Integrand size = 79, antiderivative size = 25 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=e^2+\frac {\log (2)}{5 x \left (\frac {5}{4}+4 e^2+x\right )^2} \]
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Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(25)=50\).
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6, 12, 2099} \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=-\frac {64 \log (2)}{5 \left (5+16 e^2\right )^2 \left (4 x+16 e^2+5\right )}-\frac {64 \log (2)}{5 \left (5+16 e^2\right ) \left (4 x+16 e^2+5\right )^2}+\frac {16 \log (2)}{5 \left (5+16 e^2\right )^2 x} \]
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Rule 6
Rule 12
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{\left (625+20480 e^6\right ) x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx \\ & = \log (2) \int \frac {-80-256 e^2-192 x}{\left (625+20480 e^6\right ) x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx \\ & = \log (2) \int \left (-\frac {16}{5 \left (5+16 e^2\right )^2 x^2}+\frac {512}{5 \left (5+16 e^2\right ) \left (5+16 e^2+4 x\right )^3}+\frac {256}{5 \left (5+16 e^2\right )^2 \left (5+16 e^2+4 x\right )^2}\right ) \, dx \\ & = \frac {16 \log (2)}{5 \left (5+16 e^2\right )^2 x}-\frac {64 \log (2)}{5 \left (5+16 e^2\right ) \left (5+16 e^2+4 x\right )^2}-\frac {64 \log (2)}{5 \left (5+16 e^2\right )^2 \left (5+16 e^2+4 x\right )} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \log (2)}{5 x \left (5+16 e^2+4 x\right )^2} \]
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Time = 0.93 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76
method | result | size |
norman | \(\frac {16 \ln \left (2\right )}{5 x \left (16 \,{\mathrm e}^{2}+4 x +5\right )^{2}}\) | \(19\) |
risch | \(\frac {16 \ln \left (2\right )}{5 x \left (256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} x +16 x^{2}+160 \,{\mathrm e}^{2}+40 x +25\right )}\) | \(33\) |
gosper | \(\frac {16 \ln \left (2\right )}{5 x \left (256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} x +16 x^{2}+160 \,{\mathrm e}^{2}+40 x +25\right )}\) | \(35\) |
parallelrisch | \(\frac {16 \ln \left (2\right )}{5 x \left (256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} x +16 x^{2}+160 \,{\mathrm e}^{2}+40 x +25\right )}\) | \(35\) |
default | \(16 \ln \left (2\right ) \left (\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (64 \textit {\_Z}^{3}+\left (768 \,{\mathrm e}^{2}+240\right ) \textit {\_Z}^{2}+\left (1920 \,{\mathrm e}^{2}+3072 \,{\mathrm e}^{4}+300\right ) \textit {\_Z} +1200 \,{\mathrm e}^{2}+3840 \,{\mathrm e}^{4}+4096 \,{\mathrm e}^{6}+125\right )}{\sum }\frac {\left (9375+32000 \,{\mathrm e}^{2} \textit {\_R} +153600 \textit {\_R} \,{\mathrm e}^{4}+327680 \textit {\_R} \,{\mathrm e}^{6}+262144 \textit {\_R} \,{\mathrm e}^{8}+150000 \,{\mathrm e}^{2}+960000 \,{\mathrm e}^{4}+3072000 \,{\mathrm e}^{6}+4915200 \,{\mathrm e}^{8}+3145728 \,{\mathrm e}^{10}+2500 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{25+256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} \textit {\_R} +16 \textit {\_R}^{2}+160 \,{\mathrm e}^{2}+40 \textit {\_R}}\right )}{15 \left (1200 \,{\mathrm e}^{2}+3840 \,{\mathrm e}^{4}+4096 \,{\mathrm e}^{6}+125\right )^{2}}-\frac {-38400 \,{\mathrm e}^{4}-81920 \,{\mathrm e}^{6}-65536 \,{\mathrm e}^{8}-8000 \,{\mathrm e}^{2}-625}{5 \left (1200 \,{\mathrm e}^{2}+3840 \,{\mathrm e}^{4}+4096 \,{\mathrm e}^{6}+125\right )^{2} x}\right )\) | \(185\) |
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Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \, \log \left (2\right )}{5 \, {\left (16 \, x^{3} + 40 \, x^{2} + 256 \, x e^{4} + 32 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{2} + 25 \, x\right )}} \]
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Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \log {\left (2 \right )}}{80 x^{3} + x^{2} \cdot \left (200 + 640 e^{2}\right ) + x \left (125 + 800 e^{2} + 1280 e^{4}\right )} \]
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Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \, \log \left (2\right )}{5 \, {\left (16 \, x^{3} + 8 \, x^{2} {\left (16 \, e^{2} + 5\right )} + x {\left (256 \, e^{4} + 160 \, e^{2} + 25\right )}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (21) = 42\).
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=-\frac {16}{5} \, {\left (\frac {8 \, {\left (2 \, x + 16 \, e^{2} + 5\right )}}{{\left (4 \, x + 16 \, e^{2} + 5\right )}^{2} {\left (256 \, e^{4} + 160 \, e^{2} + 25\right )}} - \frac {1}{x {\left (256 \, e^{4} + 160 \, e^{2} + 25\right )}}\right )} \log \left (2\right ) \]
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Time = 8.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16\,\ln \left (2\right )}{5\,x\,{\left (4\,x+16\,{\mathrm {e}}^2+5\right )}^2} \]
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