Integrand size = 56, antiderivative size = 34 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=-1+2 x+\frac {\left (\frac {3}{x}-x\right ) (-x+\log (4))}{2 x (5+3 x)} \]
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Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1608, 27, 12, 1634} \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=\frac {3 \log (4)}{10 x^2}+2 x+\frac {5+\log (64)}{75 (3 x+5)}-\frac {3 (5+\log (64))}{50 x} \]
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Rule 12
Rule 27
Rule 1608
Rule 1634
Rubi steps \begin{align*} \text {integral}& = \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{x^3 \left (50+60 x+18 x^2\right )} \, dx \\ & = \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{2 x^3 (5+3 x)^2} \, dx \\ & = \frac {1}{2} \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{x^3 (5+3 x)^2} \, dx \\ & = \frac {1}{2} \int \left (4-\frac {6 \log (4)}{5 x^3}+\frac {3 (5+\log (64))}{25 x^2}-\frac {2 (5+\log (64))}{25 (5+3 x)^2}\right ) \, dx \\ & = 2 x+\frac {3 \log (4)}{10 x^2}-\frac {3 (5+\log (64))}{50 x}+\frac {5+\log (64)}{75 (5+3 x)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=\frac {-9 x+60 x^3+36 x^4+9 \log (4)-x^2 (5+\log (64))}{6 x^2 (5+3 x)} \]
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Time = 0.42 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97
method | result | size |
risch | \(2 x +\frac {3 \left (-\frac {\ln \left (2\right )}{3}-\frac {5}{18}\right ) x^{2}-\frac {3 x}{2}+3 \ln \left (2\right )}{x^{2} \left (3 x +5\right )}\) | \(33\) |
norman | \(\frac {\left (-\frac {35}{2}-\ln \left (2\right )\right ) x^{2}-\frac {3 x}{2}+6 x^{4}+3 \ln \left (2\right )}{x^{2} \left (3 x +5\right )}\) | \(35\) |
gosper | \(-\frac {-12 x^{4}+2 x^{2} \ln \left (2\right )+35 x^{2}-6 \ln \left (2\right )+3 x}{2 x^{2} \left (3 x +5\right )}\) | \(38\) |
default | \(2 x +\frac {3 \ln \left (2\right )}{5 x^{2}}-\frac {3 \left (\frac {6 \ln \left (2\right )}{25}+\frac {1}{5}\right )}{2 x}-\frac {3 \left (-\frac {4 \ln \left (2\right )}{75}-\frac {2}{45}\right )}{2 \left (3 x +5\right )}\) | \(38\) |
parallelrisch | \(-\frac {-36 x^{4}+6 x^{2} \ln \left (2\right )+105 x^{2}-18 \ln \left (2\right )+9 x}{6 x^{2} \left (3 x +5\right )}\) | \(38\) |
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Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=\frac {36 \, x^{4} + 60 \, x^{3} - 5 \, x^{2} - 6 \, {\left (x^{2} - 3\right )} \log \left (2\right ) - 9 \, x}{6 \, {\left (3 \, x^{3} + 5 \, x^{2}\right )}} \]
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Time = 0.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2 x + \frac {x^{2} \left (-5 - 6 \log {\left (2 \right )}\right ) - 9 x + 18 \log {\left (2 \right )}}{18 x^{3} + 30 x^{2}} \]
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Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2 \, x - \frac {x^{2} {\left (6 \, \log \left (2\right ) + 5\right )} + 9 \, x - 18 \, \log \left (2\right )}{6 \, {\left (3 \, x^{3} + 5 \, x^{2}\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2 \, x + \frac {6 \, \log \left (2\right ) + 5}{75 \, {\left (3 \, x + 5\right )}} - \frac {3 \, {\left (6 \, x \log \left (2\right ) + 5 \, x - 10 \, \log \left (2\right )\right )}}{50 \, x^{2}} \]
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Time = 9.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.21 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2\,x-\frac {3\,x}{6\,x^3+10\,x^2}-\frac {5\,x^2}{3\,\left (6\,x^3+10\,x^2\right )}+\frac {6\,\ln \left (2\right )}{6\,x^3+10\,x^2}-\frac {2\,x^2\,\ln \left (2\right )}{6\,x^3+10\,x^2} \]
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