Integrand size = 32, antiderivative size = 16 \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5}{2} e^{2 x} x \left (-5+e^x+x\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(46\) vs. \(2(16)=32\).
Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.88, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2207, 2225, 2227} \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5}{2} e^{2 x} x^2-\frac {25}{2} e^{2 x} x-\frac {5 e^{3 x}}{6}+\frac {5}{6} e^{3 x} (3 x+1) \]
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Rule 12
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx \\ & = \frac {1}{2} \int e^{3 x} (5+15 x) \, dx+\frac {1}{2} \int e^{2 x} \left (-25-40 x+10 x^2\right ) \, dx \\ & = \frac {5}{6} e^{3 x} (1+3 x)+\frac {1}{2} \int \left (-25 e^{2 x}-40 e^{2 x} x+10 e^{2 x} x^2\right ) \, dx-\frac {5}{2} \int e^{3 x} \, dx \\ & = -\frac {5 e^{3 x}}{6}+\frac {5}{6} e^{3 x} (1+3 x)+5 \int e^{2 x} x^2 \, dx-\frac {25}{2} \int e^{2 x} \, dx-20 \int e^{2 x} x \, dx \\ & = -\frac {25 e^{2 x}}{4}-\frac {5 e^{3 x}}{6}-10 e^{2 x} x+\frac {5}{2} e^{2 x} x^2+\frac {5}{6} e^{3 x} (1+3 x)-5 \int e^{2 x} x \, dx+10 \int e^{2 x} \, dx \\ & = -\frac {5 e^{2 x}}{4}-\frac {5 e^{3 x}}{6}-\frac {25}{2} e^{2 x} x+\frac {5}{2} e^{2 x} x^2+\frac {5}{6} e^{3 x} (1+3 x)+\frac {5}{2} \int e^{2 x} \, dx \\ & = -\frac {5 e^{3 x}}{6}-\frac {25}{2} e^{2 x} x+\frac {5}{2} e^{2 x} x^2+\frac {5}{6} e^{3 x} (1+3 x) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5}{2} e^{2 x} x \left (-5+e^x+x\right ) \]
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Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50
method | result | size |
risch | \(\frac {5 x \,{\mathrm e}^{3 x}}{2}+\frac {\left (5 x^{2}-25 x \right ) {\mathrm e}^{2 x}}{2}\) | \(24\) |
default | \(\frac {5 x \,{\mathrm e}^{3 x}}{2}+\frac {5 \,{\mathrm e}^{2 x} x^{2}}{2}-\frac {25 x \,{\mathrm e}^{2 x}}{2}\) | \(25\) |
norman | \(\frac {5 x \,{\mathrm e}^{3 x}}{2}+\frac {5 \,{\mathrm e}^{2 x} x^{2}}{2}-\frac {25 x \,{\mathrm e}^{2 x}}{2}\) | \(25\) |
parallelrisch | \(\frac {5 x \,{\mathrm e}^{3 x}}{2}+\frac {5 \,{\mathrm e}^{2 x} x^{2}}{2}-\frac {25 x \,{\mathrm e}^{2 x}}{2}\) | \(25\) |
parts | \(\frac {5 x \,{\mathrm e}^{3 x}}{2}+\frac {5 \,{\mathrm e}^{2 x} x^{2}}{2}-\frac {25 x \,{\mathrm e}^{2 x}}{2}\) | \(25\) |
meijerg | \(\frac {5 \,{\mathrm e}^{3 x}}{6}-\frac {5 \left (-6 x +2\right ) {\mathrm e}^{3 x}}{12}-\frac {25 \,{\mathrm e}^{2 x}}{4}+\frac {5 \left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{24}+\frac {5 \left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\) | \(52\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5}{2} \, x e^{\left (3 \, x\right )} + \frac {5}{2} \, {\left (x^{2} - 5 \, x\right )} e^{\left (2 \, x\right )} \]
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Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5 x e^{3 x}}{2} + \frac {\left (10 x^{2} - 50 x\right ) e^{2 x}}{4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (12) = 24\).
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.56 \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5}{2} \, x e^{\left (3 \, x\right )} + \frac {5}{4} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - 5 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - \frac {25}{4} \, e^{\left (2 \, x\right )} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5}{2} \, x e^{\left (3 \, x\right )} + \frac {5}{2} \, {\left (x^{2} - 5 \, x\right )} e^{\left (2 \, x\right )} \]
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Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{2} \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx=\frac {5\,x\,{\mathrm {e}}^{2\,x}\,\left (x+{\mathrm {e}}^x-5\right )}{2} \]
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