Integrand size = 31, antiderivative size = 22 \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=5 \left (2-\left (-4+\frac {3}{2 x}+x\right ) \log \left (4 x^2\right )\right ) \]
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Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2372} \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=-5 x \log \left (4 x^2\right )-\frac {15 \log \left (4 x^2\right )}{2 x}+40 \log (x) \]
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Rule 12
Rule 14
Rule 2372
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {10 \left (3-8 x+2 x^2\right )}{x^2}-\frac {5 \left (-3+2 x^2\right ) \log \left (4 x^2\right )}{x^2}\right ) \, dx \\ & = -\left (\frac {5}{2} \int \frac {\left (-3+2 x^2\right ) \log \left (4 x^2\right )}{x^2} \, dx\right )-5 \int \frac {3-8 x+2 x^2}{x^2} \, dx \\ & = -\frac {15 \log \left (4 x^2\right )}{2 x}-5 x \log \left (4 x^2\right )+5 \int \left (2+\frac {3}{x^2}\right ) \, dx-5 \int \left (2+\frac {3}{x^2}-\frac {8}{x}\right ) \, dx \\ & = 40 \log (x)-\frac {15 \log \left (4 x^2\right )}{2 x}-5 x \log \left (4 x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=40 \log (x)-\frac {15 \log \left (4 x^2\right )}{2 x}-5 x \log \left (4 x^2\right ) \]
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Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {5 \left (2 x^{2}+3\right ) \ln \left (4 x^{2}\right )}{2 x}+40 \ln \left (x \right )\) | \(24\) |
norman | \(\frac {-5 x^{2} \ln \left (4 x^{2}\right )-\frac {15 \ln \left (4 x^{2}\right )}{2}}{x}+40 \ln \left (x \right )\) | \(30\) |
parallelrisch | \(-\frac {10 x^{2} \ln \left (4 x^{2}\right )-80 x \ln \left (x \right )+15 \ln \left (4 x^{2}\right )}{2 x}\) | \(31\) |
default | \(40 \ln \left (x \right )-5 \ln \left (2\right ) \left (2 x +\frac {3}{x}\right )-5 x \ln \left (x^{2}\right )-\frac {15 \ln \left (x^{2}\right )}{2 x}\) | \(35\) |
parts | \(40 \ln \left (x \right )-5 \ln \left (2\right ) \left (2 x +\frac {3}{x}\right )-5 x \ln \left (x^{2}\right )-\frac {15 \ln \left (x^{2}\right )}{2 x}\) | \(35\) |
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Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=-\frac {5 \, {\left (2 \, x^{2} - 8 \, x + 3\right )} \log \left (4 \, x^{2}\right )}{2 \, x} \]
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Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=40 \log {\left (x \right )} + \frac {\left (- 10 x^{2} - 15\right ) \log {\left (4 x^{2} \right )}}{2 x} \]
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Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=-5 \, x \log \left (4 \, x^{2}\right ) - \frac {15 \, \log \left (4 \, x^{2}\right )}{2 \, x} + 40 \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=-\frac {5}{2} \, {\left (2 \, x + \frac {3}{x}\right )} \log \left (4 \, x^{2}\right ) + 40 \, \log \left (x\right ) \]
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Time = 8.57 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{2 x^2} \, dx=-\frac {5\,\ln \left (4\,x^2\right )\,\left (2\,x^2-8\,x+3\right )}{2\,x} \]
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