\(\int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log (\frac {1+x}{5})+\log (\frac {x^2}{3}) (-2 x-2 x^2+e^x (1+x+x^2)+e^x (-1-2 x-x^2) \log (\frac {1+x}{5})) \log (\log (\frac {x^2}{3}))}{(1+x) \log (\frac {x^2}{3})} \, dx\) [1741]

Optimal result

Integrand size = 110, antiderivative size = 31 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=x \left (e^x-x-e^x \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \] Output:

(exp(x)-x-exp(x)*ln(1/5*x+1/5))*ln(ln(1/3*x^2))*x
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-x \left (-e^x+x+e^x \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \] Input:

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

Output:

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

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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 11.55 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39

method	result	size
parallelrisch	\(-\ln \left (\ln \left (\frac {x^{2}}{3}\right )\right ) {\mathrm e}^{x} \ln \left (\frac {x}{5}+\frac {1}{5}\right ) x -\ln \left (\ln \left (\frac {x^{2}}{3}\right )\right ) x^{2}+\ln \left (\ln \left (\frac {x^{2}}{3}\right )\right ) x \,{\mathrm e}^{x}\)	\(43\)
risch	\(\left (-{\mathrm e}^{x} \ln \left (\frac {x}{5}+\frac {1}{5}\right ) x -x^{2}+{\mathrm e}^{x} x \right ) \ln \left (-\ln \left (3\right )+2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )\)	\(61\)

Input:

int((((-x^2-2*x-1)*exp(x)*ln(1/5*x+1/5)+(x^2+x+1)*exp(x)-2*x^2-2*x)*ln(1/3 
*x^2)*ln(ln(1/3*x^2))+(-2-2*x)*exp(x)*ln(1/5*x+1/5)+(2+2*x)*exp(x)-2*x^2-2 
*x)/(1+x)/ln(1/3*x^2),x,method=_RETURNVERBOSE)
 

Output:

-ln(ln(1/3*x^2))*exp(x)*ln(1/5*x+1/5)*x-ln(ln(1/3*x^2))*x^2+ln(ln(1/3*x^2) 
)*x*exp(x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-{\left (x e^{x} \log \left (\frac {1}{5} \, x + \frac {1}{5}\right ) + x^{2} - x e^{x}\right )} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) \] Input:

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

Output:

-(x*e^x*log(1/5*x + 1/5) + x^2 - x*e^x)*log(log(1/3*x^2))
 

Sympy [A] (verification not implemented)

Time = 34.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=- x^{2} \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )} + \left (- x \log {\left (\frac {x}{5} + \frac {1}{5} \right )} \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )} + x \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )}\right ) e^{x} \] Input:

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

Output:

-x**2*log(log(x**2/3)) + (-x*log(x/5 + 1/5)*log(log(x**2/3)) + x*log(log(x 
**2/3)))*exp(x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx={\left (x {\left (\log \left (5\right ) + 1\right )} e^{x} - x e^{x} \log \left (x + 1\right ) - x^{2}\right )} \log \left (-\log \left (3\right ) + 2 \, \log \left (x\right )\right ) \] Input:

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

Output:

(x*(log(5) + 1)*e^x - x*e^x*log(x + 1) - x^2)*log(-log(3) + 2*log(x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).

Time = 0.37 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=x e^{x} \log \left (5\right ) \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) - x e^{x} \log \left (x + 1\right ) \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) - x^{2} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) + x e^{x} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) \] Input:

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

Output:

x*e^x*log(5)*log(log(1/3*x^2)) - x*e^x*log(x + 1)*log(log(1/3*x^2)) - x^2* 
log(log(1/3*x^2)) + x*e^x*log(log(1/3*x^2))
 

Mupad [B] (verification not implemented)

Time = 4.45 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.74 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-\ln \left (\ln \left (\frac {x^2}{3}\right )\right )\,\left (\frac {x^4+x^3}{x\,\left (x+1\right )}+\frac {{\mathrm {e}}^x\,\left (x-x^3\right )}{x\,\left (x+1\right )}-\frac {{\mathrm {e}}^x\,\left (x^2+x\right )}{x\,\left (x+1\right )}+\frac {{\mathrm {e}}^x\,\ln \left (\frac {x}{5}+\frac {1}{5}\right )\,\left (x^3+x^2\right )}{x\,\left (x+1\right )}\right ) \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {x^{2}}{3}\right )\right ) x \left (-e^{x} \mathrm {log}\left (\frac {x}{5}+\frac {1}{5}\right )+e^{x}-x \right ) \] Input:

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

Output:

log(log(x**2/3))*x*( - e**x*log((x + 1)/5) + e**x - x)