Integrand size = 93, antiderivative size = 22 \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx=\log ^2(x)-\log \left (\frac {1}{9} x \log (4+\log (x (5+x)))\right ) \]
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Time = 0.64 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6820, 6874, 2338, 6816} \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx=\frac {1}{4} (1-2 \log (x))^2-\log (\log (\log (x (x+5))+4)) \]
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Rule 2338
Rule 6816
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-5-2 x+(5+x) (-1+2 \log (x)) (4+\log (x (5+x))) \log (4+\log (x (5+x)))}{x (5+x) (4+\log (x (5+x))) \log (4+\log (x (5+x)))} \, dx \\ & = \int \left (\frac {-1+2 \log (x)}{x}+\frac {-5-2 x}{x (5+x) (4+\log (x (5+x))) \log (4+\log (x (5+x)))}\right ) \, dx \\ & = \int \frac {-1+2 \log (x)}{x} \, dx+\int \frac {-5-2 x}{x (5+x) (4+\log (x (5+x))) \log (4+\log (x (5+x)))} \, dx \\ & = \frac {1}{4} (1-2 \log (x))^2-\log (\log (4+\log (x (5+x)))) \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx=-\log (x)+\log ^2(x)-\log (\log (4+\log (x (5+x)))) \]
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Time = 1.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09
method | result | size |
default | \(\ln \left (x \right )^{2}-\ln \left (x \right )-\ln \left (\ln \left (\ln \left (x^{2}+5 x \right )+4\right )\right )\) | \(24\) |
parallelrisch | \(\ln \left (x \right )^{2}-\ln \left (x \right )-\ln \left (\ln \left (\ln \left (x^{2}+5 x \right )+4\right )\right )\) | \(24\) |
parts | \(\ln \left (x \right )^{2}-\ln \left (x \right )-\ln \left (\ln \left (\ln \left (x^{2}+5 x \right )+4\right )\right )\) | \(24\) |
risch | \(\ln \left (x \right )^{2}-\ln \left (x \right )-\ln \left (\ln \left (\ln \left (x \right )+\ln \left (5+x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x \left (5+x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (5+x \right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (5+x \right )\right )+\operatorname {csgn}\left (i \left (5+x \right )\right )\right )}{2}+4\right )\right )\) | \(68\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx=\log \left (x\right )^{2} - \log \left (x\right ) - \log \left (\log \left (\log \left (x^{2} + 5 \, x\right ) + 4\right )\right ) \]
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Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx=\log {\left (x \right )}^{2} - \log {\left (x \right )} - \log {\left (\log {\left (\log {\left (x^{2} + 5 x \right )} + 4 \right )} \right )} \]
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Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx=\log \left (x\right )^{2} - \log \left (x\right ) - \log \left (\log \left (\log \left (x + 5\right ) + \log \left (x\right ) + 4\right )\right ) \]
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Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx=\log \left (x\right )^{2} - \log \left (x\right ) - \log \left (\log \left (\log \left (x + 5\right ) + \log \left (x\right ) + 4\right )\right ) \]
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Time = 13.51 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-5-2 x+\left (-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )}{\left (20 x+4 x^2+\left (5 x+x^2\right ) \log \left (5 x+x^2\right )\right ) \log \left (4+\log \left (5 x+x^2\right )\right )} \, dx={\ln \left (x\right )}^2-\ln \left (x\right )-\ln \left (\ln \left (\ln \left (x^2+5\,x\right )+4\right )\right ) \]
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