Integrand size = 31, antiderivative size = 20 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-2^{2 \left (1-e^{50}\right ) x}+e^{x^2} \]
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Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2240, 2259, 2225} \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=e^{x^2}-2^{2 \left (1-e^{50}\right ) x} \]
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Rule 2225
Rule 2240
Rule 2259
Rubi steps \begin{align*} \text {integral}& = 2 \int e^{x^2} x \, dx-\left (2 \left (1-e^{50}\right ) \log (2)\right ) \int 2^{2 x-2 e^{50} x} \, dx \\ & = e^{x^2}-\left (2 \left (1-e^{50}\right ) \log (2)\right ) \int 2^{2 \left (1-e^{50}\right ) x} \, dx \\ & = -2^{2 \left (1-e^{50}\right ) x}+e^{x^2} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-4^{x-e^{50} x}+e^{x^2} \]
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Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
method | result | size |
risch | \(-2^{-2 \left ({\mathrm e}^{50}-1\right ) x}+{\mathrm e}^{x^{2}}\) | \(18\) |
norman | \(-{\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}+{\mathrm e}^{x^{2}}\) | \(23\) |
parallelrisch | \(-{\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}+{\mathrm e}^{x^{2}}\) | \(23\) |
default | \(\frac {\left ({\mathrm e}^{50}-1\right ) {\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}}{-{\mathrm e}^{50}+1}+{\mathrm e}^{x^{2}}\) | \(38\) |
parts | \(\frac {\left ({\mathrm e}^{50}-1\right ) {\mathrm e}^{2 \left (-x \,{\mathrm e}^{50}+x \right ) \ln \left (2\right )}}{-{\mathrm e}^{50}+1}+{\mathrm e}^{x^{2}}\) | \(38\) |
meijerg | \(-1+{\mathrm e}^{x^{2}}+\frac {{\mathrm e}^{50} \left (1-{\mathrm e}^{-2 x \left ({\mathrm e}^{50}-1\right ) \ln \left (2\right )}\right )}{{\mathrm e}^{50}-1}-\frac {1-{\mathrm e}^{-2 x \left ({\mathrm e}^{50}-1\right ) \ln \left (2\right )}}{{\mathrm e}^{50}-1}\) | \(52\) |
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-2^{-2 \, x e^{50} + 2 \, x} + e^{\left (x^{2}\right )} \]
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Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=e^{x^{2}} - e^{2 \left (- x e^{50} + x\right ) \log {\left (2 \right )}} \]
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Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-\frac {1}{2} \cdot 2^{-2 \, x e^{50} + 2 \, x + 1} + e^{\left (x^{2}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx=-\frac {2^{-2 \, x e^{50} + 2 \, x} {\left (e^{50} - 1\right )} \log \left (2\right )}{e^{50} \log \left (2\right ) - \log \left (2\right )} + e^{\left (x^{2}\right )} \]
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Time = 7.60 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \left (2 e^{x^2} x+2^{2 x-2 e^{50} x} \left (-2+2 e^{50}\right ) \log (2)\right ) \, dx={\mathrm {e}}^{x^2}-\frac {2^{2\,x}}{2^{2\,x\,{\mathrm {e}}^{50}}} \]
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