Integrand size = 34, antiderivative size = 18 \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx=e^{5+x} \left (2+e^{25+\frac {x}{5}}\right ) x \]
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Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(18)=36\).
Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.72, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 6874, 2207, 2225} \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx=2 e^{x+5} (x+1)-2 e^{x+5}-\frac {5}{6} e^{\frac {6 x}{5}+30}+\frac {1}{6} e^{\frac {6 x}{5}+30} (6 x+5) \]
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Rule 12
Rule 2207
Rule 2225
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx \\ & = \frac {1}{5} \int \left (10 e^{5+x} (1+x)+e^{30+\frac {6 x}{5}} (5+6 x)\right ) \, dx \\ & = \frac {1}{5} \int e^{30+\frac {6 x}{5}} (5+6 x) \, dx+2 \int e^{5+x} (1+x) \, dx \\ & = 2 e^{5+x} (1+x)+\frac {1}{6} e^{30+\frac {6 x}{5}} (5+6 x)-2 \int e^{5+x} \, dx-\int e^{30+\frac {6 x}{5}} \, dx \\ & = -2 e^{5+x}-\frac {5}{6} e^{30+\frac {6 x}{5}}+2 e^{5+x} (1+x)+\frac {1}{6} e^{30+\frac {6 x}{5}} (5+6 x) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx=2 e^{5+x} x+e^{30+\frac {6 x}{5}} x \]
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Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94
method | result | size |
risch | \(x \,{\mathrm e}^{30+\frac {6 x}{5}}+2 x \,{\mathrm e}^{5+x}\) | \(17\) |
parallelrisch | \({\mathrm e}^{5} {\mathrm e}^{\frac {x}{5}+25} {\mathrm e}^{x +\ln \left (x \right )}+2 \,{\mathrm e}^{5} {\mathrm e}^{x +\ln \left (x \right )}\) | \(25\) |
norman | \({\mathrm e}^{-125} {\mathrm e}^{5} x \,{\mathrm e}^{\frac {6 x}{5}+150}+2 \,{\mathrm e}^{-125} {\mathrm e}^{5} x \,{\mathrm e}^{x +125}\) | \(35\) |
default | \(2 \,{\mathrm e}^{5} {\mathrm e}^{x}+2 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+\frac {5 \,{\mathrm e}^{30} {\mathrm e}^{\frac {6 x}{5}}}{6}+\frac {6 \,{\mathrm e}^{30} \left (\frac {5 x \,{\mathrm e}^{\frac {6 x}{5}}}{6}-\frac {25 \,{\mathrm e}^{\frac {6 x}{5}}}{36}\right )}{5}\) | \(47\) |
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Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx={\left (x e^{\left (\frac {6}{5} \, x + 180\right )} + 2 \, x e^{\left (x + 155\right )}\right )} e^{\left (-150\right )} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx=x e^{30} \left (e^{x}\right )^{\frac {6}{5}} + 2 x e^{5} e^{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (15) = 30\).
Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.44 \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx=\frac {1}{6} \, {\left (6 \, x e^{30} - 5 \, e^{30}\right )} e^{\left (\frac {6}{5} \, x\right )} + 2 \, {\left (x e^{5} - e^{5}\right )} e^{x} + \frac {5}{6} \, e^{\left (\frac {6}{5} \, x + 30\right )} + 2 \, e^{\left (x + 5\right )} \]
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Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx=x e^{\left (\frac {6}{5} \, x + 30\right )} + 2 \, x e^{\left (x + 5\right )} \]
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Time = 12.53 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{5} e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx=x\,\left (2\,{\mathrm {e}}^{x+5}+{\mathrm {e}}^{\frac {6\,x}{5}+30}\right ) \]
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