Optimal. Leaf size=27 \[ 1+\frac {\log (25)}{4}-\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right ) \]
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Rubi [A]
time = 0.17, antiderivative size = 19, normalized size of antiderivative = 0.70, number of steps
used = 3, number of rules used = 3, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {12, 6820, 6816}
\begin {gather*} -\log \left (\log \left (\log \left (\frac {\left (x+e^2+5\right )^2}{(x+1)^2}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 6816
Rule 6820
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\left (2 \left (4+e^2\right )\right ) \int \frac {1}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx\\ &=\left (2 \left (4+e^2\right )\right ) \int \frac {1}{(1+x) \left (5+e^2+x\right ) \log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right ) \log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )} \, dx\\ &=-\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.08, size = 19, normalized size = 0.70 \begin {gather*} -\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right ) \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. \(57\) vs.
\(2(26)=52\).
time = 3.48, size = 58, normalized size = 2.15
method | result | size |
norman | \(-\ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4}+\left (2 x +10\right ) {\mathrm e}^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )\) | \(39\) |
default | \(-\frac {\left (2 \,{\mathrm e}^{2}+8\right ) \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4}+2 \,{\mathrm e}^{2} x +10 \,{\mathrm e}^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )}{2 \left (4+{\mathrm e}^{2}\right )}\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 20, normalized size = 0.74 \begin {gather*} -\log \left (\log \left (2\right ) + \log \left (\log \left (x + e^{2} + 5\right ) - \log \left (x + 1\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 33, normalized size = 1.22 \begin {gather*} -\log \left (\log \left (\log \left (\frac {x^{2} + 2 \, {\left (x + 5\right )} e^{2} + 10 \, x + e^{4} + 25}{x^{2} + 2 \, x + 1}\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.24, size = 34, normalized size = 1.26 \begin {gather*} - \log {\left (\log {\left (\log {\left (\frac {x^{2} + 10 x + \left (2 x + 10\right ) e^{2} + 25 + e^{4}}{x^{2} + 2 x + 1} \right )} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 30.87, size = 34, normalized size = 1.26 \begin {gather*} -\ln \left (\ln \left (\ln \left (\frac {10\,x+{\mathrm {e}}^4+x^2+{\mathrm {e}}^2\,\left (2\,x+10\right )+25}{x^2+2\,x+1}\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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