Optimal. Leaf size=20 \[ \frac {e^{2 x} \left (-e+\log ^2(x)\right )}{8 x} \]
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Rubi [F] time = 0.41, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \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 \text {Ei}(2 x)}{4}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \text {Ei}(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 \text {Ei}(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} \text {Ei}(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 \text {Ei}(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} \text {Ei}(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 {\text {Ei}(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} \text {Ei}(2 x) \log (x)-\frac {1}{2} (E_1(-2 x)+\text {Ei}(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 {E_1(-2 x)}{x} \, dx\\ &=-\frac {e^{2 x}}{4 x}-\frac {e^{1+2 x}}{8 x}+\frac {\text {Ei}(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} \text {Ei}(2 x) \log (x)-\frac {1}{2} (E_1(-2 x)+\text {Ei}(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 {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 20, normalized size = 1.00 \begin {gather*} -\frac {e^{2 x} \left (e-\log ^2(x)\right )}{8 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.29, size = 27, normalized size = 1.35 \begin {gather*} \frac {{\left (e^{\left (2 \, x + 1\right )} \log \relax (x)^{2} - e^{\left (2 \, x + 2\right )}\right )} e^{\left (-1\right )}}{8 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.38, size = 23, normalized size = 1.15 \begin {gather*} \frac {e^{\left (2 \, x\right )} \log \relax (x)^{2} - e^{\left (2 \, x + 1\right )}}{8 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 26, normalized size = 1.30
method | result | size |
default | \(\frac {{\mathrm e}^{2 x} \ln \relax (x )^{2}}{8 x}-\frac {{\mathrm e}^{2 x +1}}{8 x}\) | \(26\) |
risch | \(\frac {{\mathrm e}^{2 x} \ln \relax (x )^{2}}{8 x}-\frac {{\mathrm e}^{2 x +1}}{8 x}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.64, size = 31, normalized size = 1.55 \begin {gather*} -\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 \relax (x)^{2}}{8 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.62, size = 18, normalized size = 0.90 \begin {gather*} -\frac {{\mathrm {e}}^{2\,x}\,\left (\mathrm {e}-{\ln \relax (x)}^2\right )}{8\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.28, size = 15, normalized size = 0.75 \begin {gather*} \frac {\left (\log {\relax (x )}^{2} - e\right ) e^{2 x}}{8 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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