Optimal. Leaf size=23 \[ -e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right ) \]
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Rubi [A] time = 0.56, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6741, 6725, 2194, 2475, 2390, 2302, 29} \begin {gather*} 5 \log \left (5-\log \left (\left (x^2+5\right )^2\right )\right )-e^{2 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 2194
Rule 2302
Rule 2390
Rule 2475
Rule 6725
Rule 6741
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20 x-e^{2 x} \left (50+10 x^2\right )-e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{\left (5+x^2\right ) \left (5-\log \left (\left (5+x^2\right )^2\right )\right )} \, dx\\ &=\int \left (-2 e^{2 x}+\frac {20 x}{\left (5+x^2\right ) \left (-5+\log \left (\left (5+x^2\right )^2\right )\right )}\right ) \, dx\\ &=-\left (2 \int e^{2 x} \, dx\right )+20 \int \frac {x}{\left (5+x^2\right ) \left (-5+\log \left (\left (5+x^2\right )^2\right )\right )} \, dx\\ &=-e^{2 x}+10 \operatorname {Subst}\left (\int \frac {1}{(5+x) \left (-5+\log \left ((5+x)^2\right )\right )} \, dx,x,x^2\right )\\ &=-e^{2 x}+10 \operatorname {Subst}\left (\int \frac {1}{x \left (-5+\log \left (x^2\right )\right )} \, dx,x,5+x^2\right )\\ &=-e^{2 x}+5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-5+\log \left (\left (5+x^2\right )^2\right )\right )\\ &=-e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 23, normalized size = 1.00 \begin {gather*} -e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 23, normalized size = 1.00 \begin {gather*} -e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{4} + 10 \, x^{2} + 25\right ) - 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 23, normalized size = 1.00 \begin {gather*} -e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{4} + 10 \, x^{2} + 25\right ) - 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 24, normalized size = 1.04
method | result | size |
default | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\) | \(24\) |
norman | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\) | \(24\) |
risch | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{2}+5\right )-\frac {i \left (\pi \mathrm {csgn}\left (i \left (x^{2}+5\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+5\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (x^{2}+5\right )\right ) \mathrm {csgn}\left (i \left (x^{2}+5\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (x^{2}+5\right )^{2}\right )^{3}-10 i\right )}{4}\right )\) | \(88\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 18, normalized size = 0.78 \begin {gather*} -e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{2} + 5\right ) - \frac {5}{2}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.26, size = 23, normalized size = 1.00 \begin {gather*} 5\,\ln \left (\ln \left (x^4+10\,x^2+25\right )-5\right )-{\mathrm {e}}^{2\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.39, size = 20, normalized size = 0.87 \begin {gather*} - e^{2 x} + 5 \log {\left (\log {\left (x^{4} + 10 x^{2} + 25 \right )} - 5 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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