Integrand size = 51, antiderivative size = 17 \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=4+e^{\frac {x}{e^{12} (2+2 x)^2}}+x \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {12, 21, 6820, 6838} \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=x+e^{\frac {x}{4 e^{12} (x+1)^2}} \]
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Rule 12
Rule 21
Rule 6820
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{(1+x) (2+2 x)^2} \, dx}{e^{12}} \\ & = \frac {\int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{(1+x)^3} \, dx}{4 e^{12}} \\ & = \frac {\int \left (4 e^{12}-\frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (-1+x)}{(1+x)^3}\right ) \, dx}{4 e^{12}} \\ & = x-\frac {\int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (-1+x)}{(1+x)^3} \, dx}{4 e^{12}} \\ & = e^{\frac {x}{4 e^{12} (1+x)^2}}+x \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=e^{\frac {x}{4 e^{12} (1+x)^2}}+x \]
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Time = 4.77 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82
method | result | size |
risch | \(x +{\mathrm e}^{\frac {x \,{\mathrm e}^{-12}}{4 \left (1+x \right )^{2}}}\) | \(14\) |
parts | \(x +{\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}}\) | \(21\) |
default | \({\mathrm e}^{-2} \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}} {\mathrm e}^{2}+{\mathrm e}^{2} x \right )\) | \(32\) |
norman | \(\frac {\left (x^{3} {\mathrm e}^{5}-3 x \,{\mathrm e}^{5}+{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}}+x^{2} {\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}}+2 x \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}}-2 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{\left (1+x \right )^{2}}\) | \(95\) |
parallelrisch | \(\frac {{\mathrm e}^{-2} \left ({\mathrm e}^{2} x^{2} \left (2+2 x \right )^{4} {\mathrm e}^{20} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}}+2 \,{\mathrm e}^{2} x \left (2+2 x \right )^{4} {\mathrm e}^{20} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}}+{\mathrm e}^{2} \left (2+2 x \right )^{4} {\mathrm e}^{20} {\mathrm e}^{\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{-10}}{\left (2+2 x \right )^{2}}}+{\mathrm e}^{2} \left (2+2 x \right )^{4} {\mathrm e}^{20} x^{3}-3 \left (2+2 x \right )^{4} {\mathrm e}^{20} x \,{\mathrm e}^{2}-2 \left (2+2 x \right )^{4} {\mathrm e}^{20} {\mathrm e}^{2}\right ) {\mathrm e}^{-20}}{\left (2+2 x \right )^{4} \left (1+x \right )^{2}}\) | \(172\) |
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=x + e^{\left (\frac {x e^{\left (-12\right )}}{4 \, {\left (x^{2} + 2 \, x + 1\right )}}\right )} \]
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Time = 0.15 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=x + e^{\frac {x}{\left (2 x + 2\right )^{2} e^{12}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (17) = 34\).
Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 7.59 \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=\frac {1}{2} \, {\left ({\left (2 \, x - \frac {6 \, x + 5}{x^{2} + 2 \, x + 1} - 6 \, \log \left (x + 1\right )\right )} e^{12} + 3 \, {\left (\frac {4 \, x + 3}{x^{2} + 2 \, x + 1} + 2 \, \log \left (x + 1\right )\right )} e^{12} - \frac {3 \, {\left (2 \, x + 1\right )} e^{12}}{x^{2} + 2 \, x + 1} - \frac {e^{12}}{x^{2} + 2 \, x + 1} + 2 \, e^{\left (-\frac {1}{4 \, {\left (x^{2} e^{12} + 2 \, x e^{12} + e^{12}\right )}} + \frac {1}{4 \, {\left (x e^{12} + e^{12}\right )}} + 12\right )}\right )} e^{\left (-12\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (17) = 34\).
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 5.18 \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=x + e^{\left (-\frac {12 \, x^{2} e^{12}}{x^{2} e^{12} + 2 \, x e^{12} + e^{12}} - \frac {24 \, x e^{12}}{x^{2} e^{12} + 2 \, x e^{12} + e^{12}} + \frac {x}{4 \, {\left (x^{2} e^{12} + 2 \, x e^{12} + e^{12}\right )}} - \frac {12 \, e^{12}}{x^{2} e^{12} + 2 \, x e^{12} + e^{12}} + 12\right )} \]
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Time = 0.73 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {x}{e^{12} (2+2 x)^2}} (1-x)+e^{12} (1+x) (2+2 x)^2}{e^{12} (1+x) (2+2 x)^2} \, dx=x+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-12}}{4\,x^2+8\,x+4}} \]
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