Integrand size = 78, antiderivative size = 28 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\log (16) \left (e^{2 x/5} \log (4-x)+\frac {\log \left (2 x^3\right )}{x}\right ) \]
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Time = 0.41 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1607, 6874, 2326, 2340} \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {\log (16) \log \left (2 x^3\right )}{x}+\frac {e^{2 x/5} \log (16) (4 \log (4-x)-x \log (4-x))}{4-x} \]
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Rule 1607
Rule 2326
Rule 2340
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{x^2 (-20+5 x)} \, dx \\ & = \int \left (\frac {e^{2 x/5} \log (16) (5-8 \log (4-x)+2 x \log (4-x))}{5 (-4+x)}-\frac {\log (16) \left (-3+\log \left (2 x^3\right )\right )}{x^2}\right ) \, dx \\ & = \frac {1}{5} \log (16) \int \frac {e^{2 x/5} (5-8 \log (4-x)+2 x \log (4-x))}{-4+x} \, dx-\log (16) \int \frac {-3+\log \left (2 x^3\right )}{x^2} \, dx \\ & = \frac {e^{2 x/5} \log (16) (4 \log (4-x)-x \log (4-x))}{4-x}+\frac {\log (16) \log \left (2 x^3\right )}{x} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {\log (16) \left (e^{2 x/5} x \log (4-x)+\log \left (2 x^3\right )\right )}{x} \]
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Time = 8.55 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {60 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right ) x +60 \ln \left (2\right ) \ln \left (2 x^{3}\right )}{15 x}\) | \(34\) |
default | \(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )-\frac {12 \ln \left (2\right )}{x}-4 \ln \left (2\right ) \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}-\frac {\ln \left (2\right )}{x}\right )\) | \(51\) |
parts | \(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )-\frac {12 \ln \left (2\right )}{x}-4 \ln \left (2\right ) \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}-\frac {\ln \left (2\right )}{x}\right )\) | \(51\) |
risch | \(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )+\frac {2 \ln \left (2\right ) \left (-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}+2 \ln \left (2\right )+6 \ln \left (x \right )\right )}{x}\) | \(155\) |
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {4 \, {\left (x e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) + \log \left (2\right ) \log \left (2 \, x^{3}\right )\right )}}{x} \]
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Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=4 e^{\frac {2 x}{5}} \log {\left (2 \right )} \log {\left (4 - x \right )} + \frac {4 \log {\left (2 \right )} \log {\left (2 x^{3} \right )}}{x} \]
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\[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\int { \frac {4 \, {\left (5 \, x^{2} e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) + 2 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) - 5 \, {\left (x - 4\right )} \log \left (2\right ) \log \left (2 \, x^{3}\right ) + 15 \, {\left (x - 4\right )} \log \left (2\right )\right )}}{5 \, {\left (x^{3} - 4 \, x^{2}\right )}} \,d x } \]
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Time = 13.68 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=4 \, e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) + \frac {4 \, \log \left (2\right )^{2}}{x} + \frac {4 \, \log \left (2\right ) \log \left ({\left (x - 4\right )}^{3} + 12 \, {\left (x - 4\right )}^{2} + 48 \, x - 128\right )}{x} \]
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Time = 13.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {4\,\ln \left (2\right )\,\ln \left (2\,x^3\right )}{x}+4\,{\mathrm {e}}^{\frac {2\,x}{5}}\,\ln \left (2\right )\,\ln \left (4-x\right ) \]
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