Integrand size = 39, antiderivative size = 19 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=e^{x/5} x \left (4+x \log \left (2 x^2\right )\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(19)=38\).
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.68, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 2207, 2225, 1607, 2227, 2634} \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=e^{x/5} x^2 \log \left (2 x^2\right )-10 e^{x/5} x-20 e^{x/5}+2 e^{x/5} (7 x+10) \]
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Rule 12
Rule 1607
Rule 2207
Rule 2225
Rule 2227
Rule 2634
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx \\ & = \frac {1}{5} \int e^{x/5} (20+14 x) \, dx+\frac {1}{5} \int e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right ) \, dx \\ & = 2 e^{x/5} (10+7 x)+\frac {1}{5} \int e^{x/5} x (10+x) \log \left (2 x^2\right ) \, dx-14 \int e^{x/5} \, dx \\ & = -70 e^{x/5}+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right )-\frac {1}{5} \int 10 e^{x/5} x \, dx \\ & = -70 e^{x/5}+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right )-2 \int e^{x/5} x \, dx \\ & = -70 e^{x/5}-10 e^{x/5} x+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right )+10 \int e^{x/5} \, dx \\ & = -20 e^{x/5}-10 e^{x/5} x+2 e^{x/5} (10+7 x)+e^{x/5} x^2 \log \left (2 x^2\right ) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=e^{x/5} x \left (4+x \log \left (2 x^2\right )\right ) \]
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Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21
method | result | size |
norman | \({\mathrm e}^{\frac {x}{5}} \ln \left (2 x^{2}\right ) x^{2}+4 x \,{\mathrm e}^{\frac {x}{5}}\) | \(23\) |
parallelrisch | \({\mathrm e}^{\frac {x}{5}} \ln \left (2 x^{2}\right ) x^{2}+4 x \,{\mathrm e}^{\frac {x}{5}}\) | \(23\) |
default | \(4 x \,{\mathrm e}^{\frac {x}{5}}+\left (\ln \left (2 x^{2}\right )-2 \ln \left (x \right )\right ) x^{2} {\mathrm e}^{\frac {x}{5}}+2 \ln \left (x \right ) x^{2} {\mathrm e}^{\frac {x}{5}}\) | \(39\) |
risch | \(2 \ln \left (x \right ) x^{2} {\mathrm e}^{\frac {x}{5}}+\frac {\left (100-20 x -i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x^{2} \ln \left (2\right )\right ) {\mathrm e}^{\frac {x}{5}}}{2}+\frac {\left (-250+70 x \right ) {\mathrm e}^{\frac {x}{5}}}{5}\) | \(100\) |
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x^{2} e^{\left (\frac {1}{5} \, x\right )} \log \left (2 \, x^{2}\right ) + 4 \, x e^{\left (\frac {1}{5} \, x\right )} \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=\left (x^{2} \log {\left (2 x^{2} \right )} + 4 x\right ) e^{\frac {x}{5}} \]
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Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x^{2} e^{\left (\frac {1}{5} \, x\right )} \log \left (2 \, x^{2}\right ) + 4 \, {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} + 20 \, e^{\left (\frac {1}{5} \, x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x^{2} e^{\left (\frac {1}{5} \, x\right )} \log \left (2 \, x^{2}\right ) + 2 \, {\left (7 \, x - 25\right )} e^{\left (\frac {1}{5} \, x\right )} - 10 \, {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} \]
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Time = 8.83 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{5} \left (e^{x/5} (20+14 x)+e^{x/5} \left (10 x+x^2\right ) \log \left (2 x^2\right )\right ) \, dx=x\,{\mathrm {e}}^{x/5}\,\left (x\,\ln \left (2\,x^2\right )+4\right ) \]
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