Integrand size = 46, antiderivative size = 25 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=x+\frac {5 \left (-3-x+\log \left (\frac {x}{\log (5)}\right )\right )}{(4+x) \log (19)} \]
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Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 1608, 27, 6874, 1634, 2351, 31} \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=x-\frac {5 x \log \left (\frac {x}{\log (5)}\right )}{4 (x+4) \log (19)}+\frac {5 \log (x)}{4 \log (19)}+\frac {5}{(x+4) \log (19)} \]
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Rule 12
Rule 27
Rule 31
Rule 1608
Rule 1634
Rule 2351
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{16 x+8 x^2+x^3} \, dx}{\log (19)} \\ & = \frac {\int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{x \left (16+8 x+x^2\right )} \, dx}{\log (19)} \\ & = \frac {\int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{x (4+x)^2} \, dx}{\log (19)} \\ & = \frac {\int \left (\frac {20+16 x \log (19)+8 x^2 \log (19)+x^3 \log (19)}{x (4+x)^2}-\frac {5 \log \left (\frac {x}{\log (5)}\right )}{(4+x)^2}\right ) \, dx}{\log (19)} \\ & = \frac {\int \frac {20+16 x \log (19)+8 x^2 \log (19)+x^3 \log (19)}{x (4+x)^2} \, dx}{\log (19)}-\frac {5 \int \frac {\log \left (\frac {x}{\log (5)}\right )}{(4+x)^2} \, dx}{\log (19)} \\ & = -\frac {5 x \log \left (\frac {x}{\log (5)}\right )}{4 (4+x) \log (19)}+\frac {\int \left (\frac {5}{4 x}-\frac {5}{(4+x)^2}-\frac {5}{4 (4+x)}+\log (19)\right ) \, dx}{\log (19)}+\frac {5 \int \frac {1}{4+x} \, dx}{4 \log (19)} \\ & = x+\frac {5}{(4+x) \log (19)}+\frac {5 \log (x)}{4 \log (19)}-\frac {5 x \log \left (\frac {x}{\log (5)}\right )}{4 (4+x) \log (19)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {x \log (19)+\frac {5 \left (1+\log \left (\frac {x}{\log (5)}\right )\right )}{4+x}}{\log (19)} \]
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Time = 0.74 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28
method | result | size |
parallelrisch | \(\frac {5+\ln \left (19\right ) x^{2}-16 \ln \left (19\right )+5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{\ln \left (19\right ) \left (4+x \right )}\) | \(32\) |
norman | \(\frac {x^{2}+\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{\ln \left (19\right )}-\frac {16 \ln \left (19\right )-5}{\ln \left (19\right )}}{4+x}\) | \(36\) |
risch | \(\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{\ln \left (19\right ) \left (4+x \right )}+\frac {\ln \left (19\right ) x^{2}+4 x \ln \left (19\right )+5}{\ln \left (19\right ) \left (4+x \right )}\) | \(43\) |
parts | \(\frac {x \ln \left (19\right )+\frac {5 \ln \left (x \right )}{4}+\frac {5}{4+x}-\frac {5 \ln \left (4+x \right )}{4}}{\ln \left (19\right )}+\frac {5 \ln \left (4+x \right )}{4 \ln \left (19\right )}-\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right ) x}{4 \ln \left (19\right ) \left (4+x \right )}\) | \(58\) |
derivativedivides | \(\frac {\ln \left (5\right ) \left (\frac {5 \ln \left (4+x \right )}{4 \ln \left (5\right )}-\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right ) x}{4 \ln \left (5\right ) \left (4+x \right )}+\frac {\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{4}+\frac {20}{4 x +16}-\frac {5 \ln \left (4+x \right )}{4}}{\ln \left (5\right )}+\frac {x \ln \left (19\right )}{\ln \left (5\right )}\right )}{\ln \left (19\right )}\) | \(75\) |
default | \(\frac {\ln \left (5\right ) \left (\frac {5 \ln \left (4+x \right )}{4 \ln \left (5\right )}-\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right ) x}{4 \ln \left (5\right ) \left (4+x \right )}+\frac {\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{4}+\frac {20}{4 x +16}-\frac {5 \ln \left (4+x \right )}{4}}{\ln \left (5\right )}+\frac {x \ln \left (19\right )}{\ln \left (5\right )}\right )}{\ln \left (19\right )}\) | \(75\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {{\left (x^{2} + 4 \, x\right )} \log \left (19\right ) + 5 \, \log \left (\frac {x}{\log \left (5\right )}\right ) + 5}{{\left (x + 4\right )} \log \left (19\right )} \]
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Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=x + \frac {5 \log {\left (\frac {x}{\log {\left (5 \right )}} \right )}}{x \log {\left (19 \right )} + 4 \log {\left (19 \right )}} + \frac {5}{x \log {\left (19 \right )} + 4 \log {\left (19 \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.80 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {{\left (x - \frac {16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (19\right ) + 8 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (19\right ) - \frac {16 \, \log \left (19\right )}{x + 4} + \frac {5 \, \log \left (\frac {x}{\log \left (5\right )}\right )}{x + 4} + \frac {5}{x + 4}}{\log \left (19\right )} \]
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {x \log \left (19\right ) - \frac {5 \, {\left (\log \left (\log \left (5\right )\right ) - 1\right )}}{x + 4} + \frac {5 \, \log \left (x\right )}{x + 4}}{\log \left (19\right )} \]
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Time = 16.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=x+\frac {5\,\ln \left (\frac {x}{\ln \left (5\right )}\right )+5}{\ln \left (19\right )\,\left (x+4\right )} \]
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