Integrand size = 34, antiderivative size = 23 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=1-\frac {2+x}{e}-\frac {x}{5 (-4+3 \log (3))} \]
[Out]
Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {192, 21, 8} \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {1}{5} x \left (\frac {5}{e}+\frac {1}{\log (27)-4}\right ) \]
[In]
[Out]
Rule 8
Rule 21
Rule 192
Rubi steps \begin{align*} \text {integral}& = \int \frac {\frac {2 (20-e-5 \log (27))}{e}+\frac {x (20-e-5 \log (27))}{e}}{-10 (4-3 \log (3))-5 x (4-\log (27))} \, dx \\ & = -\left (\frac {1}{5} \left (\frac {5}{e}+\frac {1}{-4+\log (27)}\right ) \int 1 \, dx\right ) \\ & = -\frac {1}{5} x \left (\frac {5}{e}+\frac {1}{-4+\log (27)}\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=x \left (-\frac {1}{e}-\frac {1}{5 (-4+\log (27))}\right ) \]
[In]
[Out]
Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {x}{5 \left (3 \ln \left (3\right )-4\right )}-{\mathrm e}^{\ln \left (2+x \right )-1}\) | \(22\) |
parts | \(-\frac {x}{5 \left (3 \ln \left (3\right )-4\right )}-{\mathrm e}^{\ln \left (2+x \right )-1}\) | \(22\) |
norman | \(-\frac {{\mathrm e}^{-1} \left ({\mathrm e}+15 \ln \left (3\right )-20\right ) x}{5 \left (3 \ln \left (3\right )-4\right )}\) | \(24\) |
risch | \(-\frac {15 x \ln \left (3\right ) {\mathrm e}^{-1}}{15 \ln \left (3\right )-20}+\frac {20 x \,{\mathrm e}^{-1}}{15 \ln \left (3\right )-20}-\frac {x}{15 \ln \left (3\right )-20}\) | \(41\) |
parallelrisch | \(\frac {16+12 x -15 \ln \left (3\right ) x^{2} {\mathrm e}^{\ln \left (2+x \right )-1}-80 \,{\mathrm e}^{\ln \left (2+x \right )-1}-x^{3}+20 x^{2} {\mathrm e}^{\ln \left (2+x \right )-1}+60 \,{\mathrm e}^{\ln \left (2+x \right )-1} \ln \left (3\right )}{5 \left (2+x \right )^{2} \left (3 \ln \left (3\right )-4\right )}\) | \(72\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {x e + 15 \, x \log \left (3\right ) - 20 \, x}{5 \, {\left (3 \, e \log \left (3\right ) - 4 \, e\right )}} \]
[In]
[Out]
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=\frac {x \left (- 15 \log {\left (3 \right )} - e + 20\right )}{- 20 e + 15 e \log {\left (3 \right )}} \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {x {\left (e + 15 \, \log \left (3\right ) - 20\right )}}{5 \, {\left (3 \, e \log \left (3\right ) - 4 \, e\right )}} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {x e + 15 \, x \log \left (3\right ) - 20 \, x}{5 \, {\left (3 \, e \log \left (3\right ) - 4 \, e\right )}} \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=\frac {x\,\left (\mathrm {e}+15\,\ln \left (3\right )-20\right )}{20\,\mathrm {e}-15\,\mathrm {e}\,\ln \left (3\right )} \]
[In]
[Out]