Integrand size = 28, antiderivative size = 13 \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx=(5+x) \left (\frac {36}{e^4}+x\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(27\) vs. \(2(13)=26\).
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.08, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {12} \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx=x^3+5 x^2+\frac {1296 x}{e^8}+\frac {18 (2 x+5)^2}{e^4} \]
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Rule 12
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )\right ) \, dx}{e^8} \\ & = \frac {1296 x}{e^8}+\frac {18 (5+2 x)^2}{e^4}+\int \left (10 x+3 x^2\right ) \, dx \\ & = \frac {1296 x}{e^8}+5 x^2+x^3+\frac {18 (5+2 x)^2}{e^4} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(29\) vs. \(2(13)=26\).
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 2.23 \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx=\frac {1296 x}{e^8}+\frac {360 x}{e^4}+5 x^2+\frac {72 x^2}{e^4}+x^3 \]
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Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.08
method | result | size |
risch | \(72 \,{\mathrm e}^{-4} x^{2}+5 x^{2}+x^{3}+1296 \,{\mathrm e}^{-8} x +360 \,{\mathrm e}^{-4} x\) | \(27\) |
gosper | \(x \left (x^{2} {\mathrm e}^{8}+5 x \,{\mathrm e}^{8}+72 x \,{\mathrm e}^{4}+360 \,{\mathrm e}^{4}+1296\right ) {\mathrm e}^{-8}\) | \(37\) |
parallelrisch | \({\mathrm e}^{-8} \left ({\mathrm e}^{8} x^{3}+5 x^{2} {\mathrm e}^{8}+72 x^{2} {\mathrm e}^{4}+360 x \,{\mathrm e}^{4}+1296 x \right )\) | \(43\) |
norman | \(\left ({\mathrm e}^{7} x^{3}+{\mathrm e}^{3} \left (5 \,{\mathrm e}^{4}+72\right ) x^{2}+72 \left (5 \,{\mathrm e}^{4}+18\right ) {\mathrm e}^{-1} x \right ) {\mathrm e}^{-7}\) | \(46\) |
default | \({\mathrm e}^{-8} \left ({\mathrm e}^{8} x^{3}+\frac {\left (\left (10 \,{\mathrm e}^{4}+36\right ) {\mathrm e}^{4}+108 \,{\mathrm e}^{4}\right ) x^{2}}{2}+360 x \,{\mathrm e}^{4}+1296 x \right )\) | \(50\) |
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Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (12) = 24\).
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.31 \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx={\left ({\left (x^{3} + 5 \, x^{2}\right )} e^{8} + 72 \, {\left (x^{2} + 5 \, x\right )} e^{4} + 1296 \, x\right )} e^{\left (-8\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (10) = 20\).
Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.08 \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx=x^{3} + \frac {x^{2} \cdot \left (72 + 5 e^{4}\right )}{e^{4}} + \frac {x \left (1296 + 360 e^{4}\right )}{e^{8}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (12) = 24\).
Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.31 \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx={\left ({\left (x^{3} + 5 \, x^{2}\right )} e^{8} + 72 \, {\left (x^{2} + 5 \, x\right )} e^{4} + 1296 \, x\right )} e^{\left (-8\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (12) = 24\).
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.31 \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx={\left ({\left (x^{3} + 5 \, x^{2}\right )} e^{8} + 72 \, {\left (x^{2} + 5 \, x\right )} e^{4} + 1296 \, x\right )} e^{\left (-8\right )} \]
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Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.00 \[ \int \frac {1296+e^4 (360+144 x)+e^8 \left (10 x+3 x^2\right )}{e^8} \, dx=x^3+{\mathrm {e}}^{-4}\,\left (5\,{\mathrm {e}}^4+72\right )\,x^2+{\mathrm {e}}^{-8}\,\left (360\,{\mathrm {e}}^4+1296\right )\,x \]
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