Optimal. Leaf size=23 \[ x \left (x-\frac {-2+x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \]
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Rubi [A] time = 0.34, antiderivative size = 28, normalized size of antiderivative = 1.22, number of steps used = 18, number of rules used = 7, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6741, 6742, 2320, 2330, 2300, 2178, 2310} \begin {gather*} x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 2178
Rule 2300
Rule 2310
Rule 2320
Rule 2330
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+38 x+(2-18 x) \log \left (\frac {25}{x^2 \log ^2(2)}\right )+2 x \log ^2\left (\frac {25}{x^2 \log ^2(2)}\right )}{\left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )^2} \, dx\\ &=\int \left (2 x-\frac {2 (-2+x)}{\left (-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )^2}-\frac {2 (-1+x)}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \, dx\\ &=x^2-2 \int \frac {-2+x}{\left (-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )^2} \, dx-2 \int \frac {-1+x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx\\ &=x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}-2 \int \left (-\frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+\frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \, dx+2 \int \frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx+2 \int \frac {-2+x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx\\ &=x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+2 \int \left (-\frac {2}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+\frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \, dx+2 \int \frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx-2 \int \frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx-\frac {\left (5 \sqrt {\frac {1}{x^2}} x\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log (2)}\\ &=x^2-\frac {5 \sqrt {\frac {1}{x^2}} x \text {Ei}\left (\frac {1}{2} \left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )\right )}{e^2 \log (2)}-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+2 \int \frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx-4 \int \frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx+\frac {25 \operatorname {Subst}\left (\int \frac {e^{-x}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log ^2(2)}-\frac {\left (5 \sqrt {\frac {1}{x^2}} x\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log (2)}\\ &=x^2+\frac {25 \text {Ei}\left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{e^4 \log ^2(2)}-\frac {10 \sqrt {\frac {1}{x^2}} x \text {Ei}\left (\frac {1}{2} \left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )\right )}{e^2 \log (2)}-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}-\frac {25 \operatorname {Subst}\left (\int \frac {e^{-x}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log ^2(2)}+\frac {\left (10 \sqrt {\frac {1}{x^2}} x\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log (2)}\\ &=x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 24, normalized size = 1.04 \begin {gather*} x \left (x+\frac {2-x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 38, normalized size = 1.65 \begin {gather*} \frac {x^{2} \log \left (\frac {25}{x^{2} \log \relax (2)^{2}}\right ) - 5 \, x^{2} + 2 \, x}{\log \left (\frac {25}{x^{2} \log \relax (2)^{2}}\right ) - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 32, normalized size = 1.39 \begin {gather*} x^{2} - \frac {x^{2} - 2 \, x}{2 \, \log \relax (5) - \log \left (x^{2} \log \relax (2)^{2}\right ) - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 25, normalized size = 1.09
method | result | size |
risch | \(x^{2}-\frac {\left (x -2\right ) x}{\ln \left (\frac {25}{x^{2} \ln \relax (2)^{2}}\right )-4}\) | \(25\) |
norman | \(\frac {x^{2} \ln \left (\frac {25}{x^{2} \ln \relax (2)^{2}}\right )+2 x -5 x^{2}}{\ln \left (\frac {25}{x^{2} \ln \relax (2)^{2}}\right )-4}\) | \(39\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 43, normalized size = 1.87 \begin {gather*} \frac {x^{2} {\left (2 \, \log \relax (5) - 2 \, \log \left (\log \relax (2)\right ) - 5\right )} - 2 \, x^{2} \log \relax (x) + 2 \, x}{2 \, {\left (\log \relax (5) - \log \relax (x) - \log \left (\log \relax (2)\right ) - 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.43, size = 24, normalized size = 1.04 \begin {gather*} x^2-\frac {x\,\left (x-2\right )}{\ln \left (\frac {25}{x^2\,{\ln \relax (2)}^2}\right )-4} \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 = 0.96 \begin {gather*} x^{2} + \frac {- x^{2} + 2 x}{\log {\left (\frac {25}{x^{2} \log {\relax (2 )}^{2}} \right )} - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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