Integrand size = 51, antiderivative size = 28 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=2+\frac {1}{15} \left (-e^2-x\right )-e^x \log \left (-1+\frac {4}{x^2}\right ) \]
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Time = 0.33 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1607, 6857, 2326} \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-\frac {e^x \left (4 x \log \left (\frac {4}{x^2}-1\right )-x^3 \log \left (\frac {4}{x^2}-1\right )\right )}{\left (4-x^2\right ) x}-\frac {x}{15} \]
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Rule 1607
Rule 2326
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{x \left (-60+15 x^2\right )} \, dx \\ & = \int \left (-\frac {1}{15}-\frac {e^x \left (8-4 x \log \left (-1+\frac {4}{x^2}\right )+x^3 \log \left (-1+\frac {4}{x^2}\right )\right )}{x \left (-4+x^2\right )}\right ) \, dx \\ & = -\frac {x}{15}-\int \frac {e^x \left (8-4 x \log \left (-1+\frac {4}{x^2}\right )+x^3 \log \left (-1+\frac {4}{x^2}\right )\right )}{x \left (-4+x^2\right )} \, dx \\ & = -\frac {x}{15}-\frac {e^x \left (4 x \log \left (-1+\frac {4}{x^2}\right )-x^3 \log \left (-1+\frac {4}{x^2}\right )\right )}{x \left (4-x^2\right )} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=\frac {1}{15} \left (-x-15 e^x \log \left (-1+\frac {4}{x^2}\right )\right ) \]
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Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71
method | result | size |
parallelrisch | \(-{\mathrm e}^{x} \ln \left (-\frac {x^{2}-4}{x^{2}}\right )-\frac {x}{15}\) | \(20\) |
default | \(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\) | \(21\) |
norman | \(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\) | \(21\) |
parts | \(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\) | \(21\) |
risch | \(-{\mathrm e}^{x} \ln \left (x^{2}-4\right )+2 \,{\mathrm e}^{x} \ln \left (x \right )-\frac {x}{15}-\frac {i {\mathrm e}^{x} \pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{2}}{2}+\frac {i {\mathrm e}^{x} \pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{2}-\frac {i {\mathrm e}^{x} \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{3}}{2}+i {\mathrm e}^{x} \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{2}-\frac {i {\mathrm e}^{x} \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{2}-\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}-i {\mathrm e}^{x} \pi \) | \(212\) |
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Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-e^{x} \log \left (-\frac {x^{2} - 4}{x^{2}}\right ) - \frac {1}{15} \, x \]
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Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=- \frac {x}{15} - e^{x} \log {\left (\frac {4 - x^{2}}{x^{2}} \right )} \]
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-e^{x} \log \left (x + 2\right ) + 2 \, e^{x} \log \left (x\right ) - e^{x} \log \left (-x + 2\right ) - \frac {1}{15} \, x \]
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Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-e^{x} \log \left (-\frac {x^{2} - 4}{x^{2}}\right ) - \frac {1}{15} \, x \]
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Time = 13.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-\frac {x}{15}-{\mathrm {e}}^x\,\ln \left (-\frac {x^2-4}{x^2}\right ) \]
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