Integrand size = 36, antiderivative size = 16 \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)} \]
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Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 6874, 45, 2579, 31, 8} \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)} \]
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Rule 8
Rule 12
Rule 31
Rule 45
Rule 2579
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{-1+5 x} \, dx}{\log (2)} \\ & = \frac {\int \left (\frac {-1+15 x}{-1+5 x}+\log \left (x (-1+5 x)^2\right )\right ) \, dx}{\log (2)} \\ & = \frac {\int \frac {-1+15 x}{-1+5 x} \, dx}{\log (2)}+\frac {\int \log \left (x (-1+5 x)^2\right ) \, dx}{\log (2)} \\ & = \frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)}+\frac {\int \left (3+\frac {2}{-1+5 x}\right ) \, dx}{\log (2)}-\frac {2 \int \frac {1}{-1+5 x} \, dx}{\log (2)}-\frac {3 \int 1 \, dx}{\log (2)} \\ & = \frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {x \log \left (x (-1+5 x)^2\right )}{\log (2)} \]
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Time = 1.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25
method | result | size |
default | \(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x}{\ln \left (2\right )}\) | \(20\) |
norman | \(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x}{\ln \left (2\right )}\) | \(20\) |
risch | \(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x}{\ln \left (2\right )}\) | \(20\) |
parallelrisch | \(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x}{\ln \left (2\right )}\) | \(20\) |
parts | \(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x -3 x -\frac {2 \ln \left (5 x -1\right )}{5}}{\ln \left (2\right )}+\frac {3 x +\frac {2 \ln \left (5 x -1\right )}{5}}{\ln \left (2\right )}\) | \(51\) |
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {x \log \left (25 \, x^{3} - 10 \, x^{2} + x\right )}{\log \left (2\right )} \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {x \log {\left (25 x^{3} - 10 x^{2} + x \right )}}{\log {\left (2 \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.06 \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {2 \, {\left (5 \, x - 1\right )} \log \left (5 \, x - 1\right ) + 5 \, x \log \left (x\right ) + 2 \, \log \left (5 \, x - 1\right )}{5 \, \log \left (2\right )} \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {x \log \left (25 \, x^{3} - 10 \, x^{2} + x\right )}{\log \left (2\right )} \]
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Time = 9.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{(-1+5 x) \log (2)} \, dx=\frac {x\,\ln \left (x\,{\left (5\,x-1\right )}^2\right )}{\ln \left (2\right )} \]
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