Integrand size = 45, antiderivative size = 20 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {e^{2 x} \left (-e+\log ^2(x)\right )}{8 x} \]
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\[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{x^2} \, dx \\ & = \frac {1}{8} \int \left (\frac {e^{1+2 x}}{x^2}-\frac {2 e^{1+2 x}}{x}+\frac {2 e^{2 x} \log (x)}{x^2}-\frac {e^{2 x} \log ^2(x)}{x^2}+\frac {2 e^{2 x} \log ^2(x)}{x}\right ) \, dx \\ & = \frac {1}{8} \int \frac {e^{1+2 x}}{x^2} \, dx-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx-\frac {1}{4} \int \frac {e^{1+2 x}}{x} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log (x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ & = -\frac {e^{1+2 x}}{8 x}-\frac {e \operatorname {ExpIntegralEi}(2 x)}{4}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{1+2 x}}{x} \, dx-\frac {1}{4} \int \frac {-e^{2 x}+2 x \operatorname {ExpIntegralEi}(2 x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ & = -\frac {e^{1+2 x}}{8 x}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx-\frac {1}{4} \int \left (-\frac {e^{2 x}}{x^2}+\frac {2 \operatorname {ExpIntegralEi}(2 x)}{x}\right ) \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ & = -\frac {e^{1+2 x}}{8 x}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x}}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx-\frac {1}{2} \int \frac {\operatorname {ExpIntegralEi}(2 x)}{x} \, dx \\ & = -\frac {e^{2 x}}{4 x}-\frac {e^{1+2 x}}{8 x}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{2} (\operatorname {ExpIntegralE}(1,-2 x)+\operatorname {ExpIntegralEi}(2 x)) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx+\frac {1}{2} \int \frac {e^{2 x}}{x} \, dx+\frac {1}{2} \int \frac {\operatorname {ExpIntegralE}(1,-2 x)}{x} \, dx \\ & = -\frac {e^{2 x}}{4 x}-\frac {e^{1+2 x}}{8 x}+\frac {\operatorname {ExpIntegralEi}(2 x)}{2}-x \, _3F_3(1,1,1;2,2,2;2 x)-\frac {1}{4} \log ^2(-2 x)-\frac {1}{2} \gamma \log (x)-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{2} (\operatorname {ExpIntegralE}(1,-2 x)+\operatorname {ExpIntegralEi}(2 x)) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=-\frac {e^{2 x} \left (e-\log ^2(x)\right )}{8 x} \]
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Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20
method | result | size |
parallelrisch | \(-\frac {-{\mathrm e}^{2 x} \ln \left (x \right )^{2}+{\mathrm e} \,{\mathrm e}^{2 x}}{8 x}\) | \(24\) |
default | \(\frac {{\mathrm e}^{2 x} \ln \left (x \right )^{2}}{8 x}-\frac {{\mathrm e}^{1+2 x}}{8 x}\) | \(26\) |
risch | \(\frac {{\mathrm e}^{2 x} \ln \left (x \right )^{2}}{8 x}-\frac {{\mathrm e}^{1+2 x}}{8 x}\) | \(26\) |
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Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {{\left (e^{\left (2 \, x + 1\right )} \log \left (x\right )^{2} - e^{\left (2 \, x + 2\right )}\right )} e^{\left (-1\right )}}{8 \, x} \]
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Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {\left (\log {\left (x \right )}^{2} - e\right ) e^{2 x}}{8 x} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=-\frac {1}{4} \, {\rm Ei}\left (2 \, x\right ) e + \frac {1}{4} \, e \Gamma \left (-1, -2 \, x\right ) + \frac {e^{\left (2 \, x\right )} \log \left (x\right )^{2}}{8 \, x} \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {e^{\left (2 \, x\right )} \log \left (x\right )^{2} - e^{\left (2 \, x + 1\right )}}{8 \, x} \]
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Time = 8.44 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=-\frac {{\mathrm {e}}^{2\,x}\,\left (\mathrm {e}-{\ln \left (x\right )}^2\right )}{8\,x} \]
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