\(\int \frac {e^{-4+\frac {-2 x+e^4 \log (x+x^2) \log (\log (x))}{e^4 \log (x+x^2)}} ((12 x+24 x^2) \log (x)+(-12 x-12 x^2) \log (x) \log (x+x^2)+e^4 (6+6 x) \log ^2(x+x^2))}{(x+x^2) \log (x) \log ^2(x+x^2)} \, dx\) [1476]

Optimal result

Integrand size = 101, antiderivative size = 20 \[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=6 e^{-\frac {2 x}{e^4 \log \left (x+x^2\right )}} \log (x) \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=6 e^{-\frac {2 x}{e^4 \log (x (1+x))}} \log (x) \] Input:

Integrate[(E^(-4 + (-2*x + E^4*Log[x + x^2]*Log[Log[x]])/(E^4*Log[x + x^2] 
))*((12*x + 24*x^2)*Log[x] + (-12*x - 12*x^2)*Log[x]*Log[x + x^2] + E^4*(6 
 + 6*x)*Log[x + x^2]^2))/((x + x^2)*Log[x]*Log[x + x^2]^2),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\left (e^4 (6 x+6) \log ^2\left (x^2+x\right )+\left (-12 x^2-12 x\right ) \log (x) \log \left (x^2+x\right )+\left (24 x^2+12 x\right ) \log (x)\right ) \exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{\left (x^2+x\right ) \log (x) \log ^2\left (x^2+x\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (e^4 (6 x+6) \log ^2\left (x^2+x\right )+\left (-12 x^2-12 x\right ) \log (x) \log \left (x^2+x\right )+\left (24 x^2+12 x\right ) \log (x)\right ) \exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{x (x+1) \log (x) \log ^2\left (x^2+x\right )}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (e^4 (6 x+6) \log ^2\left (x^2+x\right )+\left (-12 x^2-12 x\right ) \log (x) \log \left (x^2+x\right )+\left (24 x^2+12 x\right ) \log (x)\right ) \exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{x (x+1) \log (x) \log ^2(x (x+1))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {12 (2 x+1) \exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{(x+1) \log ^2(x (x+1))}+\frac {6 \exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}\right )}{x \log (x)}-\frac {12 \exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{\log (x (x+1))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 24 \int \frac {\exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{\log ^2(x (x+1))}dx-12 \int \frac {\exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{(x+1) \log ^2(x (x+1))}dx-12 \int \frac {\exp \left (\frac {e^4 \log \left (x^2+x\right ) \log (\log (x))-2 x}{e^4 \log \left (x^2+x\right )}-4\right )}{\log (x (x+1))}dx+6 \int \frac {e^{-\frac {2 x}{e^4 \log (x (x+1))}}}{x}dx\)

Input:

Int[(E^(-4 + (-2*x + E^4*Log[x + x^2]*Log[Log[x]])/(E^4*Log[x + x^2]))*((1 
2*x + 24*x^2)*Log[x] + (-12*x - 12*x^2)*Log[x]*Log[x + x^2] + E^4*(6 + 6*x 
)*Log[x + x^2]^2))/((x + x^2)*Log[x]*Log[x + x^2]^2),x]
 

Output:

$Aborted
 

Maple [F]

\[\int \frac {\left (\left (6+6 x \right ) {\mathrm e}^{4} \ln \left (x^{2}+x \right )^{2}+\left (-12 x^{2}-12 x \right ) \ln \left (x \right ) \ln \left (x^{2}+x \right )+\left (24 x^{2}+12 x \right ) \ln \left (x \right )\right ) {\mathrm e}^{\frac {\left ({\mathrm e}^{4} \ln \left (x^{2}+x \right ) \ln \left (\ln \left (x \right )\right )-2 x \right ) {\mathrm e}^{-4}}{\ln \left (x^{2}+x \right )}} {\mathrm e}^{-4}}{\left (x^{2}+x \right ) \ln \left (x \right ) \ln \left (x^{2}+x \right )^{2}}d x\]

Input:

int(((6+6*x)*exp(4)*ln(x^2+x)^2+(-12*x^2-12*x)*ln(x)*ln(x^2+x)+(24*x^2+12* 
x)*ln(x))*exp((exp(4)*ln(x^2+x)*ln(ln(x))-2*x)/exp(4)/ln(x^2+x))/(x^2+x)/e 
xp(4)/ln(x)/ln(x^2+x)^2,x)
 

Output:

int(((6+6*x)*exp(4)*ln(x^2+x)^2+(-12*x^2-12*x)*ln(x)*ln(x^2+x)+(24*x^2+12* 
x)*ln(x))*exp((exp(4)*ln(x^2+x)*ln(ln(x))-2*x)/exp(4)/ln(x^2+x))/(x^2+x)/e 
xp(4)/ln(x)/ln(x^2+x)^2,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.10 \[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=6 \, e^{\left (\frac {{\left (e^{4} \log \left (x^{2} + x\right ) \log \left (\log \left (x\right )\right ) - 4 \, e^{4} \log \left (x^{2} + x\right ) - 2 \, x\right )} e^{\left (-4\right )}}{\log \left (x^{2} + x\right )} + 4\right )} \] Input:

integrate(((6+6*x)*exp(4)*log(x^2+x)^2+(-12*x^2-12*x)*log(x)*log(x^2+x)+(2 
4*x^2+12*x)*log(x))*exp((exp(4)*log(x^2+x)*log(log(x))-2*x)/exp(4)/log(x^2 
+x))/(x^2+x)/exp(4)/log(x)/log(x^2+x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(((6+6*x)*exp(4)*ln(x**2+x)**2+(-12*x**2-12*x)*ln(x)*ln(x**2+x)+( 
24*x**2+12*x)*ln(x))*exp((exp(4)*ln(x**2+x)*ln(ln(x))-2*x)/exp(4)/ln(x**2+ 
x))/(x**2+x)/exp(4)/ln(x)/ln(x**2+x)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(((6+6*x)*exp(4)*log(x^2+x)^2+(-12*x^2-12*x)*log(x)*log(x^2+x)+(2 
4*x^2+12*x)*log(x))*exp((exp(4)*log(x^2+x)*log(log(x))-2*x)/exp(4)/log(x^2 
+x))/(x^2+x)/exp(4)/log(x)/log(x^2+x)^2,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [F]

\[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=\int { \frac {6 \, {\left ({\left (x + 1\right )} e^{4} \log \left (x^{2} + x\right )^{2} - 2 \, {\left (x^{2} + x\right )} \log \left (x^{2} + x\right ) \log \left (x\right ) + 2 \, {\left (2 \, x^{2} + x\right )} \log \left (x\right )\right )} e^{\left (\frac {{\left (e^{4} \log \left (x^{2} + x\right ) \log \left (\log \left (x\right )\right ) - 2 \, x\right )} e^{\left (-4\right )}}{\log \left (x^{2} + x\right )} - 4\right )}}{{\left (x^{2} + x\right )} \log \left (x^{2} + x\right )^{2} \log \left (x\right )} \,d x } \] Input:

integrate(((6+6*x)*exp(4)*log(x^2+x)^2+(-12*x^2-12*x)*log(x)*log(x^2+x)+(2 
4*x^2+12*x)*log(x))*exp((exp(4)*log(x^2+x)*log(log(x))-2*x)/exp(4)/log(x^2 
+x))/(x^2+x)/exp(4)/log(x)/log(x^2+x)^2,x, algorithm="giac")
 

Output:

undef
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=\int \frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-4}\,\left (2\,x-\ln \left (x^2+x\right )\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^4\right )}{\ln \left (x^2+x\right )}}\,{\mathrm {e}}^{-4}\,\left ({\mathrm {e}}^4\,\left (6\,x+6\right )\,{\ln \left (x^2+x\right )}^2-\ln \left (x\right )\,\left (12\,x^2+12\,x\right )\,\ln \left (x^2+x\right )+\ln \left (x\right )\,\left (24\,x^2+12\,x\right )\right )}{{\ln \left (x^2+x\right )}^2\,\ln \left (x\right )\,\left (x^2+x\right )} \,d x \] Input:

int((exp(-(exp(-4)*(2*x - log(x + x^2)*log(log(x))*exp(4)))/log(x + x^2))* 
exp(-4)*(log(x)*(12*x + 24*x^2) + log(x + x^2)^2*exp(4)*(6*x + 6) - log(x 
+ x^2)*log(x)*(12*x + 12*x^2)))/(log(x + x^2)^2*log(x)*(x + x^2)),x)
 

Output:

int((exp(-(exp(-4)*(2*x - log(x + x^2)*log(log(x))*exp(4)))/log(x + x^2))* 
exp(-4)*(log(x)*(12*x + 24*x^2) + log(x + x^2)^2*exp(4)*(6*x + 6) - log(x 
+ x^2)*log(x)*(12*x + 12*x^2)))/(log(x + x^2)^2*log(x)*(x + x^2)), x)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}} \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{\left (x+x^2\right ) \log (x) \log ^2\left (x+x^2\right )} \, dx=\frac {6 \,\mathrm {log}\left (x \right )}{e^{\frac {2 x}{\mathrm {log}\left (x^{2}+x \right ) e^{4}}}} \] Input:

int(((6+6*x)*exp(4)*log(x^2+x)^2+(-12*x^2-12*x)*log(x)*log(x^2+x)+(24*x^2+ 
12*x)*log(x))*exp((exp(4)*log(x^2+x)*log(log(x))-2*x)/exp(4)/log(x^2+x))/( 
x^2+x)/exp(4)/log(x)/log(x^2+x)^2,x)
 

Output:

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