Integrand size = 31, antiderivative size = 20 \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=\left (10+\frac {1}{4-x+5 (3+x)}-\log (3)\right )^2 \]
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Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {2099} \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=\frac {1}{(4 x+19)^2}+\frac {2 (10-\log (3))}{4 x+19} \]
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Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8}{(19+4 x)^3}+\frac {8 (-10+\log (3))}{(19+4 x)^2}\right ) \, dx \\ & = \frac {1}{(19+4 x)^2}+\frac {2 (10-\log (3))}{19+4 x} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=\frac {381-8 x (-10+\log (3))-38 \log (3)}{(19+4 x)^2} \]
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Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15
method | result | size |
norman | \(\frac {\left (80-8 \ln \left (3\right )\right ) x +381-38 \ln \left (3\right )}{\left (4 x +19\right )^{2}}\) | \(23\) |
default | \(\frac {1}{\left (4 x +19\right )^{2}}-\frac {8 \left (\frac {\ln \left (3\right )}{4}-\frac {5}{2}\right )}{4 x +19}\) | \(24\) |
risch | \(\frac {\left (-\frac {\ln \left (3\right )}{2}+5\right ) x +\frac {381}{16}-\frac {19 \ln \left (3\right )}{8}}{x^{2}+\frac {19}{2} x +\frac {361}{16}}\) | \(26\) |
gosper | \(-\frac {8 x \ln \left (3\right )+38 \ln \left (3\right )-80 x -381}{16 x^{2}+152 x +361}\) | \(29\) |
parallelrisch | \(-\frac {-6096+128 x \ln \left (3\right )+608 \ln \left (3\right )-1280 x}{16 \left (16 x^{2}+152 x +361\right )}\) | \(29\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=-\frac {2 \, {\left (4 \, x + 19\right )} \log \left (3\right ) - 80 \, x - 381}{16 \, x^{2} + 152 \, x + 361} \]
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Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=- \frac {x \left (-80 + 8 \log {\left (3 \right )}\right ) - 381 + 38 \log {\left (3 \right )}}{16 x^{2} + 152 x + 361} \]
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none
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=-\frac {8 \, x {\left (\log \left (3\right ) - 10\right )} + 38 \, \log \left (3\right ) - 381}{16 \, x^{2} + 152 \, x + 361} \]
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=-\frac {8 \, x \log \left (3\right ) - 80 \, x + 38 \, \log \left (3\right ) - 381}{{\left (4 \, x + 19\right )}^{2}} \]
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Time = 9.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-1528-320 x+(152+32 x) \log (3)}{6859+4332 x+912 x^2+64 x^3} \, dx=\frac {1}{{\left (4\,x+19\right )}^2}-\frac {2\,\ln \left (3\right )-20}{4\,x+19} \]
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