Integrand size = 79, antiderivative size = 30 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=\frac {8 x \left (3-x^2\right ) \log (2)}{e-(-e+5 (3-x)) x} \]
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Time = 0.39 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {12, 1694, 1828, 21, 8} \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=\frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (5 x^2-(15-e) x+e\right )}-\frac {8}{5} x \log (2) \]
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Rule 8
Rule 12
Rule 21
Rule 1694
Rule 1828
Rubi steps \begin{align*} \text {integral}& = \log (2) \int \frac {-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx \\ & = \log (2) \text {Subst}\left (\int \frac {8 \left (3 (5-e)^2 (25-e) (45-e)+80 (15-e) \left (150-45 e+e^2\right ) x+600 (5-e) (35-e) x^2-10000 x^4\right )}{5 \left ((5-e) (45-e)-100 x^2\right )^2} \, dx,x,\frac {1}{100} (-150+10 e)+x\right ) \\ & = \frac {1}{5} (8 \log (2)) \text {Subst}\left (\int \frac {3 (5-e)^2 (25-e) (45-e)+80 (15-e) \left (150-45 e+e^2\right ) x+600 (5-e) (35-e) x^2-10000 x^4}{\left ((5-e) (45-e)-100 x^2\right )^2} \, dx,x,\frac {1}{100} (-150+10 e)+x\right ) \\ & = \frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (e-(15-e) x+5 x^2\right )}-\frac {(4 \log (2)) \text {Subst}\left (\int \frac {2 \left (225-50 e+e^2\right )^2-200 (5-e) (45-e) x^2}{(5-e) (45-e)-100 x^2} \, dx,x,\frac {1}{100} (-150+10 e)+x\right )}{5 \left (225-50 e+e^2\right )} \\ & = \frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (e-(15-e) x+5 x^2\right )}-\frac {1}{5} (8 \log (2)) \text {Subst}\left (\int 1 \, dx,x,\frac {1}{100} (-150+10 e)+x\right ) \\ & = -\frac {8}{5} x \log (2)+\frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (e-(15-e) x+5 x^2\right )} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-8 \left (\frac {x}{5}+\frac {-15 e+e^2+150 x-35 e x+e^2 x}{25 \left (e-15 x+e x+5 x^2\right )}\right ) \log (2) \]
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Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
gosper | \(-\frac {8 x \left (x^{2}-3\right ) \ln \left (2\right )}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\) | \(28\) |
parallelrisch | \(\frac {\ln \left (2\right ) \left (-40 x^{3}+120 x \right )}{5 x \,{\mathrm e}+25 x^{2}+5 \,{\mathrm e}-75 x}\) | \(31\) |
norman | \(\frac {24 x \ln \left (2\right )-8 x^{3} \ln \left (2\right )}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\) | \(32\) |
risch | \(-\frac {8 x \ln \left (2\right )}{5}+\frac {\ln \left (2\right ) \left (\frac {\left (-\frac {8 \,{\mathrm e}^{2}}{5}+56 \,{\mathrm e}-240\right ) x}{5}-\frac {8 \,{\mathrm e}^{2}}{25}+\frac {24 \,{\mathrm e}}{5}\right )}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\) | \(49\) |
default | \(\ln \left (2\right ) \left (-\frac {8 x}{5}+\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+\left (10 \,{\mathrm e}-150\right ) \textit {\_Z}^{3}+\left ({\mathrm e}^{2}-20 \,{\mathrm e}+225\right ) \textit {\_Z}^{2}+\left (2 \,{\mathrm e}^{2}-30 \,{\mathrm e}\right ) \textit {\_Z} +{\mathrm e}^{2}\right )}{\sum }\frac {\left (\left ({\mathrm e}^{2}-35 \,{\mathrm e}+150\right ) \textit {\_R}^{2}+2 \left ({\mathrm e}^{2}-15 \,{\mathrm e}\right ) \textit {\_R} +{\mathrm e}^{2}+15 \,{\mathrm e}\right ) \ln \left (x -\textit {\_R} \right )}{{\mathrm e}^{2} \textit {\_R} +15 \textit {\_R}^{2} {\mathrm e}+50 \textit {\_R}^{3}+{\mathrm e}^{2}-20 \textit {\_R} \,{\mathrm e}-225 \textit {\_R}^{2}-15 \,{\mathrm e}+225 \textit {\_R}}\right )}{5}\right )\) | \(129\) |
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {8 \, {\left (25 \, x^{3} - 75 \, x^{2} + {\left (x + 1\right )} e^{2} + 5 \, {\left (x^{2} - 6 \, x - 3\right )} e + 150 \, x\right )} \log \left (2\right )}{25 \, {\left (5 \, x^{2} + {\left (x + 1\right )} e - 15 \, x\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (24) = 48\).
Time = 0.68 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=- \frac {8 x \log {\left (2 \right )}}{5} - \frac {x \left (- 280 e \log {\left (2 \right )} + 8 e^{2} \log {\left (2 \right )} + 1200 \log {\left (2 \right )}\right ) - 120 e \log {\left (2 \right )} + 8 e^{2} \log {\left (2 \right )}}{125 x^{2} + x \left (-375 + 25 e\right ) + 25 e} \]
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Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {8}{25} \, {\left (5 \, x + \frac {x {\left (e^{2} - 35 \, e + 150\right )} + e^{2} - 15 \, e}{5 \, x^{2} + x {\left (e - 15\right )} + e}\right )} \log \left (2\right ) \]
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Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {8}{25} \, {\left (5 \, x + \frac {x e^{2} - 35 \, x e + 150 \, x + e^{2} - 15 \, e}{5 \, x^{2} + x e - 15 \, x + e}\right )} \log \left (2\right ) \]
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Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {\left (-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )\right ) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx=-\frac {\frac {8\,{\mathrm {e}}^2\,\ln \left (2\right )}{5}-24\,\mathrm {e}\,\ln \left (2\right )+x\,\left (240\,\ln \left (2\right )-56\,\mathrm {e}\,\ln \left (2\right )+\frac {8\,{\mathrm {e}}^2\,\ln \left (2\right )}{5}\right )}{25\,x^2+\left (5\,\mathrm {e}-75\right )\,x+5\,\mathrm {e}}-\frac {8\,x\,\ln \left (2\right )}{5} \]
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