Integrand size = 43, antiderivative size = 29 \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=\frac {\left (5+\frac {e^x}{2}+5 \left (e^4-x\right )^2 x^2\right ) \log (4)}{x} \]
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Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2228} \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=5 x^3 \log (4)-10 e^4 x^2 \log (4)+5 e^8 x \log (4)+\frac {e^x \log (4)}{2 x}+\frac {5 \log (4)}{x} \]
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Rule 12
Rule 14
Rule 2228
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {e^x (-1+x) \log (4)}{x^2}+\frac {10 \left (-1+e^8 x^2-4 e^4 x^3+3 x^4\right ) \log (4)}{x^2}\right ) \, dx \\ & = \frac {1}{2} \log (4) \int \frac {e^x (-1+x)}{x^2} \, dx+(5 \log (4)) \int \frac {-1+e^8 x^2-4 e^4 x^3+3 x^4}{x^2} \, dx \\ & = \frac {e^x \log (4)}{2 x}+(5 \log (4)) \int \left (e^8-\frac {1}{x^2}-4 e^4 x+3 x^2\right ) \, dx \\ & = \frac {5 \log (4)}{x}+\frac {e^x \log (4)}{2 x}+5 e^8 x \log (4)-10 e^4 x^2 \log (4)+5 x^3 \log (4) \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=\frac {1}{2} \left (\frac {10}{x}+\frac {e^x}{x}+10 e^8 x-20 e^4 x^2+10 x^3\right ) \log (4) \]
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Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10
method | result | size |
parts | \(\frac {\ln \left (2\right ) {\mathrm e}^{x}}{x}+10 \ln \left (2\right ) \left (x^{3}-2 x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{8}+\frac {1}{x}\right )\) | \(32\) |
risch | \(-20 x^{2} {\mathrm e}^{4} \ln \left (2\right )+10 x^{3} \ln \left (2\right )+10 \ln \left (2\right ) {\mathrm e}^{8} x +\frac {10 \ln \left (2\right )}{x}+\frac {\ln \left (2\right ) {\mathrm e}^{x}}{x}\) | \(40\) |
norman | \(\frac {{\mathrm e}^{x} \ln \left (2\right )+10 x^{4} \ln \left (2\right )+10 x^{2} {\mathrm e}^{8} \ln \left (2\right )-20 x^{3} {\mathrm e}^{4} \ln \left (2\right )+10 \ln \left (2\right )}{x}\) | \(42\) |
parallelrisch | \(\frac {20 \ln \left (2\right )+20 x^{2} {\mathrm e}^{8} \ln \left (2\right )-40 x^{3} {\mathrm e}^{4} \ln \left (2\right )+20 x^{4} \ln \left (2\right )+2 \,{\mathrm e}^{x} \ln \left (2\right )}{2 x}\) | \(44\) |
default | \(-\ln \left (2\right ) \operatorname {Ei}_{1}\left (-x \right )+10 x^{3} \ln \left (2\right )+10 \ln \left (2\right ) {\mathrm e}^{8} x +\frac {10 \ln \left (2\right )}{x}-\ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )-20 x^{2} {\mathrm e}^{4} \ln \left (2\right )\) | \(62\) |
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=\frac {10 \, {\left (x^{4} - 2 \, x^{3} e^{4} + x^{2} e^{8} + 1\right )} \log \left (2\right ) + e^{x} \log \left (2\right )}{x} \]
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Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=10 x^{3} \log {\left (2 \right )} - 20 x^{2} e^{4} \log {\left (2 \right )} + 10 x e^{8} \log {\left (2 \right )} + \frac {e^{x} \log {\left (2 \right )}}{x} + \frac {10 \log {\left (2 \right )}}{x} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=10 \, x^{3} \log \left (2\right ) - 20 \, x^{2} e^{4} \log \left (2\right ) + 10 \, x e^{8} \log \left (2\right ) + {\rm Ei}\left (x\right ) \log \left (2\right ) - \Gamma \left (-1, -x\right ) \log \left (2\right ) + \frac {10 \, \log \left (2\right )}{x} \]
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Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=\frac {10 \, x^{4} \log \left (2\right ) - 20 \, x^{3} e^{4} \log \left (2\right ) + 10 \, x^{2} e^{8} \log \left (2\right ) + e^{x} \log \left (2\right ) + 10 \, \log \left (2\right )}{x} \]
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Time = 13.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{2 x^2} \, dx=10\,x\,\ln \left (2\right )\,{\left (x-{\mathrm {e}}^4\right )}^2+\frac {\ln \left (2\right )\,\left ({\mathrm {e}}^x+10\right )}{x} \]
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