Integrand size = 42, antiderivative size = 23 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=-x+x \left (2 x^2+\log \left (\frac {1}{2} x (-1+4 x)\right )\right ) \]
[Out]
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6874, 785, 2579, 31, 8} \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 x^3-x+x \log \left (-\frac {1}{2} (1-4 x) x\right ) \]
[In]
[Out]
Rule 8
Rule 31
Rule 785
Rule 2579
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 x \left (2-3 x+12 x^2\right )}{-1+4 x}+\log \left (\frac {1}{2} x (-1+4 x)\right )\right ) \, dx \\ & = 2 \int \frac {x \left (2-3 x+12 x^2\right )}{-1+4 x} \, dx+\int \log \left (\frac {1}{2} x (-1+4 x)\right ) \, dx \\ & = x \log \left (-\frac {1}{2} (1-4 x) x\right )-2 \int 1 \, dx+2 \int \left (\frac {1}{2}+3 x^2+\frac {1}{2 (-1+4 x)}\right ) \, dx-\int \frac {1}{-1+4 x} \, dx \\ & = -x+2 x^3+x \log \left (-\frac {1}{2} (1-4 x) x\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=-\frac {9}{32}-x+2 x^3-\frac {1}{4} \log (1-4 x)+\frac {1}{4} \log (-1+4 x)+x \log \left (\frac {1}{2} x (-1+4 x)\right ) \]
[In]
[Out]
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\ln \left (2 x^{2}-\frac {1}{2} x \right ) x -x +2 x^{3}\) | \(22\) |
risch | \(\ln \left (2 x^{2}-\frac {1}{2} x \right ) x -x +2 x^{3}\) | \(22\) |
parallelrisch | \(2 x^{3}-\frac {1}{2}+\ln \left (2 x^{2}-\frac {1}{2} x \right ) x -x\) | \(23\) |
default | \(2 x^{3}-x -x \ln \left (2\right )+x \ln \left (4 x^{2}-x \right )\) | \(27\) |
parts | \(2 x^{3}-x -x \ln \left (2\right )+x \ln \left (4 x^{2}-x \right )\) | \(27\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 \, x^{3} + x \log \left (2 \, x^{2} - \frac {1}{2} \, x\right ) - x \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 x^{3} + x \log {\left (2 x^{2} - \frac {x}{2} \right )} - x \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 \, x^{3} - x {\left (\log \left (2\right ) + 2\right )} + \frac {1}{4} \, {\left (4 \, x - 1\right )} \log \left (4 \, x - 1\right ) + x \log \left (x\right ) + x + \frac {1}{4} \, \log \left (4 \, x - 1\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=2 \, x^{3} + x \log \left (2 \, x^{2} - \frac {1}{2} \, x\right ) - x \]
[In]
[Out]
Time = 9.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {4 x-6 x^2+24 x^3+(-1+4 x) \log \left (\frac {1}{2} \left (-x+4 x^2\right )\right )}{-1+4 x} \, dx=x\,\left (\ln \left (2\,x^2-\frac {x}{2}\right )-1\right )+2\,x^3 \]
[In]
[Out]