Integrand size = 109, antiderivative size = 32 \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=\frac {3 \left (5+\frac {3-x}{5}\right )}{(-5+2 x) (-4+x (x-\log (4)))} \]
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Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.78, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2099, 632, 212, 652} \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=\frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (-x^2+x \log (4)+4\right )}-\frac {306}{5 (5-2 x) (9-10 \log (4))} \]
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Rule 212
Rule 632
Rule 652
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {612}{5 (-5+2 x)^2 (-9+10 \log (4))}+\frac {153}{5 (-9+10 \log (4)) \left (4-x^2+x \log (4)\right )}+\frac {3 \left (x (264-61 \log (4))+4 \left (102-33 \log (4)+14 \log ^2(4)\right )\right )}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )^2}\right ) \, dx \\ & = -\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 \int \frac {x (264-61 \log (4))+4 \left (102-33 \log (4)+14 \log ^2(4)\right )}{\left (4-x^2+x \log (4)\right )^2} \, dx}{5 (9-10 \log (4))}-\frac {153 \int \frac {1}{4-x^2+x \log (4)} \, dx}{5 (9-10 \log (4))} \\ & = -\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )}+\frac {153 \int \frac {1}{4-x^2+x \log (4)} \, dx}{5 (9-10 \log (4))}+\frac {306 \text {Subst}\left (\int \frac {1}{16-x^2+\log ^2(4)} \, dx,x,-2 x+\log (4)\right )}{5 (9-10 \log (4))} \\ & = -\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )}-\frac {306 \text {arctanh}\left (\frac {2 x-\log (4)}{\sqrt {16+\log ^2(4)}}\right )}{5 (9-10 \log (4)) \sqrt {16+\log ^2(4)}}-\frac {306 \text {Subst}\left (\int \frac {1}{16-x^2+\log ^2(4)} \, dx,x,-2 x+\log (4)\right )}{5 (9-10 \log (4))} \\ & = -\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(132\) vs. \(2(32)=64\).
Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 4.12 \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=\frac {3 \left (x \left (-2240 \log ^4(4)+72 (-18+107 \log (16))+2 \log ^3(4) (2007+535 \log (16))+48 \log (4) (-261+589 \log (16))-\log ^2(4) (58225+1917 \log (16))\right )+4 \left (9072-5640 \log ^3(4)+700 \log ^4(4)+39280 \log (16)-5 \log (4) (19744+1129 \log (16))+\log ^2(4) (23057+2190 \log (16))\right )\right )}{5 (-5+2 x) (9-10 \log (4))^2 \left (-4+x^2-x \log (4)\right ) \left (16+\log ^2(4)\right )} \]
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Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {\frac {3 x}{5}-\frac {84}{5}}{\left (-5+2 x \right ) \left (2 x \ln \left (2\right )-x^{2}+4\right )}\) | \(28\) |
gosper | \(\frac {\frac {3 x}{5}-\frac {84}{5}}{4 x^{2} \ln \left (2\right )-2 x^{3}-10 x \ln \left (2\right )+5 x^{2}+8 x -20}\) | \(35\) |
risch | \(\frac {\frac {3 x}{20}-\frac {21}{5}}{x^{2} \ln \left (2\right )-\frac {x^{3}}{2}-\frac {5 x \ln \left (2\right )}{2}+\frac {5 x^{2}}{4}+2 x -5}\) | \(35\) |
parallelrisch | \(-\frac {168-6 x}{10 \left (4 x^{2} \ln \left (2\right )-2 x^{3}-10 x \ln \left (2\right )+5 x^{2}+8 x -20\right )}\) | \(37\) |
default | \(-\frac {306}{5 \left (20 \ln \left (2\right )-9\right ) \left (-5+2 x \right )}-\frac {3 \left (\frac {51 x}{2}-56 \ln \left (2\right )+66\right )}{5 \left (20 \ln \left (2\right )-9\right ) \left (x \ln \left (2\right )-\frac {x^{2}}{2}+2\right )}\) | \(51\) |
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Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=-\frac {3 \, {\left (x - 28\right )}}{5 \, {\left (2 \, x^{3} - 5 \, x^{2} - 2 \, {\left (2 \, x^{2} - 5 \, x\right )} \log \left (2\right ) - 8 \, x + 20\right )}} \]
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Time = 1.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=\frac {84 - 3 x}{10 x^{3} + x^{2} \left (-25 - 20 \log {\left (2 \right )}\right ) + x \left (-40 + 50 \log {\left (2 \right )}\right ) + 100} \]
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Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=-\frac {3 \, {\left (x - 28\right )}}{5 \, {\left (2 \, x^{3} - x^{2} {\left (4 \, \log \left (2\right ) + 5\right )} + 2 \, x {\left (5 \, \log \left (2\right ) - 4\right )} + 20\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=-\frac {3 \, {\left (x - 28\right )}}{5 \, {\left (2 \, x^{3} - 4 \, x^{2} \log \left (2\right ) - 5 \, x^{2} + 10 \, x \log \left (2\right ) - 8 \, x + 20\right )}} \]
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Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {612+840 x-519 x^2+12 x^3+\left (-420+336 x-6 x^2\right ) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+\left (1000 x-800 x^2-90 x^3+200 x^4-40 x^5\right ) \log (4)+\left (125 x^2-100 x^3+20 x^4\right ) \log ^2(4)} \, dx=\frac {3\,\left (x-28\right )}{5\,\left (2\,x-5\right )\,\left (-x^2+2\,\ln \left (2\right )\,x+4\right )} \]
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