Integrand size = 36, antiderivative size = 16 \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx=\left (28+e^{5+x^2}\right ) \left (-5+\frac {1}{e^4}+x\right ) \]
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Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.75, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 2258, 2235, 2240, 2243} \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx=e^{x^2+5} x+\left (1-5 e^4\right ) e^{x^2+1}+28 x \]
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Rule 12
Rule 2235
Rule 2240
Rule 2243
Rule 2258
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )\right ) \, dx}{e^4} \\ & = 28 x+\frac {\int e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right ) \, dx}{e^4} \\ & = 28 x+\frac {\int \left (e^{9+x^2}-2 e^{5+x^2} \left (-1+5 e^4\right ) x+2 e^{9+x^2} x^2\right ) \, dx}{e^4} \\ & = 28 x-\left (2 \left (5-\frac {1}{e^4}\right )\right ) \int e^{5+x^2} x \, dx+\frac {\int e^{9+x^2} \, dx}{e^4}+\frac {2 \int e^{9+x^2} x^2 \, dx}{e^4} \\ & = -\left (\left (5-\frac {1}{e^4}\right ) e^{5+x^2}\right )+28 x+e^{5+x^2} x+\frac {1}{2} e^5 \sqrt {\pi } \text {erfi}(x)-\frac {\int e^{9+x^2} \, dx}{e^4} \\ & = -\left (\left (5-\frac {1}{e^4}\right ) e^{5+x^2}\right )+28 x+e^{5+x^2} x \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.75 \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx=e^{1+x^2} \left (1-5 e^4\right )+28 x+e^{5+x^2} x \]
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Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38
method | result | size |
risch | \(28 x +\left (x \,{\mathrm e}^{4}-5 \,{\mathrm e}^{4}+1\right ) {\mathrm e}^{x^{2}+1}\) | \(22\) |
norman | \(x \,{\mathrm e}^{x^{2}+5}+28 x -\left (5 \,{\mathrm e}^{4}-1\right ) {\mathrm e}^{-4} {\mathrm e}^{x^{2}+5}\) | \(31\) |
parallelrisch | \({\mathrm e}^{-4} \left ({\mathrm e}^{4} {\mathrm e}^{x^{2}+5} x -5 \,{\mathrm e}^{4} {\mathrm e}^{x^{2}+5}+{\mathrm e}^{x^{2}+5}+28 x \,{\mathrm e}^{4}\right )\) | \(38\) |
parts | \(28 x +{\mathrm e}^{-4} \left (\frac {{\mathrm e}^{4} {\mathrm e}^{5} \sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2}+{\mathrm e}^{5} {\mathrm e}^{x^{2}}-5 \,{\mathrm e}^{x^{2}} {\mathrm e}^{5} {\mathrm e}^{4}+2 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{x^{2}} x}{2}-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{4}\right )\right )\) | \(60\) |
default | \({\mathrm e}^{-4} \left (\frac {{\mathrm e}^{4} {\mathrm e}^{5} \sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2}+{\mathrm e}^{5} {\mathrm e}^{x^{2}}-5 \,{\mathrm e}^{x^{2}} {\mathrm e}^{5} {\mathrm e}^{4}+2 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{x^{2}} x}{2}-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{4}\right )+28 x \,{\mathrm e}^{4}\right )\) | \(61\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx={\left (28 \, x e^{4} + {\left ({\left (x - 5\right )} e^{4} + 1\right )} e^{\left (x^{2} + 5\right )}\right )} e^{\left (-4\right )} \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx=28 x + \frac {\left (x e^{4} - 5 e^{4} + 1\right ) e^{x^{2} + 5}}{e^{4}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).
Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.94 \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx={\left (28 \, x e^{4} + x e^{\left (x^{2} + 9\right )} - 5 \, e^{\left (x^{2} + 9\right )} + e^{\left (x^{2} + 5\right )}\right )} e^{\left (-4\right )} \]
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx={\left (28 \, x e^{4} + {\left (x - 5\right )} e^{\left (x^{2} + 9\right )} + e^{\left (x^{2} + 5\right )}\right )} e^{\left (-4\right )} \]
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Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )}{e^4} \, dx=28\,x+x\,{\mathrm {e}}^{x^2+5}+{\mathrm {e}}^{x^2+5}\,\left ({\mathrm {e}}^{-4}-5\right ) \]
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