Integrand size = 18, antiderivative size = 15 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=-\frac {3}{10}+\frac {5 x}{4}+\log (-2 e+x) \]
[Out]
Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {45} \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5 x}{4}+\log (2 e-x) \]
[In]
[Out]
Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5}{4}+\frac {1}{-2 e+x}\right ) \, dx \\ & = \frac {5 x}{4}+\log (2 e-x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=-\frac {5}{16} (8 e-4 x)+\log (8 e-4 x) \]
[In]
[Out]
Time = 0.74 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {5 x}{4}+\ln \left (-2 \,{\mathrm e}+x \right )\) | \(12\) |
risch | \(\frac {5 x}{4}+\ln \left (-2 \,{\mathrm e}+x \right )\) | \(12\) |
parallelrisch | \(\frac {5 x}{4}+\ln \left (-2 \,{\mathrm e}+x \right )\) | \(12\) |
norman | \(\frac {5 x}{4}+\ln \left (8 \,{\mathrm e}-4 x \right )\) | \(14\) |
meijerg | \(-\frac {5 \,{\mathrm e} \ln \left (1-\frac {{\mathrm e}^{-1} x}{2}\right )}{2}+\ln \left (1-\frac {{\mathrm e}^{-1} x}{2}\right )-\frac {5 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-1} x}{2}-\ln \left (1-\frac {{\mathrm e}^{-1} x}{2}\right )\right )}{2}\) | \(42\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5}{4} \, x + \log \left (x - 2 \, e\right ) \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5 x}{4} + \log {\left (x - 2 e \right )} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5}{4} \, x + \log \left (x - 2 \, e\right ) \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5}{4} \, x + \log \left ({\left | x - 2 \, e \right |}\right ) \]
[In]
[Out]
Time = 14.83 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5\,x}{4}+\ln \left (x-2\,\mathrm {e}\right ) \]
[In]
[Out]