Optimal. Leaf size=27 \[ \frac {2-e^{2 x}}{24 x \left (-2-\frac {3}{x}+\log (5)\right )} \]
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Rubi [A] time = 0.58, antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 8, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6, 6741, 27, 12, 6742, 2197} \begin {gather*} \frac {e^{2 x}}{24 (x (2-\log (5))+3)}-\frac {1}{12 (x (2-\log (5))+3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 27
Rule 2197
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+\left (-144 x-96 x^2\right ) \log (5)+x^2 \left (96+24 \log ^2(5)\right )} \, dx\\ &=\int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+144 x (2-\log (5))+24 x^2 (2-\log (5))^2} \, dx\\ &=\int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{24 (-3-2 x+x \log (5))^2} \, dx\\ &=\int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{24 (-3+x (-2+\log (5)))^2} \, dx\\ &=\frac {1}{24} \int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{(-3+x (-2+\log (5)))^2} \, dx\\ &=\frac {1}{24} \int \left (\frac {2 (2-\log (5))}{(3+x (2-\log (5)))^2}+\frac {e^{2 x} (4+2 x (2-\log (5))+\log (5))}{(3+x (2-\log (5)))^2}\right ) \, dx\\ &=-\frac {1}{12 (3+x (2-\log (5)))}+\frac {1}{24} \int \frac {e^{2 x} (4+2 x (2-\log (5))+\log (5))}{(3+x (2-\log (5)))^2} \, dx\\ &=-\frac {1}{12 (3+x (2-\log (5)))}+\frac {e^{2 x}}{24 (3+x (2-\log (5)))}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.31, size = 23, normalized size = 0.85 \begin {gather*} \frac {2-e^{2 x}}{24 (-3+x (-2+\log (5)))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 19, normalized size = 0.70 \begin {gather*} -\frac {e^{\left (2 \, x\right )} - 2}{24 \, {\left (x \log \relax (5) - 2 \, x - 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.90, size = 19, normalized size = 0.70 \begin {gather*} -\frac {e^{\left (2 \, x\right )} - 2}{24 \, {\left (x \log \relax (5) - 2 \, x - 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.56, size = 28, normalized size = 1.04
method | result | size |
norman | \(\frac {\left (-\frac {1}{18}+\frac {\ln \relax (5)}{36}\right ) x -\frac {{\mathrm e}^{2 x}}{24}}{x \ln \relax (5)-2 x -3}\) | \(28\) |
risch | \(\frac {1}{12 x \ln \relax (5)-24 x -36}-\frac {{\mathrm e}^{2 x}}{24 \left (x \ln \relax (5)-2 x -3\right )}\) | \(32\) |
derivativedivides | \(-\frac {\ln \relax (5) {\mathrm e}^{2 x}}{12 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {1}{3 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{3 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{2}}+\frac {\ln \relax (5)}{6 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{\left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{\left (\ln \relax (5)-2\right )^{3}}+\frac {\ln \relax (5) {\mathrm e}^{2 x}}{2 \left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}+\frac {\ln \relax (5) {\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{3}}\) | \(251\) |
default | \(-\frac {\ln \relax (5) {\mathrm e}^{2 x}}{12 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {1}{3 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{3 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{2}}+\frac {\ln \relax (5)}{6 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{\left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{\left (\ln \relax (5)-2\right )^{3}}+\frac {\ln \relax (5) {\mathrm e}^{2 x}}{2 \left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}+\frac {\ln \relax (5) {\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{3}}\) | \(251\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.59, size = 63, normalized size = 2.33 \begin {gather*} -\frac {e^{\left (2 \, x\right )}}{24 \, {\left (x {\left (\log \relax (5) - 2\right )} - 3\right )}} + \frac {\log \relax (5)}{12 \, {\left ({\left (\log \relax (5)^{2} - 4 \, \log \relax (5) + 4\right )} x - 3 \, \log \relax (5) + 6\right )}} - \frac {1}{6 \, {\left ({\left (\log \relax (5)^{2} - 4 \, \log \relax (5) + 4\right )} x - 3 \, \log \relax (5) + 6\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.26, size = 20, normalized size = 0.74 \begin {gather*} -\frac {\frac {{\mathrm {e}}^{2\,x}}{24}-\frac {1}{12}}{x\,\left (\ln \relax (5)-2\right )-3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.39, size = 44, normalized size = 1.63 \begin {gather*} - \frac {2 - \log {\relax (5 )}}{x \left (- 48 \log {\relax (5 )} + 12 \log {\relax (5 )}^{2} + 48\right ) - 36 \log {\relax (5 )} + 72} - \frac {e^{2 x}}{- 48 x + 24 x \log {\relax (5 )} - 72} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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