Integrand size = 16, antiderivative size = 28 \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=\frac {1}{2} \left (-\frac {10-4 x}{20 x}-x-\log (2)+\log (4)\right ) \log (5) \]
[Out]
Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14} \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=-\frac {1}{2} x \log (5)-\frac {\log (5)}{4 x} \]
[In]
[Out]
Rule 12
Rule 14
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \log (5) \int \frac {1-2 x^2}{x^2} \, dx \\ & = \frac {1}{4} \log (5) \int \left (-2+\frac {1}{x^2}\right ) \, dx \\ & = -\frac {\log (5)}{4 x}-\frac {1}{2} x \log (5) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=\frac {1}{4} \left (-\frac {1}{x}-2 x\right ) \log (5) \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50
method | result | size |
default | \(\frac {\ln \left (5\right ) \left (-2 x -\frac {1}{x}\right )}{4}\) | \(14\) |
risch | \(-\frac {x \ln \left (5\right )}{2}-\frac {\ln \left (5\right )}{4 x}\) | \(14\) |
gosper | \(-\frac {\ln \left (5\right ) \left (2 x^{2}+1\right )}{4 x}\) | \(15\) |
parallelrisch | \(-\frac {\ln \left (5\right ) \left (2 x^{2}+1\right )}{4 x}\) | \(15\) |
norman | \(\frac {-\frac {x^{2} \ln \left (5\right )}{2}-\frac {\ln \left (5\right )}{4}}{x}\) | \(17\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50 \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=-\frac {{\left (2 \, x^{2} + 1\right )} \log \left (5\right )}{4 \, x} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50 \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=- \frac {x \log {\left (5 \right )}}{2} - \frac {\log {\left (5 \right )}}{4 x} \]
[In]
[Out]
none
Time = 0.17 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.39 \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=-\frac {1}{4} \, {\left (2 \, x + \frac {1}{x}\right )} \log \left (5\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.39 \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=-\frac {1}{4} \, {\left (2 \, x + \frac {1}{x}\right )} \log \left (5\right ) \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50 \[ \int \frac {\left (1-2 x^2\right ) \log (5)}{4 x^2} \, dx=-\frac {\ln \left (5\right )\,\left (2\,x^2+1\right )}{4\,x} \]
[In]
[Out]