Integrand size = 16, antiderivative size = 36 \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=-\frac {7}{3 (2-x)^3}+\frac {2}{(2-x)^2}+\frac {2}{2-x}+\log (2-x) \]
[Out]
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1864} \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=\frac {2}{2-x}+\frac {2}{(2-x)^2}-\frac {7}{3 (2-x)^3}+\log (2-x) \]
[In]
[Out]
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {7}{(-2+x)^4}-\frac {4}{(-2+x)^3}+\frac {2}{(-2+x)^2}+\frac {1}{-2+x}\right ) \, dx \\ & = -\frac {7}{3 (2-x)^3}+\frac {2}{(2-x)^2}+\frac {2}{2-x}+\log (2-x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=\frac {-29+30 x-6 x^2}{3 (-2+x)^3}+\log (-2+x) \]
[In]
[Out]
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.61
method | result | size |
norman | \(\frac {-2 x^{2}+10 x -\frac {29}{3}}{\left (-2+x \right )^{3}}+\ln \left (-2+x \right )\) | \(22\) |
risch | \(\frac {-2 x^{2}+10 x -\frac {29}{3}}{\left (-2+x \right )^{3}}+\ln \left (-2+x \right )\) | \(22\) |
default | \(-\frac {2}{-2+x}+\frac {2}{\left (-2+x \right )^{2}}+\frac {7}{3 \left (-2+x \right )^{3}}+\ln \left (-2+x \right )\) | \(27\) |
parallelrisch | \(\frac {3 \ln \left (-2+x \right ) x^{3}-29-18 \ln \left (-2+x \right ) x^{2}+36 \ln \left (-2+x \right ) x -6 x^{2}-24 \ln \left (-2+x \right )+30 x}{3 \left (-2+x \right )^{3}}\) | \(49\) |
meijerg | \(\frac {x \left (\frac {1}{4} x^{2}-\frac {3}{2} x +3\right )}{48 \left (1-\frac {x}{2}\right )^{3}}+\frac {x \left (\frac {11}{2} x^{2}-15 x +12\right )}{24 \left (1-\frac {x}{2}\right )^{3}}+\ln \left (1-\frac {x}{2}\right )-\frac {x^{3}}{12 \left (1-\frac {x}{2}\right )^{3}}\) | \(60\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=-\frac {6 \, x^{2} - 3 \, {\left (x^{3} - 6 \, x^{2} + 12 \, x - 8\right )} \log \left (x - 2\right ) - 30 \, x + 29}{3 \, {\left (x^{3} - 6 \, x^{2} + 12 \, x - 8\right )}} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=\frac {- 6 x^{2} + 30 x - 29}{3 x^{3} - 18 x^{2} + 36 x - 24} + \log {\left (x - 2 \right )} \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=-\frac {6 \, x^{2} - 30 \, x + 29}{3 \, {\left (x^{3} - 6 \, x^{2} + 12 \, x - 8\right )}} + \log \left (x - 2\right ) \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.64 \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=-\frac {6 \, x^{2} - 30 \, x + 29}{3 \, {\left (x - 2\right )}^{3}} + \log \left ({\left | x - 2 \right |}\right ) \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.61 \[ \int \frac {1-4 x^2+x^3}{(-2+x)^4} \, dx=\ln \left (x-2\right )-\frac {2\,x^2-10\,x+\frac {29}{3}}{{\left (x-2\right )}^3} \]
[In]
[Out]