Integrand size = 45, antiderivative size = 16 \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4 \left (1+\log \left (3 (5+x)^2\right )\right )} \]
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Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6820, 12, 2437, 2339, 30} \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4 \left (\log \left (3 (x+5)^2\right )+1\right )} \]
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Rule 12
Rule 30
Rule 2339
Rule 2437
Rule 6820
Rubi steps \begin{align*} \text {integral}& = -\int \frac {1}{2 (5+x) \left (1+\log \left (3 (5+x)^2\right )\right )^2} \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{(5+x) \left (1+\log \left (3 (5+x)^2\right )\right )^2} \, dx\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \left (1+\log \left (3 x^2\right )\right )^2} \, dx,x,5+x\right )\right ) \\ & = -\left (\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,1+\log \left (3 (5+x)^2\right )\right )\right ) \\ & = \frac {1}{4 \left (1+\log \left (3 (5+x)^2\right )\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4 \left (1+\log \left (3 (5+x)^2\right )\right )} \]
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Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {1}{4 \ln \left (3\right )+4 \ln \left (x^{2}+10 x +25\right )+4}\) | \(18\) |
norman | \(\frac {1}{4 \ln \left (3 x^{2}+30 x +75\right )+4}\) | \(18\) |
risch | \(\frac {1}{4 \ln \left (3 x^{2}+30 x +75\right )+4}\) | \(18\) |
parallelrisch | \(\frac {1}{4 \ln \left (3 x^{2}+30 x +75\right )+4}\) | \(18\) |
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Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4 \, {\left (\log \left (3 \, x^{2} + 30 \, x + 75\right ) + 1\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4 \log {\left (3 x^{2} + 30 x + 75 \right )} + 4} \]
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none
Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4 \, {\left (\log \left (3\right ) + 2 \, \log \left (x + 5\right ) + 1\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4 \, {\left (\log \left (3 \, x^{2} + 30 \, x + 75\right ) + 1\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int -\frac {1}{10+2 x+(20+4 x) \log \left (75+30 x+3 x^2\right )+(10+2 x) \log ^2\left (75+30 x+3 x^2\right )} \, dx=\frac {1}{4\,\left (\ln \left (3\,x^2+30\,x+75\right )+1\right )} \]
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