Integrand size = 110, antiderivative size = 31 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=x \left (e^x-x-e^x \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \]
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-x \left (-e^x+x+e^x \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right ) \]
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]
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 |
\(\displaystyle -2 \int \frac {e^x \log \left (\frac {x}{5}+\frac {1}{5}\right )}{\log \left (\frac {x^2}{3}\right )}dx-2 \int \frac {e^x}{\log (3)-\log \left (x^2\right )}dx+\int e^x x \log \left (\log \left (\frac {x^2}{3}\right )\right )dx+\int \frac {e^x \log \left (\log \left (\frac {x^2}{3}\right )\right )}{x+1}dx-\int e^x \log \left (\frac {x}{5}+\frac {1}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )dx-2 \int \frac {e^x \log \left (\frac {x}{5}+\frac {1}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{-x-1}dx-\int e^x x \log \left (\frac {x}{5}+\frac {1}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )dx-2 \int \frac {e^x \log \left (\frac {x}{5}+\frac {1}{5}\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{x+1}dx+x^2 \left (-\log \left (\log \left (\frac {x^2}{3}\right )\right )\right )\) |
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]
3.18.41.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
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 | \(\left (-{\mathrm e}^{x} \ln \left (\frac {x}{5}+\frac {1}{5}\right ) x -x^{2}+{\mathrm e}^{x} x \right ) \ln \left (-\ln \left (3\right )+2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )\) | \(61\) |
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)
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-{\left (x e^{x} \log \left (\frac {1}{5} \, x + \frac {1}{5}\right ) + x^{2} - x e^{x}\right )} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) \]
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=\
Time = 32.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=- x^{2} \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )} + \left (- x \log {\left (\frac {x}{5} + \frac {1}{5} \right )} \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )} + x \log {\left (\log {\left (\frac {x^{2}}{3} \right )} \right )}\right ) e^{x} \]
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)
Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx={\left (x {\left (\log \left (5\right ) + 1\right )} e^{x} - x e^{x} \log \left (x + 1\right ) - x^{2}\right )} \log \left (-\log \left (3\right ) + 2 \, \log \left (x\right )\right ) \]
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=\
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 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=x e^{x} \log \left (5\right ) \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) - x e^{x} \log \left (x + 1\right ) \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) - x^{2} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) + x e^{x} \log \left (\log \left (\frac {1}{3} \, x^{2}\right )\right ) \]
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=\
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))
Time = 10.45 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.74 \[ \int \frac {-2 x-2 x^2+e^x (2+2 x)+e^x (-2-2 x) \log \left (\frac {1+x}{5}\right )+\log \left (\frac {x^2}{3}\right ) \left (-2 x-2 x^2+e^x \left (1+x+x^2\right )+e^x \left (-1-2 x-x^2\right ) \log \left (\frac {1+x}{5}\right )\right ) \log \left (\log \left (\frac {x^2}{3}\right )\right )}{(1+x) \log \left (\frac {x^2}{3}\right )} \, dx=-\ln \left (\ln \left (\frac {x^2}{3}\right )\right )\,\left (\frac {x^4+x^3}{x\,\left (x+1\right )}+\frac {{\mathrm {e}}^x\,\left (x-x^3\right )}{x\,\left (x+1\right )}-\frac {{\mathrm {e}}^x\,\left (x^2+x\right )}{x\,\left (x+1\right )}+\frac {{\mathrm {e}}^x\,\ln \left (\frac {x}{5}+\frac {1}{5}\right )\,\left (x^3+x^2\right )}{x\,\left (x+1\right )}\right ) \]
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)