Integrand size = 102, antiderivative size = 24 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {x}{1+e^{16-x}-x+\log (x)-\log \left (x^2\right )} \]
Time = 2.51 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {e^x x}{e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )} \]
Integrate[(2 + E^(16 - x)*(1 + x) + Log[x] - Log[x^2])/(1 + E^(32 - 2*x) + E^(16 - x)*(2 - 2*x) - 2*x + x^2 + (2 + 2*E^(16 - x) - 2*x)*Log[x] + Log[ x]^2 + (-2 - 2*E^(16 - x) + 2*x - 2*Log[x])*Log[x^2] + Log[x^2]^2),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.
\(\displaystyle \int \frac {-\log \left (x^2\right )+e^{16-x} (x+1)+\log (x)+2}{x^2+\log ^2\left (x^2\right )+\left (2 x-2 e^{16-x}-2 \log (x)-2\right ) \log \left (x^2\right )-2 x+e^{32-2 x}+e^{16-x} (2-2 x)+\log ^2(x)+\left (-2 x+2 e^{16-x}+2\right ) \log (x)+1} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x} \left (-\log \left (x^2\right )+e^{16-x} (x+1)+\log (x)+2\right )}{\left (-e^x \log \left (x^2\right )-e^x x+e^x+e^x \log (x)+e^{16}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{2 x-16} (x+1) \left (\log \left (x^2\right )+x-\log (x)-1\right )}{-e^x \log \left (x^2\right )-e^x x+e^x+e^x \log (x)+e^{16}}+\frac {e^{2 x} \left (x^2+x \log \left (x^2\right )-x \log (x)+1\right )}{\left (e^x \log \left (x^2\right )+e^x x-e^x-e^x \log (x)-e^{16}\right )^2}+e^{x-16} (x+1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{2 x-16} \log (x)}{-e^x x+e^x+e^x \log (x)-e^x \log \left (x^2\right )+e^{16}}dx+\int \frac {e^{2 x}}{\left (e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}\right )^2}dx+\int \frac {e^{2 x} x^2}{\left (e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}\right )^2}dx-\int \frac {e^{2 x} x \log (x)}{\left (e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}\right )^2}dx+\int \frac {e^{2 x} x \log \left (x^2\right )}{\left (e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}\right )^2}dx+\int \frac {e^{2 x-16}}{e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}}dx-\int \frac {e^{2 x-16} x^2}{e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}}dx+\int \frac {e^{2 x-16} x \log (x)}{e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}}dx-\int \frac {e^{2 x-16} \log \left (x^2\right )}{e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}}dx-\int \frac {e^{2 x-16} x \log \left (x^2\right )}{e^x x-e^x-e^x \log (x)+e^x \log \left (x^2\right )-e^{16}}dx+e^{x-16} (x+1)-e^{x-16}\) |
Int[(2 + E^(16 - x)*(1 + x) + Log[x] - Log[x^2])/(1 + E^(32 - 2*x) + E^(16 - x)*(2 - 2*x) - 2*x + x^2 + (2 + 2*E^(16 - x) - 2*x)*Log[x] + Log[x]^2 + (-2 - 2*E^(16 - x) + 2*x - 2*Log[x])*Log[x^2] + Log[x^2]^2),x]
3.20.25.3.1 Defintions of rubi rules used
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(-\frac {x}{x -{\mathrm e}^{16-x}-\ln \left (x \right )+\ln \left (x^{2}\right )-1}\) | \(25\) |
risch | \(-\frac {2 x}{-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x -2 \,{\mathrm e}^{16-x}+2 \ln \left (x \right )-2}\) | \(72\) |
int((-ln(x^2)+ln(x)+(1+x)*exp(16-x)+2)/(ln(x^2)^2+(-2*ln(x)-2*exp(16-x)+2* x-2)*ln(x^2)+ln(x)^2+(2*exp(16-x)-2*x+2)*ln(x)+exp(16-x)^2+(2-2*x)*exp(16- x)+x^2-2*x+1),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=-\frac {x}{x - e^{\left (-x + 16\right )} + \log \left (x\right ) - 1} \]
integrate((-log(x^2)+log(x)+(1+x)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*ex p(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(2 -2*x)*exp(16-x)+x^2-2*x+1),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.50 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {x}{- x + e^{16 - x} - \log {\left (x \right )} + 1} \]
integrate((-ln(x**2)+ln(x)+(1+x)*exp(16-x)+2)/(ln(x**2)**2+(-2*ln(x)-2*exp (16-x)+2*x-2)*ln(x**2)+ln(x)**2+(2*exp(16-x)-2*x+2)*ln(x)+exp(16-x)**2+(2- 2*x)*exp(16-x)+x**2-2*x+1),x)
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=-\frac {x e^{x}}{{\left (x + \log \left (x\right ) - 1\right )} e^{x} - e^{16}} \]
integrate((-log(x^2)+log(x)+(1+x)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*ex p(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(2 -2*x)*exp(16-x)+x^2-2*x+1),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=-\frac {x}{x - e^{\left (-x + 16\right )} + \log \left (x\right ) - 1} \]
integrate((-log(x^2)+log(x)+(1+x)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*ex p(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(2 -2*x)*exp(16-x)+x^2-2*x+1),x, algorithm=\
Time = 13.79 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {x\,{\mathrm {e}}^{x-16}}{{\mathrm {e}}^{x-16}-x\,{\mathrm {e}}^{x-16}+{\mathrm {e}}^{x-16}\,\ln \left (x\right )-\ln \left (x^2\right )\,{\mathrm {e}}^{x-16}+1} \]
int((log(x) - log(x^2) + exp(16 - x)*(x + 1) + 2)/(exp(32 - 2*x) - 2*x - l og(x^2)*(2*exp(16 - x) - 2*x + 2*log(x) + 2) - exp(16 - x)*(2*x - 2) + log (x)^2 + log(x)*(2*exp(16 - x) - 2*x + 2) + log(x^2)^2 + x^2 + 1),x)