Integrand size = 55, antiderivative size = 21 \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=16 \left (-e^x+x\right )^2 (2-\log (2)-\log (15)) \]
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Leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(21)=42\).
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.62, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6, 2225, 2218, 2207} \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=16 x^2 (2-\log (30))-32 e^x (x (2-\log (30))+2-\log (30))+32 e^x (2-\log (30))+16 e^{2 x} (2-\log (30)) \]
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Rule 6
Rule 2207
Rule 2218
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \int \left (x (64-32 \log (2))+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx \\ & = \int \left (e^{2 x} (64-32 \log (2)-32 \log (15))+x (64-32 \log (2)-32 \log (15))+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx \\ & = 16 x^2 (2-\log (30))+(32 (2-\log (30))) \int e^{2 x} \, dx+\int e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15)) \, dx \\ & = 16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))+\int e^x (-64-64 x+(32+32 x) (\log (2)+\log (15))) \, dx \\ & = 16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))+\int e^x (-32 (2-\log (30))-32 x (2-\log (30))) \, dx \\ & = 16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))-32 e^x (2+x (2-\log (30))-\log (30))+(32 (2-\log (30))) \int e^x \, dx \\ & = 32 e^x (2-\log (30))+16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))-32 e^x (2+x (2-\log (30))-\log (30)) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=-16 \left (e^x-x\right )^2 (-2+\log (30)) \]
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Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14
method | result | size |
norman | \(\left (-16 \ln \left (15\right )-16 \ln \left (2\right )+32\right ) x^{2}+\left (-16 \ln \left (15\right )-16 \ln \left (2\right )+32\right ) {\mathrm e}^{2 x}+\left (32 \ln \left (15\right )+32 \ln \left (2\right )-64\right ) x \,{\mathrm e}^{x}\) | \(45\) |
default | \(\frac {\left (-32 \ln \left (15\right )-32 \ln \left (2\right )+64\right ) {\mathrm e}^{2 x}}{2}-64 \,{\mathrm e}^{x} x +32 x \ln \left (2\right ) {\mathrm e}^{x}+32 \ln \left (15\right ) {\mathrm e}^{x} x +32 x^{2}-16 x^{2} \ln \left (2\right )-16 x^{2} \ln \left (15\right )\) | \(56\) |
parts | \(\frac {\left (-32 \ln \left (15\right )-32 \ln \left (2\right )+64\right ) {\mathrm e}^{2 x}}{2}-64 \,{\mathrm e}^{x} x +32 x \ln \left (2\right ) {\mathrm e}^{x}+32 \ln \left (15\right ) {\mathrm e}^{x} x +32 x^{2}-16 x^{2} \ln \left (2\right )-16 x^{2} \ln \left (15\right )\) | \(56\) |
parallelrisch | \(-16 x^{2} \ln \left (2\right )+32 x \ln \left (2\right ) {\mathrm e}^{x}-16 \ln \left (2\right ) {\mathrm e}^{2 x}-16 x^{2} \ln \left (15\right )+32 \ln \left (15\right ) {\mathrm e}^{x} x -16 \,{\mathrm e}^{2 x} \ln \left (15\right )+32 x^{2}-64 \,{\mathrm e}^{x} x +32 \,{\mathrm e}^{2 x}\) | \(62\) |
risch | \(-16 \ln \left (2\right ) {\mathrm e}^{2 x}-16 \,{\mathrm e}^{2 x} \ln \left (5\right )-16 \,{\mathrm e}^{2 x} \ln \left (3\right )+32 \,{\mathrm e}^{2 x}+32 \left (\ln \left (2\right )+\ln \left (5\right )+\ln \left (3\right )-2\right ) x \,{\mathrm e}^{x}-16 x^{2} \ln \left (5\right )-16 x^{2} \ln \left (3\right )-16 x^{2} \ln \left (2\right )+32 x^{2}\) | \(71\) |
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (16) = 32\).
Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=-16 \, x^{2} \log \left (15\right ) - 16 \, x^{2} \log \left (2\right ) + 32 \, x^{2} - 16 \, {\left (\log \left (15\right ) + \log \left (2\right ) - 2\right )} e^{\left (2 \, x\right )} + 32 \, {\left (x \log \left (15\right ) + x \log \left (2\right ) - 2 \, x\right )} e^{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (17) = 34\).
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43 \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=x^{2} \left (- 16 \log {\left (15 \right )} - 16 \log {\left (2 \right )} + 32\right ) + \left (- 64 x + 32 x \log {\left (2 \right )} + 32 x \log {\left (15 \right )}\right ) e^{x} + \left (- 16 \log {\left (15 \right )} - 16 \log {\left (2 \right )} + 32\right ) e^{2 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (16) = 32\).
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=32 \, x {\left (\log \left (5\right ) + \log \left (3\right ) + \log \left (2\right ) - 2\right )} e^{x} - 16 \, x^{2} \log \left (15\right ) - 16 \, x^{2} \log \left (2\right ) + 32 \, x^{2} - 16 \, {\left (\log \left (15\right ) + \log \left (2\right ) - 2\right )} e^{\left (2 \, x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (16) = 32\).
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=-16 \, x^{2} \log \left (15\right ) - 16 \, x^{2} \log \left (2\right ) + 32 \, x^{2} - 16 \, {\left (\log \left (15\right ) + \log \left (2\right ) - 2\right )} e^{\left (2 \, x\right )} + 32 \, {\left (x \log \left (15\right ) + x \log \left (2\right ) - 2 \, x\right )} e^{x} \]
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Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \left (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx=-{\left (x-{\mathrm {e}}^x\right )}^2\,\left (16\,\ln \left (30\right )-32\right ) \]
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