Optimal. Leaf size=29 \[ \frac {25 \left (-2+4 x^4\right )}{8 x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \]
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Rubi [F] time = 1.60, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 \left (-4 x-32 x^5+30 x^4 \log (2)+\log (32)+\left (1+6 x^4\right ) (x-\log (2)) \log \left (\frac {1}{5} (-x+\log (2))\right )\right )}{4 x^2 (x-\log (2)) \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx\\ &=\frac {25}{4} \int \frac {-4 x-32 x^5+30 x^4 \log (2)+\log (32)+\left (1+6 x^4\right ) (x-\log (2)) \log \left (\frac {1}{5} (-x+\log (2))\right )}{x^2 (x-\log (2)) \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx\\ &=\frac {25}{4} \int \left (\frac {1-2 x^4}{x (x-\log (2)) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}+\frac {1+6 x^4}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}\right ) \, dx\\ &=\frac {25}{4} \int \frac {1-2 x^4}{x (x-\log (2)) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx+\frac {25}{4} \int \frac {1+6 x^4}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx\\ &=\frac {25}{4} \int \left (-\frac {2 x^2}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}-\frac {1}{x \log (2) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}-\frac {2 \log ^2(2)}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}+\frac {1-2 \log ^4(2)}{(x-\log (2)) \log (2) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}-\frac {x \log (4)}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}\right ) \, dx+\frac {25}{4} \int \left (\frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {6 x^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx\\ &=\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx-\frac {25}{2} \int \frac {x^2}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx+\frac {75}{2} \int \frac {x^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}-\frac {1}{2} \left (25 \log ^2(2)\right ) \int \frac {1}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx+\frac {\left (25 \left (1-2 \log ^4(2)\right )\right ) \int \frac {1}{(x-\log (2)) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}-\frac {1}{4} (25 \log (4)) \int \frac {x}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx\\ &=-\frac {25 x^2 (x-\log (2))}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 x (x-\log (2)) \log (4)}{4 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx+\frac {75}{2} \int \left (\frac {\log ^2(2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}+\frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {(-x+\log (2)) \log (4)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx-\frac {75}{2} \int \frac {x^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}+(25 \log (2)) \int \frac {x}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{2} \left (25 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-5+\log \left (\frac {x}{5}\right )\right )^2} \, dx,x,-x+\log (2)\right )+\frac {\left (25 \left (1-2 \log ^4(2)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-5+\log \left (\frac {x}{5}\right )\right )^2} \, dx,x,-x+\log (2)\right )}{4 \log (2)}-\frac {1}{2} (25 \log (4)) \int \frac {x}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{4} (25 \log (2) \log (4)) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx\\ &=-\frac {25 x^2 (x-\log (2))}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 (x-\log (2)) \log ^2(2)}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 x (x-\log (2)) \log (4)}{4 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx-\frac {75}{2} \int \left (\frac {\log ^2(2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}+\frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {(-x+\log (2)) \log (4)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx+\frac {75}{2} \int \frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}+(25 \log (2)) \int \left (\frac {\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx+\frac {1}{2} \left (25 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )+\frac {1}{2} \left (75 \log ^2(2)\right ) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {\left (25 \left (1-2 \log ^4(2)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}{4 \log (2)}-\frac {1}{2} (25 \log (4)) \int \left (\frac {\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx-\frac {1}{2} (75 \log (4)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{4} (25 \log (2) \log (4)) \operatorname {Subst}\left (\int \frac {1}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )\\ &=-\frac {25 x^2 (x-\log (2))}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 (x-\log (2)) \log ^2(2)}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25 \left (1-2 \log ^4(2)\right )}{4 \log (2) \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 x (x-\log (2)) \log (4)}{4 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx-\frac {75}{2} \int \frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {75}{2} \operatorname {Subst}\left (\int \frac {x^2}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}-(25 \log (2)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\left (25 \log ^2(2)\right ) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{2} \left (75 \log ^2(2)\right ) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{2} \left (75 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )+\frac {1}{2} \left (125 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {e^x}{-5+x} \, dx,x,\log \left (\frac {1}{5} (-x+\log (2))\right )\right )+\frac {1}{2} (25 \log (4)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{2} (75 \log (4)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{2} (75 \log (4)) \operatorname {Subst}\left (\int \frac {x}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )-\frac {1}{2} (25 \log (2) \log (4)) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{4} (125 \log (2) \log (4)) \operatorname {Subst}\left (\int \frac {e^x}{-5+x} \, dx,x,\log \left (\frac {1}{5} (-x+\log (2))\right )\right )\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 2.30, size = 29, normalized size = 1.00 \begin {gather*} \frac {25 \left (-1+2 x^4\right )}{4 x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 26, normalized size = 0.90 \begin {gather*} \frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, \log \relax (2)\right ) - 5 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 29, normalized size = 1.00 \begin {gather*} -\frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x \log \relax (5) - x \log \left (-x + \log \relax (2)\right ) + 5 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.22, size = 25, normalized size = 0.86
method | result | size |
norman | \(\frac {-\frac {25}{4}+\frac {25 x^{4}}{2}}{x \left (\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )-5\right )}\) | \(25\) |
risch | \(\frac {-\frac {25}{4}+\frac {25 x^{4}}{2}}{x \left (\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )-5\right )}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 28, normalized size = 0.97 \begin {gather*} -\frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x {\left (\log \relax (5) + 5\right )} - x \log \left (-x + \log \relax (2)\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {100\,x-\ln \relax (2)\,\left (750\,x^4+125\right )-\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )\,\left (25\,x-\ln \relax (2)\,\left (150\,x^4+25\right )+150\,x^5\right )+800\,x^5}{\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )\,\left (40\,x^2\,\ln \relax (2)-40\,x^3\right )-{\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )}^2\,\left (4\,x^2\,\ln \relax (2)-4\,x^3\right )-100\,x^2\,\ln \relax (2)+100\,x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 22, normalized size = 0.76 \begin {gather*} \frac {50 x^{4} - 25}{4 x \log {\left (- \frac {x}{5} + \frac {\log {\relax (2 )}}{5} \right )} - 20 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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