Integrand size = 37, antiderivative size = 20 \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=2 \left (1+\frac {5 \log ^2(x)}{\left (9+\frac {1}{x}\right )^2 x}\right ) \]
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Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {6820, 12, 6874, 2398, 90, 2351, 31, 2404, 2338} \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=\frac {5 \log ^2(x)}{18}-\frac {5 (1-9 x)^2 \log ^2(x)}{18 (9 x+1)^2} \]
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Rule 12
Rule 31
Rule 90
Rule 2338
Rule 2351
Rule 2398
Rule 2404
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 \log (x) (2+18 x+\log (x)-9 x \log (x))}{(1+9 x)^3} \, dx \\ & = 10 \int \frac {\log (x) (2+18 x+\log (x)-9 x \log (x))}{(1+9 x)^3} \, dx \\ & = 10 \int \left (\frac {2 \log (x)}{(1+9 x)^2}-\frac {(-1+9 x) \log ^2(x)}{(1+9 x)^3}\right ) \, dx \\ & = -\left (10 \int \frac {(-1+9 x) \log ^2(x)}{(1+9 x)^3} \, dx\right )+20 \int \frac {\log (x)}{(1+9 x)^2} \, dx \\ & = \frac {20 x \log (x)}{1+9 x}-\frac {5 (1-9 x)^2 \log ^2(x)}{18 (1+9 x)^2}+\frac {5}{9} \int \frac {(-1+9 x)^2 \log (x)}{x (1+9 x)^2} \, dx-20 \int \frac {1}{1+9 x} \, dx \\ & = \frac {20 x \log (x)}{1+9 x}-\frac {5 (1-9 x)^2 \log ^2(x)}{18 (1+9 x)^2}-\frac {20}{9} \log (1+9 x)+\frac {5}{9} \int \left (\frac {\log (x)}{x}-\frac {36 \log (x)}{(1+9 x)^2}\right ) \, dx \\ & = \frac {20 x \log (x)}{1+9 x}-\frac {5 (1-9 x)^2 \log ^2(x)}{18 (1+9 x)^2}-\frac {20}{9} \log (1+9 x)+\frac {5}{9} \int \frac {\log (x)}{x} \, dx-20 \int \frac {\log (x)}{(1+9 x)^2} \, dx \\ & = \frac {5 \log ^2(x)}{18}-\frac {5 (1-9 x)^2 \log ^2(x)}{18 (1+9 x)^2}-\frac {20}{9} \log (1+9 x)+20 \int \frac {1}{1+9 x} \, dx \\ & = \frac {5 \log ^2(x)}{18}-\frac {5 (1-9 x)^2 \log ^2(x)}{18 (1+9 x)^2} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=\frac {10 x \log ^2(x)}{(1+9 x)^2} \]
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Time = 0.66 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75
method | result | size |
norman | \(\frac {10 x \ln \left (x \right )^{2}}{\left (9 x +1\right )^{2}}\) | \(15\) |
risch | \(\frac {10 x \ln \left (x \right )^{2}}{81 x^{2}+18 x +1}\) | \(20\) |
parallelrisch | \(\frac {10 x \ln \left (x \right )^{2}}{81 x^{2}+18 x +1}\) | \(20\) |
default | \(\frac {20 \ln \left (x \right ) x}{9 x +1}+\frac {-180 x^{2} \ln \left (x \right )-20 x \ln \left (x \right )+10 x \ln \left (x \right )^{2}}{\left (9 x +1\right )^{2}}\) | \(42\) |
parts | \(\frac {20 \ln \left (x \right ) x}{9 x +1}+\frac {-180 x^{2} \ln \left (x \right )-20 x \ln \left (x \right )+10 x \ln \left (x \right )^{2}}{\left (9 x +1\right )^{2}}\) | \(42\) |
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=\frac {10 \, x \log \left (x\right )^{2}}{81 \, x^{2} + 18 \, x + 1} \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=\frac {10 x \log {\left (x \right )}^{2}}{81 x^{2} + 18 x + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (18) = 36\).
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.45 \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=-\frac {10 \, {\left (18 \, x + 1\right )} \log \left (x\right )}{9 \, {\left (81 \, x^{2} + 18 \, x + 1\right )}} + \frac {10 \, {\left (9 \, x \log \left (x\right )^{2} + 2 \, {\left (9 \, x + 1\right )} \log \left (x\right )\right )}}{9 \, {\left (81 \, x^{2} + 18 \, x + 1\right )}} - \frac {10 \, \log \left (x\right )}{9 \, {\left (81 \, x^{2} + 18 \, x + 1\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=\frac {10 \, x \log \left (x\right )^{2}}{81 \, x^{2} + 18 \, x + 1} \]
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Time = 9.79 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {(20+180 x) \log (x)+(10-90 x) \log ^2(x)}{1+27 x+243 x^2+729 x^3} \, dx=\frac {10\,x\,{\ln \left (x\right )}^2}{{\left (9\,x+1\right )}^2} \]
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