Integrand size = 34, antiderivative size = 20 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (-\frac {1}{4}+\left (-\frac {21}{4}-\frac {3}{x}\right )^2-\log (x)\right ) \]
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\[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {504}{-144-504 x-437 x^2+16 x^2 \log (x)}+\frac {288}{x \left (-144-504 x-437 x^2+16 x^2 \log (x)\right )}+\frac {16 x}{-144-504 x-437 x^2+16 x^2 \log (x)}\right ) \, dx \\ & = 16 \int \frac {x}{-144-504 x-437 x^2+16 x^2 \log (x)} \, dx+288 \int \frac {1}{x \left (-144-504 x-437 x^2+16 x^2 \log (x)\right )} \, dx+504 \int \frac {1}{-144-504 x-437 x^2+16 x^2 \log (x)} \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=8 \left (-\frac {\log (x)}{4}+\frac {1}{8} \log \left (144+504 x+437 x^2-16 x^2 \log (x)\right )\right ) \]
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Time = 2.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\ln \left (\ln \left (x \right )-\frac {437 x^{2}+504 x +144}{16 x^{2}}\right )\) | \(20\) |
parallelrisch | \(\ln \left (x^{2} \ln \left (x \right )-\frac {437 x^{2}}{16}-\frac {63 x}{2}-9\right )-2 \ln \left (x \right )\) | \(23\) |
default | \(-2 \ln \left (x \right )+\ln \left (16 x^{2} \ln \left (x \right )-437 x^{2}-504 x -144\right )\) | \(24\) |
norman | \(-2 \ln \left (x \right )+\ln \left (16 x^{2} \ln \left (x \right )-437 x^{2}-504 x -144\right )\) | \(24\) |
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (\frac {16 \, x^{2} \log \left (x\right ) - 437 \, x^{2} - 504 \, x - 144}{x^{2}}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log {\left (\log {\left (x \right )} + \frac {- 437 x^{2} - 504 x - 144}{16 x^{2}} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (\frac {16 \, x^{2} \log \left (x\right ) - 437 \, x^{2} - 504 \, x - 144}{16 \, x^{2}}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (16 \, x^{2} \log \left (x\right ) - 437 \, x^{2} - 504 \, x - 144\right ) - 2 \, \log \left (x\right ) \]
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Time = 10.51 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\ln \left (504\,x-16\,x^2\,\ln \left (x\right )+437\,x^2+144\right )-2\,\ln \left (x\right ) \]
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