(x)+(-x^4+3*x^3+x^2+x)*exp(x)^2+1)/(((x*log(x)+x*exp(x)^2)*log(log(x)+e 
xp(x)^2)+(x^4-x^2)*log(x)+(x^4-x^2)*exp(x)^2)*log(1/2*log(log(x)+exp(x)^2) 
+1/2*x^3-1/2*x)+(-x^2*log(x)-exp(x)^2*x^2)*log(log(x)+exp(x)^2)+(-x^5+x^3) 
*log(x)+(-x^5+x^3)*exp(x)^2),x, algorithm="giac")
 

Output:

log(x + log(2) - log(x^3 - x + log(e^(2*x) + log(x))))
 

Mupad [B] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\ln \left (\ln \left (\frac {\ln \left ({\mathrm {e}}^{2\,x}+\ln \left (x\right )\right )}{2}-\frac {x}{2}+\frac {x^3}{2}\right )-x\right ) \] Input:

int(-(log(x)*(x - x^2 - 3*x^3 + x^4) + log(exp(2*x) + log(x))*(x*exp(2*x) 
+ x*log(x)) - exp(2*x)*(x + x^2 + 3*x^3 - x^4) - 1)/(exp(2*x)*(x^3 - x^5) 
- log(log(exp(2*x) + log(x))/2 - x/2 + x^3/2)*(exp(2*x)*(x^2 - x^4) - log( 
exp(2*x) + log(x))*(x*exp(2*x) + x*log(x)) + log(x)*(x^2 - x^4)) - log(exp 
(2*x) + log(x))*(x^2*log(x) + x^2*exp(2*x)) + log(x)*(x^3 - x^5)),x)
 

Output:

log(log(log(exp(2*x) + log(x))/2 - x/2 + x^3/2) - x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1+e^{2 x} \left (x+x^2+3 x^3-x^4\right )+\left (-x+x^2+3 x^3-x^4\right ) \log (x)+\left (-e^{2 x} x-x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )}{e^{2 x} \left (x^3-x^5\right )+\left (x^3-x^5\right ) \log (x)+\left (-e^{2 x} x^2-x^2 \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )+\left (e^{2 x} \left (-x^2+x^4\right )+\left (-x^2+x^4\right ) \log (x)+\left (e^{2 x} x+x \log (x)\right ) \log \left (e^{2 x}+\log (x)\right )\right ) \log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (e^{2 x}+\mathrm {log}\left (x \right )\right )}{2}+\frac {x^{3}}{2}-\frac {x}{2}\right )-x \right ) \] Input:

int(((-x*log(x)-x*exp(x)^2)*log(log(x)+exp(x)^2)+(-x^4+3*x^3+x^2-x)*log(x) 
+(-x^4+3*x^3+x^2+x)*exp(x)^2+1)/(((x*log(x)+x*exp(x)^2)*log(log(x)+exp(x)^ 
2)+(x^4-x^2)*log(x)+(x^4-x^2)*exp(x)^2)*log(1/2*log(log(x)+exp(x)^2)+1/2*x 
^3-1/2*x)+(-x^2*log(x)-exp(x)^2*x^2)*log(log(x)+exp(x)^2)+(-x^5+x^3)*log(x 
)+(-x^5+x^3)*exp(x)^2),x)
 

Output:

log(log((log(e**(2*x) + log(x)) + x**3 - x)/2) - x)