\(\int \frac {(4 e^2 x+8 x^2+e^x (e^2 x+2 x^2)) \log (4 x+e^x x) \log (x \log (4 x+e^x x))+e^{4-x} \log (x \log (4 x+e^x x)) (4 e^2+4 x+e^x (x+x^2+e^2 (1+x))+(4 e^2+4 x+e^x (e^2+x)) \log (4 x+e^x x)+(4 x-4 e^2 x-4 x^2+e^x (x-e^2 x-x^2)) \log (4 x+e^x x) \log (x \log (4 x+e^x x)))}{(4 x+e^x x) \log (4 x+e^x x) \log (x \log (4 x+e^x x))} \, dx\) [2657]

Optimal result

Integrand size = 222, antiderivative size = 29 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=4+\left (e^2+x\right ) \left (x+e^{4-x} \log \left (x \log \left (\left (4+e^x\right ) x\right )\right )\right ) \] Output:

4+(x+exp(2))*(exp(ln(ln(ln((exp(x)+4)*x)*x))-x+4)+x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=e^2 x+x^2+e^{4-x} \left (e^2+x\right ) \log \left (x \log \left (\left (4+e^x\right ) x\right )\right ) \] Input:

Integrate[((4*E^2*x + 8*x^2 + E^x*(E^2*x + 2*x^2))*Log[4*x + E^x*x]*Log[x* 
Log[4*x + E^x*x]] + E^(4 - x)*Log[x*Log[4*x + E^x*x]]*(4*E^2 + 4*x + E^x*( 
x + x^2 + E^2*(1 + x)) + (4*E^2 + 4*x + E^x*(E^2 + x))*Log[4*x + E^x*x] + 
(4*x - 4*E^2*x - 4*x^2 + E^x*(x - E^2*x - x^2))*Log[4*x + E^x*x]*Log[x*Log 
[4*x + E^x*x]]))/((4*x + E^x*x)*Log[4*x + E^x*x]*Log[x*Log[4*x + E^x*x]]), 
x]
 

Output:

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(62\) vs. \(2(29)=58\).

Time = 14.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {7239, 7293, 2009}

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.

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

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {e^{-x} \left (2 e^x x^2+e^{x+2} x+e^4 x-e^4 \left (x+e^2-1\right ) x \log \left (x \log \left (\left (e^x+4\right ) x\right )\right )+\frac {e^4 \left (x+e^2\right ) \left (e^x (x+1)+4\right )}{\left (e^x+4\right ) \log \left (\left (e^x+4\right ) x\right )}+e^6\right )}{x}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[((4*E^2*x + 8*x^2 + E^x*(E^2*x + 2*x^2))*Log[4*x + E^x*x]*Log[x*Log[4* 
x + E^x*x]] + E^(4 - x)*Log[x*Log[4*x + E^x*x]]*(4*E^2 + 4*x + E^x*(x + x^ 
2 + E^2*(1 + x)) + (4*E^2 + 4*x + E^x*(E^2 + x))*Log[4*x + E^x*x] + (4*x - 
 4*E^2*x - 4*x^2 + E^x*(x - E^2*x - x^2))*Log[4*x + E^x*x]*Log[x*Log[4*x + 
 E^x*x]]))/((4*x + E^x*x)*Log[4*x + E^x*x]*Log[x*Log[4*x + E^x*x]]),x]
 

Output:

E^2*x + x^2 + E^(4 - x)*Log[x*Log[4*x + E^x*x]] - E^(4 - x)*(1 - E^2 - x)* 
Log[x*Log[4*x + E^x*x]]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.10 (sec) , antiderivative size = 2626, normalized size of antiderivative = 90.55

\[\text {Expression too large to display}\]

Input:

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

Output:

ln(Pi*csgn(I*(exp(x)+4)*x)^3+Pi*csgn(I*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x 
)+4)*x)+2*I*ln(x)-Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4) 
)*csgn(I*(exp(x)+4)*x)^2+2*I*ln(exp(x)+4))*(x+exp(2))*exp(-x+4)+1/2*(-I*ex 
p(6)*Pi-I*exp(6)*Pi*csgn(I*x*(Pi*csgn(I*(exp(x)+4)*x)^3+Pi*csgn(I*x)*csgn( 
I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)-Pi*csgn(I*x)*csgn(I*(exp(x)+4 
)*x)^2-Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2+2*I*ln(exp(x)+4)))^3-I 
*exp(6)*Pi*csgn(I*Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2-I*Pi*csgn(I 
*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)-I*Pi*csgn(I*(exp(x)+4)*x)^3+I* 
Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x)^2+2*ln(exp(x)+4)+2*ln(x))*csgn(I*x*(Pi*c 
sgn(I*(exp(x)+4)*x)^3+Pi*csgn(I*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x) 
+2*I*ln(x)-Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4))*csgn( 
I*(exp(x)+4)*x)^2+2*I*ln(exp(x)+4)))^2+I*exp(4)*x*Pi*csgn(I*Pi*csgn(I*(exp 
(x)+4))*csgn(I*(exp(x)+4)*x)^2-I*Pi*csgn(I*x)*csgn(I*(exp(x)+4))*csgn(I*(e 
xp(x)+4)*x)-I*Pi*csgn(I*(exp(x)+4)*x)^3+I*Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x 
)^2+2*ln(exp(x)+4)+2*ln(x))*csgn(I*x*(Pi*csgn(I*(exp(x)+4)*x)^3+Pi*csgn(I* 
x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)-Pi*csgn(I*x)*csgn(I*( 
exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2+2*I*ln(exp(x)+ 
4)))*csgn(I*x)+I*exp(4)*x*Pi*csgn(I*x*(Pi*csgn(I*(exp(x)+4)*x)^3+Pi*csgn(I 
*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)-Pi*csgn(I*x)*csgn(I* 
(exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2+2*I*ln(exp...
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx={\left ({\left (x^{2} + x e^{2}\right )} e^{x} + {\left (x e^{4} + e^{6}\right )} \log \left (x \log \left (x e^{x} + 4 \, x\right )\right )\right )} e^{\left (-x\right )} \] Input:

integrate(((((-exp(2)*x-x^2+x)*exp(x)-4*exp(2)*x-4*x^2+4*x)*log(exp(x)*x+4 
*x)*log(x*log(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*log(exp(x)*x 
+4*x)+((1+x)*exp(2)+x^2+x)*exp(x)+4*exp(2)+4*x)*exp(log(log(x*log(exp(x)*x 
+4*x)))-x+4)+((exp(2)*x+2*x^2)*exp(x)+4*exp(2)*x+8*x^2)*log(exp(x)*x+4*x)* 
log(x*log(exp(x)*x+4*x)))/(exp(x)*x+4*x)/log(exp(x)*x+4*x)/log(x*log(exp(x 
)*x+4*x)),x, algorithm="fricas")
 

Output:

((x^2 + x*e^2)*e^x + (x*e^4 + e^6)*log(x*log(x*e^x + 4*x)))*e^(-x)
 

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=x^{2} + x e^{2} + \left (x e^{4} + e^{6}\right ) e^{- x} \log {\left (x \log {\left (x e^{x} + 4 x \right )} \right )} \] Input:

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

Output:

x**2 + x*exp(2) + (x*exp(4) + exp(6))*exp(-x)*log(x*log(x*exp(x) + 4*x))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx={\left (x e^{4} + e^{6}\right )} e^{\left (-x\right )} \log \left (x\right ) + {\left (x e^{4} + e^{6}\right )} e^{\left (-x\right )} \log \left (\log \left (x\right ) + \log \left (e^{x} + 4\right )\right ) + x^{2} + x e^{2} \] Input:

integrate(((((-exp(2)*x-x^2+x)*exp(x)-4*exp(2)*x-4*x^2+4*x)*log(exp(x)*x+4 
*x)*log(x*log(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*log(exp(x)*x 
+4*x)+((1+x)*exp(2)+x^2+x)*exp(x)+4*exp(2)+4*x)*exp(log(log(x*log(exp(x)*x 
+4*x)))-x+4)+((exp(2)*x+2*x^2)*exp(x)+4*exp(2)*x+8*x^2)*log(exp(x)*x+4*x)* 
log(x*log(exp(x)*x+4*x)))/(exp(x)*x+4*x)/log(exp(x)*x+4*x)/log(x*log(exp(x 
)*x+4*x)),x, algorithm="maxima")
 

Output:

(x*e^4 + e^6)*e^(-x)*log(x) + (x*e^4 + e^6)*e^(-x)*log(log(x) + log(e^x + 
4)) + x^2 + x*e^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx={\left (x^{2} e^{x} + x e^{4} \log \left (x\right ) + x e^{4} \log \left (\log \left (x\right ) + \log \left (e^{x} + 4\right )\right ) + x e^{\left (x + 2\right )} + e^{6} \log \left (x\right ) + e^{6} \log \left (\log \left (x\right ) + \log \left (e^{x} + 4\right )\right )\right )} e^{\left (-x\right )} \] Input:

integrate(((((-exp(2)*x-x^2+x)*exp(x)-4*exp(2)*x-4*x^2+4*x)*log(exp(x)*x+4 
*x)*log(x*log(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*log(exp(x)*x 
+4*x)+((1+x)*exp(2)+x^2+x)*exp(x)+4*exp(2)+4*x)*exp(log(log(x*log(exp(x)*x 
+4*x)))-x+4)+((exp(2)*x+2*x^2)*exp(x)+4*exp(2)*x+8*x^2)*log(exp(x)*x+4*x)* 
log(x*log(exp(x)*x+4*x)))/(exp(x)*x+4*x)/log(exp(x)*x+4*x)/log(x*log(exp(x 
)*x+4*x)),x, algorithm="giac")
 

Output:

(x^2*e^x + x*e^4*log(x) + x*e^4*log(log(x) + log(e^x + 4)) + x*e^(x + 2) + 
 e^6*log(x) + e^6*log(log(x) + log(e^x + 4)))*e^(-x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=\int \frac {{\mathrm {e}}^{\ln \left (\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\right )-x+4}\,\left (4\,x+4\,{\mathrm {e}}^2+\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\left (4\,x+4\,{\mathrm {e}}^2+{\mathrm {e}}^x\,\left (x+{\mathrm {e}}^2\right )\right )+{\mathrm {e}}^x\,\left (x+{\mathrm {e}}^2\,\left (x+1\right )+x^2\right )-\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\,\left (4\,x\,{\mathrm {e}}^2-4\,x+{\mathrm {e}}^x\,\left (x\,{\mathrm {e}}^2-x+x^2\right )+4\,x^2\right )\right )+\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\,\left ({\mathrm {e}}^x\,\left (2\,x^2+{\mathrm {e}}^2\,x\right )+4\,x\,{\mathrm {e}}^2+8\,x^2\right )}{\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\,\left (4\,x+x\,{\mathrm {e}}^x\right )} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=\frac {e^{x} e^{2} x +e^{x} x^{2}+\mathrm {log}\left (\mathrm {log}\left (e^{x} x +4 x \right ) x \right ) e^{6}+\mathrm {log}\left (\mathrm {log}\left (e^{x} x +4 x \right ) x \right ) e^{4} x}{e^{x}} \] Input:

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

Output:

(e**x*e**2*x + e**x*x**2 + log(log(e**x*x + 4*x)*x)*e**6 + log(log(e**x*x 
+ 4*x)*x)*e**4*x)/e**x