Integrand size = 180, antiderivative size = 30 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {2}{-2+\frac {3+x}{2 \log \left (3+e^{1-x}+x-\log (2)\right )}} \]
[Out]
\[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \left (-e+e^x\right ) (3+x)-4 \left (e+e^x (3+x-\log (2))\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{\left (e+e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx \\ & = \int \left (\frac {4 e (3+x) (-4-x+\log (2))}{\left (e+e^x x+3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {4 \left (3+x-x \log \left (3+e^{1-x}+x-\log (2)\right )-3 \left (1-\frac {\log (2)}{3}\right ) \log \left (3+e^{1-x}+x-\log (2)\right )\right )}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}\right ) \, dx \\ & = 4 \int \frac {3+x-x \log \left (3+e^{1-x}+x-\log (2)\right )-3 \left (1-\frac {\log (2)}{3}\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e) \int \frac {(3+x) (-4-x+\log (2))}{\left (e+e^x x+3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx \\ & = 4 \int \frac {3+x+(-3-x+\log (2)) \log \left (3+e^{1-x}+x-\log (2)\right )}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e) \int \frac {(3+x) (-4-x+\log (2))}{\left (e+e^x (3+x-\log (2))\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx \\ & = 4 \int \left (-\frac {(3+x) (-1+x-\log (2))}{4 (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {1}{4 \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )}\right ) \, dx+(4 e) \int \left (\frac {4}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {x}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {\log (2)}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}\right ) \, dx \\ & = (4 e) \int \frac {x}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(16 e) \int \frac {1}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e \log (2)) \int \frac {1}{\left (-e-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx-\int \frac {(3+x) (-1+x-\log (2))}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+\int \frac {1}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \, dx \\ & = (4 e) \int \frac {x}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(16 e) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e \log (2)) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx-\int \left (-\frac {1}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}+\frac {x}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}-\frac {4 \log (2)}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2}\right ) \, dx+\int \frac {1}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \, dx \\ & = (4 e) \int \frac {x}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(16 e) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 \log (2)) \int \frac {1}{(3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+(4 e \log (2)) \int \frac {1}{\left (-e-e^x (3+x-\log (2))\right ) (3+x-\log (2)) \left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+\int \frac {1}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx-\int \frac {x}{\left (3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )\right )^2} \, dx+\int \frac {1}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {3+x}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \]
[In]
[Out]
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87
method | result | size |
risch | \(\frac {3+x}{x -4 \ln \left ({\mathrm e}^{1-x}-\ln \left (2\right )+3+x \right )+3}\) | \(26\) |
parallelrisch | \(\frac {4 x +12}{4 x -16 \ln \left ({\mathrm e}^{1-x}-\ln \left (2\right )+3+x \right )+12}\) | \(29\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {x + 3}{x - 4 \, \log \left (x + e^{\left (-x + 1\right )} - \log \left (2\right ) + 3\right ) + 3} \]
[In]
[Out]
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {- x - 3}{- x + 4 \log {\left (x + e^{1 - x} - \log {\left (2 \right )} + 3 \right )} - 3} \]
[In]
[Out]
none
Time = 0.46 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {x + 3}{5 \, x - 4 \, \log \left ({\left (x - \log \left (2\right ) + 3\right )} e^{x} + e\right ) + 3} \]
[In]
[Out]
none
Time = 0.60 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {x + 3}{x - 4 \, \log \left (x + e^{\left (-x + 1\right )} - \log \left (2\right ) + 3\right ) + 3} \]
[In]
[Out]
Timed out. \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\text {Hanged} \]
[In]
[Out]