Optimal. Leaf size=26 \[ 5-e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}} \]
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Rubi [A]
time = 0.22, antiderivative size = 36, normalized size of antiderivative = 1.38, number of steps
used = 6, number of rules used = 5, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 12, 6820,
2262, 2240} \begin {gather*} -\exp \left (\frac {2}{5 \left (e^5-\log (4)\right )^2}-\frac {2}{(x+5) \left (e^5-\log (4)\right )^2}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 2240
Rule 2262
Rule 6820
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int -\frac {2 \exp \left (\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}\right )}{e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \left (e^{10}+\log ^2(4)\right )} \, dx\\ &=-\left (2 \int \frac {\exp \left (\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}\right )}{e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \left (e^{10}+\log ^2(4)\right )} \, dx\right )\\ &=-\left (2 \int \frac {e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}}}{(5+x)^2 \left (e^5-\log (4)\right )^2} \, dx\right )\\ &=-\frac {2 \int \frac {e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}}}{(5+x)^2} \, dx}{\left (e^5-\log (4)\right )^2}\\ &=-\frac {2 \int \frac {\exp \left (\frac {2}{5 \left (e^5-\log (4)\right )^2}-\frac {2}{(5+x) \left (e^5-\log (4)\right )^2}\right )}{(5+x)^2} \, dx}{\left (e^5-\log (4)\right )^2}\\ &=-\exp \left (\frac {2}{5 \left (e^5-\log (4)\right )^2}-\frac {2}{(5+x) \left (e^5-\log (4)\right )^2}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.01, size = 24, normalized size = 0.92 \begin {gather*} -e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(108\) vs.
\(2(24)=48\).
time = 0.31, size = 109, normalized size = 4.19
method | result | size |
risch | \(-{\mathrm e}^{\frac {2 x}{5 \left (5+x \right ) \left (-4 \,{\mathrm e}^{5} \ln \left (2\right )+4 \ln \left (2\right )^{2}+{\mathrm e}^{10}\right )}}\) | \(29\) |
gosper | \(-{\mathrm e}^{\frac {2 x}{5 \left (x \,{\mathrm e}^{10}-4 x \,{\mathrm e}^{5} \ln \left (2\right )+4 x \ln \left (2\right )^{2}+5 \,{\mathrm e}^{10}-20 \,{\mathrm e}^{5} \ln \left (2\right )+20 \ln \left (2\right )^{2}\right )}}\) | \(48\) |
derivativedivides | \(\frac {25 \left (4 \,{\mathrm e}^{5} \ln \left (2\right )-4 \ln \left (2\right )^{2}-{\mathrm e}^{10}\right ) \left (\frac {{\mathrm e}^{10}}{25}-\frac {4 \,{\mathrm e}^{5} \ln \left (2\right )}{25}+\frac {4 \ln \left (2\right )^{2}}{25}\right ) {\mathrm e}^{-\frac {2}{\left (-4 \,{\mathrm e}^{5} \ln \left (2\right )+4 \ln \left (2\right )^{2}+{\mathrm e}^{10}\right ) \left (5+x \right )}+\frac {2}{5 \left (-4 \,{\mathrm e}^{5} \ln \left (2\right )+4 \ln \left (2\right )^{2}+{\mathrm e}^{10}\right )}}}{\left (-4 \,{\mathrm e}^{5} \ln \left (2\right )+4 \ln \left (2\right )^{2}+{\mathrm e}^{10}\right )^{2}}\) | \(109\) |
default | \(\frac {25 \left (4 \,{\mathrm e}^{5} \ln \left (2\right )-4 \ln \left (2\right )^{2}-{\mathrm e}^{10}\right ) \left (\frac {{\mathrm e}^{10}}{25}-\frac {4 \,{\mathrm e}^{5} \ln \left (2\right )}{25}+\frac {4 \ln \left (2\right )^{2}}{25}\right ) {\mathrm e}^{-\frac {2}{\left (-4 \,{\mathrm e}^{5} \ln \left (2\right )+4 \ln \left (2\right )^{2}+{\mathrm e}^{10}\right ) \left (5+x \right )}+\frac {2}{5 \left (-4 \,{\mathrm e}^{5} \ln \left (2\right )+4 \ln \left (2\right )^{2}+{\mathrm e}^{10}\right )}}}{\left (-4 \,{\mathrm e}^{5} \ln \left (2\right )+4 \ln \left (2\right )^{2}+{\mathrm e}^{10}\right )^{2}}\) | \(109\) |
norman | \(\frac {\left (-5 \,{\mathrm e}^{5}+10 \ln \left (2\right )\right ) {\mathrm e}^{\frac {2 x}{4 \left (25+5 x \right ) \ln \left (2\right )^{2}+2 \left (-10 x -50\right ) {\mathrm e}^{5} \ln \left (2\right )+\left (25+5 x \right ) {\mathrm e}^{10}}}+\left (-{\mathrm e}^{5}+2 \ln \left (2\right )\right ) x \,{\mathrm e}^{\frac {2 x}{4 \left (25+5 x \right ) \ln \left (2\right )^{2}+2 \left (-10 x -50\right ) {\mathrm e}^{5} \ln \left (2\right )+\left (25+5 x \right ) {\mathrm e}^{10}}}}{\left (5+x \right ) \left ({\mathrm e}^{5}-2 \ln \left (2\right )\right )}\) | \(116\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs.
\(2 (22) = 44\).
time = 0.49, size = 65, normalized size = 2.50 \begin {gather*} -e^{\left (\frac {2}{{\left (4 \, e^{5} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{10}\right )} x + 20 \, e^{5} \log \left (2\right ) - 20 \, \log \left (2\right )^{2} - 5 \, e^{10}} - \frac {2}{5 \, {\left (4 \, e^{5} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{10}\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.52, size = 34, normalized size = 1.31 \begin {gather*} -e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, {\left (x + 5\right )} e^{5} \log \left (2\right ) - 4 \, {\left (x + 5\right )} \log \left (2\right )^{2} - {\left (x + 5\right )} e^{10}\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.20, size = 37, normalized size = 1.42 \begin {gather*} - e^{\frac {2 x}{\left (- 20 x - 100\right ) e^{5} \log {\left (2 \right )} + \left (5 x + 25\right ) e^{10} + \left (20 x + 100\right ) \log {\left (2 \right )}^{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 161 vs.
\(2 (22) = 44\).
time = 0.42, size = 161, normalized size = 6.19 \begin {gather*} \frac {4 \, e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, x e^{5} \log \left (2\right ) - 4 \, x \log \left (2\right )^{2} - x e^{10} + 20 \, e^{5} \log \left (2\right ) - 20 \, \log \left (2\right )^{2} - 5 \, e^{10}\right )}}\right )} \log \left (2\right )^{2} - 4 \, e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, x e^{5} \log \left (2\right ) - 4 \, x \log \left (2\right )^{2} - x e^{10} + 20 \, e^{5} \log \left (2\right ) - 20 \, \log \left (2\right )^{2} - 5 \, e^{10}\right )}} + 5\right )} \log \left (2\right ) + e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, x e^{5} \log \left (2\right ) - 4 \, x \log \left (2\right )^{2} - x e^{10} + 20 \, e^{5} \log \left (2\right ) - 20 \, \log \left (2\right )^{2} - 5 \, e^{10}\right )}} + 10\right )}}{4 \, e^{5} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{10}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.85, size = 44, normalized size = 1.69 \begin {gather*} -{\mathrm {e}}^{\frac {2\,x}{25\,{\mathrm {e}}^{10}-100\,{\mathrm {e}}^5\,\ln \left (2\right )+5\,x\,{\mathrm {e}}^{10}+20\,x\,{\ln \left (2\right )}^2+100\,{\ln \left (2\right )}^2-20\,x\,{\mathrm {e}}^5\,\ln \left (2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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