Integrand size = 122, antiderivative size = 19 \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {1}{16} \left (e^{x+e^2 x^3}+x\right )^4 \]
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Leaf count is larger than twice the leaf count of optimal. \(147\) vs. \(2(19)=38\).
Time = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 7.74, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12, 6838, 2326, 1608} \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {x^4}{16}+\frac {1}{16} e^{4 e^2 x^3+4 x}+\frac {e^{e^2 x^3+x} \left (3 e^2 x^3+x\right ) x^2}{4 \left (3 e^2 x^2+1\right )}+\frac {3 e^{2 e^2 x^3+2 x} \left (3 e^2 x^3+x\right ) x}{8 \left (3 e^2 x^2+1\right )}+\frac {e^{3 e^2 x^3+3 x} \left (3 e^2 x^3+x\right )}{4 \left (3 e^2 x^2+1\right )} \]
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Rule 12
Rule 1608
Rule 2326
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx \\ & = \frac {x^4}{16}+\frac {1}{4} \int e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right ) \, dx+\frac {1}{4} \int e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right ) \, dx+\frac {1}{4} \int e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right ) \, dx+\frac {1}{4} \int e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right ) \, dx \\ & = \frac {1}{16} e^{4 x+4 e^2 x^3}+\frac {x^4}{16}+\frac {e^{3 x+3 e^2 x^3} \left (x+3 e^2 x^3\right )}{4 \left (1+3 e^2 x^2\right )}+\frac {1}{4} \int e^{x+e^2 x^3} x^2 \left (3+x+3 e^2 x^3\right ) \, dx+\frac {1}{4} \int e^{2 x+2 e^2 x^3} x \left (3+3 x+9 e^2 x^3\right ) \, dx \\ & = \frac {1}{16} e^{4 x+4 e^2 x^3}+\frac {x^4}{16}+\frac {e^{3 x+3 e^2 x^3} \left (x+3 e^2 x^3\right )}{4 \left (1+3 e^2 x^2\right )}+\frac {3 e^{2 x+2 e^2 x^3} x \left (x+3 e^2 x^3\right )}{8 \left (1+3 e^2 x^2\right )}+\frac {e^{x+e^2 x^3} x^2 \left (x+3 e^2 x^3\right )}{4 \left (1+3 e^2 x^2\right )} \\ \end{align*}
Time = 1.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {1}{16} \left (e^{x+e^2 x^3}+x\right )^4 \]
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Time = 0.64 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.63
method | result | size |
risch | \(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x \left (x^{2} {\mathrm e}^{2}+1\right )}}{16}+\frac {{\mathrm e}^{3 x \left (x^{2} {\mathrm e}^{2}+1\right )} x}{4}+\frac {3 \,{\mathrm e}^{2 x \left (x^{2} {\mathrm e}^{2}+1\right )} x^{2}}{8}+\frac {{\mathrm e}^{x \left (x^{2} {\mathrm e}^{2}+1\right )} x^{3}}{4}\) | \(69\) |
default | \(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x^{3} {\mathrm e}^{2}+4 x}}{16}+\frac {{\mathrm e}^{3 x^{3} {\mathrm e}^{2}+3 x} x}{4}+\frac {3 \,{\mathrm e}^{2 x^{3} {\mathrm e}^{2}+2 x} x^{2}}{8}+\frac {{\mathrm e}^{x^{3} {\mathrm e}^{2}+x} x^{3}}{4}\) | \(72\) |
norman | \(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x^{3} {\mathrm e}^{2}+4 x}}{16}+\frac {{\mathrm e}^{3 x^{3} {\mathrm e}^{2}+3 x} x}{4}+\frac {3 \,{\mathrm e}^{2 x^{3} {\mathrm e}^{2}+2 x} x^{2}}{8}+\frac {{\mathrm e}^{x^{3} {\mathrm e}^{2}+x} x^{3}}{4}\) | \(72\) |
parallelrisch | \(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x^{3} {\mathrm e}^{2}+4 x}}{16}+\frac {{\mathrm e}^{3 x^{3} {\mathrm e}^{2}+3 x} x}{4}+\frac {3 \,{\mathrm e}^{2 x^{3} {\mathrm e}^{2}+2 x} x^{2}}{8}+\frac {{\mathrm e}^{x^{3} {\mathrm e}^{2}+x} x^{3}}{4}\) | \(72\) |
parts | \(\frac {x^{4}}{16}+\frac {{\mathrm e}^{4 x^{3} {\mathrm e}^{2}+4 x}}{16}+\frac {{\mathrm e}^{3 x^{3} {\mathrm e}^{2}+3 x} x}{4}+\frac {3 \,{\mathrm e}^{2 x^{3} {\mathrm e}^{2}+2 x} x^{2}}{8}+\frac {{\mathrm e}^{x^{3} {\mathrm e}^{2}+x} x^{3}}{4}\) | \(72\) |
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (15) = 30\).
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 3.47 \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {1}{16} \, x^{4} + \frac {1}{4} \, x^{3} e^{\left (x^{3} e^{2} + x\right )} + \frac {3}{8} \, x^{2} e^{\left (2 \, x^{3} e^{2} + 2 \, x\right )} + \frac {1}{4} \, x e^{\left (3 \, x^{3} e^{2} + 3 \, x\right )} + \frac {1}{16} \, e^{\left (4 \, x^{3} e^{2} + 4 \, x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (31) = 62\).
Time = 0.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.74 \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {x^{4}}{16} + \frac {x^{3} e^{x^{3} e^{2} + x}}{4} + \frac {3 x^{2} e^{2 x^{3} e^{2} + 2 x}}{8} + \frac {x e^{3 x^{3} e^{2} + 3 x}}{4} + \frac {e^{4 x^{3} e^{2} + 4 x}}{16} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (15) = 30\).
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 3.47 \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {1}{16} \, x^{4} + \frac {1}{4} \, x^{3} e^{\left (x^{3} e^{2} + x\right )} + \frac {3}{8} \, x^{2} e^{\left (2 \, x^{3} e^{2} + 2 \, x\right )} + \frac {1}{4} \, x e^{\left (3 \, x^{3} e^{2} + 3 \, x\right )} + \frac {1}{16} \, e^{\left (4 \, x^{3} e^{2} + 4 \, x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (15) = 30\).
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 3.47 \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {1}{16} \, x^{4} + \frac {1}{4} \, x^{3} e^{\left (x^{3} e^{2} + x\right )} + \frac {3}{8} \, x^{2} e^{\left (2 \, x^{3} e^{2} + 2 \, x\right )} + \frac {1}{4} \, x e^{\left (3 \, x^{3} e^{2} + 3 \, x\right )} + \frac {1}{16} \, e^{\left (4 \, x^{3} e^{2} + 4 \, x\right )} \]
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Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{4} \left (x^3+e^{4 x+4 e^2 x^3} \left (1+3 e^2 x^2\right )+e^{3 x+3 e^2 x^3} \left (1+3 x+9 e^2 x^3\right )+e^{2 x+2 e^2 x^3} \left (3 x+3 x^2+9 e^2 x^4\right )+e^{x+e^2 x^3} \left (3 x^2+x^3+3 e^2 x^5\right )\right ) \, dx=\frac {{\left (x+{\mathrm {e}}^{{\mathrm {e}}^2\,x^3+x}\right )}^4}{16} \]
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