\(\int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log (\frac {1+x}{5})+\log (\frac {x^2}{3}) (-2 x-2 x^2+e^x (1+x+x^2)+e^x (-1-2 x-x^2) \log (\frac {1+x}{5})) \log (\log (\frac {x^2}{3}))}{(1+x) \log (\frac {x^2}{3})} \, dx\) [4741]

Optimal result

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

[Out]

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

Rubi [F]

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

[In]

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]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \left (x+x \log \left (\frac {x^2}{3}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )\right )}{\log \left (\frac {x^2}{3}\right )}-\frac {e^x \left (-2-2 x+2 \log \left (\frac {1+x}{5}\right )+2 x \log \left (\frac {1+x}{5}\right )-\log \left (\frac {x^2}{3}\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 )-x^2 \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 {1+x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+2 x \log \left (\frac {x^2}{3}\right ) \log \left (\frac {1+x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+x^2 \log \left (\frac {x^2}{3}\right ) \log \left (\frac {1+x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )}\right ) \, dx \\ & = -\left (2 \int \frac {x+x \log \left (\frac {x^2}{3}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{\log \left (\frac {x^2}{3}\right )} \, dx\right )-\int \frac {e^x \left (-2-2 x+2 \log \left (\frac {1+x}{5}\right )+2 x \log \left (\frac {1+x}{5}\right )-\log \left (\frac {x^2}{3}\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 )-x^2 \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 {1+x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+2 x \log \left (\frac {x^2}{3}\right ) \log \left (\frac {1+x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+x^2 \log \left (\frac {x^2}{3}\right ) \log \left (\frac {1+x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx \\ & = -\left (2 \int x \left (\frac {1}{\log \left (\frac {x^2}{3}\right )}+\log \left (\log \left (\frac {x^2}{3}\right )\right )\right ) \, dx\right )-\int \frac {e^x \left (-2 (1+x)-\left (1+x+x^2\right ) \log \left (\frac {x^2}{3}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+(1+x) \log \left (\frac {1+x}{5}\right ) \left (2+(1+x) \log \left (\frac {x^2}{3}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx \\ & = -\left (2 \int \left (\frac {x}{\log \left (\frac {x^2}{3}\right )}+x \log \left (\log \left (\frac {x^2}{3}\right )\right )\right ) \, dx\right )-\int \left (\frac {2 e^x \left (-1+\log \left (\frac {1+x}{5}\right )\right )}{\log \left (\frac {x^2}{3}\right )}+\frac {e^x \left (-1-x-x^2+\log \left (\frac {1+x}{5}\right )+2 x \log \left (\frac {1+x}{5}\right )+x^2 \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}\right ) \, dx \\ & = -\left (2 \int \frac {x}{\log \left (\frac {x^2}{3}\right )} \, dx\right )-2 \int \frac {e^x \left (-1+\log \left (\frac {1+x}{5}\right )\right )}{\log \left (\frac {x^2}{3}\right )} \, dx-2 \int x \log \left (\log \left (\frac {x^2}{3}\right )\right ) \, dx-\int \frac {e^x \left (-1-x-x^2+\log \left (\frac {1+x}{5}\right )+2 x \log \left (\frac {1+x}{5}\right )+x^2 \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx \\ & = -x^2 \log \left (\log \left (\frac {x^2}{3}\right )\right )+2 \int \frac {x}{\log \left (\frac {x^2}{3}\right )} \, dx-2 \int \left (\frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right )}{\log \left (\frac {x^2}{3}\right )}+\frac {e^x}{\log (3)-\log \left (x^2\right )}\right ) \, dx-\int \frac {e^x \left (-1-x-x^2+(1+x)^2 \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx-\text {Subst}\left (\int \frac {1}{\log \left (\frac {x}{3}\right )} \, dx,x,x^2\right ) \\ & = -x^2 \log \left (\log \left (\frac {x^2}{3}\right )\right )-3 \text {li}\left (\frac {x^2}{3}\right )-2 \int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right )}{\log \left (\frac {x^2}{3}\right )} \, dx-2 \int \frac {e^x}{\log (3)-\log \left (x^2\right )} \, dx-\int \left (-\frac {e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}-\frac {e^x x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}-\frac {e^x x^2 \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}+\frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}+\frac {2 e^x x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}+\frac {e^x x^2 \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}\right ) \, dx+\text {Subst}\left (\int \frac {1}{\log \left (\frac {x}{3}\right )} \, dx,x,x^2\right ) \\ & = -x^2 \log \left (\log \left (\frac {x^2}{3}\right )\right )-2 \int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right )}{\log \left (\frac {x^2}{3}\right )} \, dx-2 \int \frac {e^x}{\log (3)-\log \left (x^2\right )} \, dx-2 \int \frac {e^x x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx+\int \frac {e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx+\int \frac {e^x x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx+\int \frac {e^x x^2 \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx-\int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx-\int \frac {e^x x^2 \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx \\ & = -x^2 \log \left (\log \left (\frac {x^2}{3}\right )\right )-2 \int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right )}{\log \left (\frac {x^2}{3}\right )} \, dx-2 \int \frac {e^x}{\log (3)-\log \left (x^2\right )} \, dx-2 \int \left (e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+\frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{-1-x}\right ) \, dx+\int \frac {e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx-\int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx+\int \left (e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )-\frac {e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}\right ) \, dx+\int \left (-e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )+e^x x \log \left (\log \left (\frac {x^2}{3}\right )\right )+\frac {e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}\right ) \, dx-\int \left (-e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+e^x x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )+\frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x}\right ) \, dx \\ & = -x^2 \log \left (\log \left (\frac {x^2}{3}\right )\right )-2 \int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right )}{\log \left (\frac {x^2}{3}\right )} \, dx-2 \int \frac {e^x}{\log (3)-\log \left (x^2\right )} \, dx-2 \int e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \, dx-2 \int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{-1-x} \, 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 )}{1+x} \, dx+\int e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \, dx-\int e^x x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \, dx-2 \int \frac {e^x \log \left (\frac {1}{5}+\frac {x}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{1+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-x \left (-e^x+x+e^x \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \]

[In]

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]

[Out]

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

Maple [A] (verified)

Time = 8.49 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39

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

[In]

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)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-{\left (x e^{x} \log \left (\frac {1}{5} \, x + \frac {1}{5}\right ) + x^{2} - x e^{x}\right )} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) \]

[In]

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")

[Out]

-(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 = 32.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=- x^{2} \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )} + \left (- x \log {\left (\frac {x}{5} + \frac {1}{5} \right )} \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )} + x \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )}\right ) e^{x} \]

[In]

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)

[Out]

-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)

none

Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx={\left (x {\left (\log \left (5\right ) + 1\right )} e^{x} - x e^{x} \log \left (x + 1\right ) - x^{2}\right )} \log \left (-\log \left (3\right ) + 2 \, \log \left (x\right )\right ) \]

[In]

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")

[Out]

(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 = 0.53 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=x e^{x} \log \left (5\right ) \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) - x e^{x} \log \left (x + 1\right ) \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) - x^{2} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) + x e^{x} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) \]

[In]

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")

[Out]

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 = 10.45 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.74 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-\ln \left (\ln \left (\frac {x^2}{3}\right )\right )\,\left (\frac {x^4+x^3}{x\,\left (x+1\right )}+\frac {{\mathrm {e}}^x\,\left (x-x^3\right )}{x\,\left (x+1\right )}-\frac {{\mathrm {e}}^x\,\left (x^2+x\right )}{x\,\left (x+1\right )}+\frac {{\mathrm {e}}^x\,\ln \left (\frac {x}{5}+\frac {1}{5}\right )\,\left (x^3+x^2\right )}{x\,\left (x+1\right )}\right ) \]

[In]

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

[Out]

-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)))