Optimal. Leaf size=18 \[ 4 x \log ^{2 e^{5 (5+\log (2))^2}}(x) \]
________________________________________________________________________________________
Rubi [C] time = 0.12, antiderivative size = 173, normalized size of antiderivative = 9.61, number of steps used = 4, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2299, 2181, 2361, 19, 6557} \begin {gather*} 4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{1+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )+4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \Gamma \left (1+2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )-4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \left (\log (x)+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}\right ) \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 19
Rule 2181
Rule 2299
Rule 2361
Rule 6557
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=-4 \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )}+\log (x)\right )+4 \int \frac {\Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x)}{x} \, dx\\ &=-4 \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )}+\log (x)\right )+\left (4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x)\right ) \int \frac {\Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )}{x} \, dx\\ &=-4 \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )}+\log (x)\right )+\left (4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x)\right ) \operatorname {Subst}\left (\int \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-x\right ) \, dx,x,\log (x)\right )\\ &=4 \Gamma \left (1+2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x)+4 \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{1+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x)-4 \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )}+\log (x)\right )\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.08, size = 18, normalized size = 1.00 \begin {gather*} 4 x \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.98, size = 21, normalized size = 1.17 \begin {gather*} 4 \, x \log \relax (x)^{2 \, e^{\left (5 \, \log \relax (2)^{2} + 50 \, \log \relax (2) + 125\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, \log \relax (x)^{2 \, e^{\left (5 \, \log \relax (2)^{2} + 50 \, \log \relax (2) + 125\right )}} {\left (2 \, e^{\left (5 \, \log \relax (2)^{2} + 50 \, \log \relax (2) + 125\right )} + \log \relax (x)\right )}}{\log \relax (x)}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F(-1)] time = 180.00, size = 0, normalized size = 0.00 hanged
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [C] time = 0.71, size = 114, normalized size = 6.33 \begin {gather*} -4 \, \left (-\log \relax (x)\right )^{-2251799813685248 \, e^{\left (5 \, \log \relax (2)^{2} + 125\right )} - 1} \log \relax (x)^{2251799813685248 \, e^{\left (5 \, \log \relax (2)^{2} + 125\right )} + 1} \Gamma \left (2251799813685248 \, e^{\left (5 \, \log \relax (2)^{2} + 125\right )} + 1, -\log \relax (x)\right ) - \frac {9007199254740992 \, \log \relax (x)^{2251799813685248 \, e^{\left (5 \, \log \relax (2)^{2} + 125\right )}} e^{\left (5 \, \log \relax (2)^{2} + 125\right )} \Gamma \left (2251799813685248 \, e^{\left (5 \, \log \relax (2)^{2} + 125\right )}, -\log \relax (x)\right )}{\left (-\log \relax (x)\right )^{2251799813685248 \, e^{\left (5 \, \log \relax (2)^{2} + 125\right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 0.78, size = 17, normalized size = 0.94 \begin {gather*} 4\,x\,{\ln \relax (x)}^{2251799813685248\,{\mathrm {e}}^{5\,{\ln \relax (2)}^2+125}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 4.80, size = 128, normalized size = 7.11 \begin {gather*} \frac {9007199254740992 e^{125} e^{5 \log {\relax (2 )}^{2}} \log {\relax (x )}^{-1 + 2251799813685248 e^{125} e^{5 \log {\relax (2 )}^{2}}} \Gamma \left (2251799813685248 e^{125} e^{5 \log {\relax (2 )}^{2}}, - \log {\relax (x )}\right )}{\left (- \log {\relax (x )}\right )^{-1 + 2251799813685248 e^{125} e^{5 \log {\relax (2 )}^{2}}}} + \frac {4 \log {\relax (x )}^{2251799813685248 e^{125} e^{5 \log {\relax (2 )}^{2}}} \Gamma \left (1 + 2251799813685248 e^{125} e^{5 \log {\relax (2 )}^{2}}, - \log {\relax (x )}\right )}{\left (- \log {\relax (x )}\right )^{2251799813685248 e^{125} e^{5 \log {\relax (2 )}^{2}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________