Optimal. Leaf size=25 \[ 25 x \log \left (\log \left (4 e^{2 e^{4+\frac {\log (5)}{e^5}}}-x\right )\right ) \]
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Rubi [B] time = 0.53, antiderivative size = 75, normalized size of antiderivative = 3.00, number of steps used = 12, number of rules used = 8, integrand size = 134, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6688, 12, 2411, 2353, 2298, 2302, 29, 2520} \begin {gather*} 100 e^{2\ 5^{\frac {1}{e^5}} e^4} \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 29
Rule 2298
Rule 2302
Rule 2353
Rule 2411
Rule 2520
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int 25 \left (-\frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )}+\log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\right ) \, dx\\ &=25 \int \left (-\frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )}+\log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\right ) \, dx\\ &=-\left (25 \int \frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )} \, dx\right )+25 \int \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \, dx\\ &=25 \operatorname {Subst}\left (\int \frac {4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x}{x \log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )-25 \operatorname {Subst}\left (\int \log (\log (x)) \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\\ &=-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+25 \operatorname {Subst}\left (\int \left (-\frac {1}{\log (x)}+\frac {4 e^{2\ 5^{\frac {1}{e^5}} e^4}}{x \log (x)}\right ) \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )+25 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\\ &=-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+25 \text {li}\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )-25 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )+\left (100 e^{2\ 5^{\frac {1}{e^5}} e^4}\right ) \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\\ &=-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+\left (100 e^{2\ 5^{\frac {1}{e^5}} e^4}\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\\ &=100 e^{2\ 5^{\frac {1}{e^5}} e^4} \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [C] time = 0.14, size = 66, normalized size = 2.64 \begin {gather*} 25 \left (\text {Ei}\left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+x \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-\text {li}\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.98, size = 25, normalized size = 1.00 \begin {gather*} 25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \relax (5)\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.41, size = 25, normalized size = 1.00 \begin {gather*} 25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \relax (5)\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 21, normalized size = 0.84
method | result | size |
risch | \(25 x \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,5^{{\mathrm e}^{-5}} {\mathrm e}^{4}}-x \right )\right )\) | \(21\) |
default | \(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \relax (5)+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\) | \(28\) |
norman | \(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \relax (5)+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 100 \, {\left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right ) - \log \left (-x + 4 \, e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right ) - \log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right )} e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )} \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )}\right )\right ) + 100 \, e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )} \log \left (x - 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right ) + 25 \, \int \frac {x \log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right ) + x}{{\left (x - 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )} \log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.94, size = 20, normalized size = 0.80 \begin {gather*} 25\,x\,\ln \left (\ln \left (4\,{\mathrm {e}}^{2\,5^{{\mathrm {e}}^{-5}}\,{\mathrm {e}}^4}-x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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