Integrand size = 32, antiderivative size = 17 \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=\frac {x}{5}+\left (3+2^{2 x}\right )^2 x \]
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Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(17)=34\).
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 3.94, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 2207, 2225} \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=\frac {46 x}{5}+\frac {3\ 2^{2 x} (2 x \log (2)+1)}{\log (2)}+\frac {2^{4 x-2} (4 x \log (2)+1)}{\log (2)}-\frac {3\ 2^{2 x}}{\log (2)}-\frac {2^{4 x-2}}{\log (2)} \]
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Rule 12
Rule 2207
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx \\ & = \frac {46 x}{5}+\frac {1}{5} \int 2^{4 x} (5+20 x \log (2)) \, dx+\frac {1}{5} \int 2^{2 x} (30+60 x \log (2)) \, dx \\ & = \frac {46 x}{5}+\frac {3\ 2^{2 x} (1+2 x \log (2))}{\log (2)}+\frac {2^{-2+4 x} (1+4 x \log (2))}{\log (2)}-6 \int 2^{2 x} \, dx-\int 2^{4 x} \, dx \\ & = \frac {46 x}{5}-\frac {3\ 2^{2 x}}{\log (2)}-\frac {2^{-2+4 x}}{\log (2)}+\frac {3\ 2^{2 x} (1+2 x \log (2))}{\log (2)}+\frac {2^{-2+4 x} (1+4 x \log (2))}{\log (2)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.65 \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=\frac {1}{5} \left (46 x+5\ 16^x x+\frac {15\ 4^{1+x} x \log (2)}{\log (4)}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06
method | result | size |
risch | \(\frac {46 x}{5}+x 4^{2 x}+6 \,4^{x} x\) | \(18\) |
default | \(\frac {46 x}{5}+x \,{\mathrm e}^{4 x \ln \left (2\right )}+6 \,{\mathrm e}^{2 x \ln \left (2\right )} x\) | \(24\) |
norman | \(\frac {46 x}{5}+x \,{\mathrm e}^{4 x \ln \left (2\right )}+6 \,{\mathrm e}^{2 x \ln \left (2\right )} x\) | \(24\) |
parallelrisch | \(\frac {46 x}{5}+x \,{\mathrm e}^{4 x \ln \left (2\right )}+6 \,{\mathrm e}^{2 x \ln \left (2\right )} x\) | \(24\) |
parts | \(\frac {46 x}{5}+x \,{\mathrm e}^{4 x \ln \left (2\right )}+6 \,{\mathrm e}^{2 x \ln \left (2\right )} x\) | \(24\) |
derivativedivides | \(\frac {92 x \ln \left (2\right )+10 \,{\mathrm e}^{4 x \ln \left (2\right )} \ln \left (2\right ) x +60 \,{\mathrm e}^{2 x \ln \left (2\right )} \ln \left (2\right ) x}{10 \ln \left (2\right )}\) | \(37\) |
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=2^{4 \, x} x + 6 \cdot 2^{2 \, x} x + \frac {46}{5} \, x \]
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Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=x e^{4 x \log {\left (2 \right )}} + 6 x e^{2 x \log {\left (2 \right )}} + \frac {46 x}{5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (15) = 30\).
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.47 \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=6 \cdot 2^{2 \, x} x + \frac {46}{5} \, x + \frac {{\left (4 \, x \log \left (2\right ) - 1\right )} 2^{4 \, x}}{4 \, \log \left (2\right )} + \frac {2^{4 \, x - 2}}{\log \left (2\right )} \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=2^{4 \, x} x + 6 \cdot 2^{2 \, x} x + \frac {46}{5} \, x \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {1}{5} \left (46+2^{4 x} (5+20 x \log (2))+2^{2 x} (30+60 x \log (2))\right ) \, dx=\frac {x\,\left (30\,2^{2\,x}+5\,2^{4\,x}+46\right )}{5} \]
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