Integrand size = 50, antiderivative size = 22 \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=2+\frac {(49-x) x \left (4+25 x^4\right )}{x+\log (5)} \]
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Leaf count is larger than twice the leaf count of optimal. \(80\) vs. \(2(22)=44\).
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.64, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {27, 1864} \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=-25 x^5+25 x^4 (49+\log (5))-25 x^3 \log (5) (49+\log (5))+25 x^2 \log ^2(5) (49+\log (5))-x \left (4+25 \log ^3(5) (49+\log (5))\right )-\frac {\log (5) \left (196+25 \log ^5(5)+1225 \log ^4(5)+\log (625)\right )}{x+\log (5)} \]
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Rule 27
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{(x+\log (5))^2} \, dx \\ & = \int \left (-125 x^4+100 x^3 (49+\log (5))-75 x^2 \log (5) (49+\log (5))+50 x \log ^2(5) (49+\log (5))-4 \left (1+\frac {25}{4} \log ^3(5) (49+\log (5))\right )+\frac {\log (5) \left (196+1225 \log ^4(5)+25 \log ^5(5)+\log (625)\right )}{(x+\log (5))^2}\right ) \, dx \\ & = -25 x^5+25 x^4 (49+\log (5))-25 x^3 \log (5) (49+\log (5))+25 x^2 \log ^2(5) (49+\log (5))-x \left (4+25 \log ^3(5) (49+\log (5))\right )-\frac {\log (5) \left (196+1225 \log ^4(5)+25 \log ^5(5)+\log (625)\right )}{x+\log (5)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(22)=44\).
Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 4.05 \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=\frac {1225 x^5-25 x^6+x^2 \left (-4-375 \log ^4(5)+125 \log ^3(5) \log (125)\right )-x \log (5) \left (4+625 \log ^4(5)-25 \log ^3(5) (-49+8 \log (125))\right )-\log (5) \left (196+625 \log ^5(5)-25 \log ^4(5) (-49+8 \log (125))+\log (625)\right )}{x+\log (5)} \]
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Time = 1.69 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27
method | result | size |
norman | \(\frac {-4 x^{2}+1225 x^{5}-25 x^{6}-196 \ln \left (5\right )}{\ln \left (5\right )+x}\) | \(28\) |
parallelrisch | \(\frac {-4 x^{2}+1225 x^{5}-25 x^{6}-196 \ln \left (5\right )}{\ln \left (5\right )+x}\) | \(28\) |
gosper | \(-\frac {25 x^{6}-1225 x^{5}+4 x^{2}+196 \ln \left (5\right )}{\ln \left (5\right )+x}\) | \(29\) |
default | \(-25 x \ln \left (5\right )^{4}+25 x^{2} \ln \left (5\right )^{3}-25 x^{3} \ln \left (5\right )^{2}+25 x^{4} \ln \left (5\right )-25 x^{5}-1225 x \ln \left (5\right )^{3}+1225 x^{2} \ln \left (5\right )^{2}-1225 x^{3} \ln \left (5\right )+1225 x^{4}-4 x -\frac {\ln \left (5\right ) \left (25 \ln \left (5\right )^{5}+1225 \ln \left (5\right )^{4}+4 \ln \left (5\right )+196\right )}{\ln \left (5\right )+x}\) | \(98\) |
risch | \(-25 x \ln \left (5\right )^{4}+25 x^{2} \ln \left (5\right )^{3}-25 x^{3} \ln \left (5\right )^{2}+25 x^{4} \ln \left (5\right )-25 x^{5}-1225 x \ln \left (5\right )^{3}+1225 x^{2} \ln \left (5\right )^{2}-1225 x^{3} \ln \left (5\right )+1225 x^{4}-4 x -\frac {25 \ln \left (5\right )^{6}}{\ln \left (5\right )+x}-\frac {1225 \ln \left (5\right )^{5}}{\ln \left (5\right )+x}-\frac {4 \ln \left (5\right )^{2}}{\ln \left (5\right )+x}-\frac {196 \ln \left (5\right )}{\ln \left (5\right )+x}\) | \(116\) |
meijerg | \(\ln \left (5\right )^{4} \left (-150 \ln \left (5\right )+4900\right ) \left (-\frac {x \left (-\frac {3 x^{4}}{\ln \left (5\right )^{4}}+\frac {5 x^{3}}{\ln \left (5\right )^{3}}-\frac {10 x^{2}}{\ln \left (5\right )^{2}}+\frac {30 x}{\ln \left (5\right )}+60\right )}{12 \ln \left (5\right ) \left (1+\frac {x}{\ln \left (5\right )}\right )}+5 \ln \left (1+\frac {x}{\ln \left (5\right )}\right )\right )+6125 \ln \left (5\right )^{4} \left (\frac {x \left (\frac {5 x^{3}}{\ln \left (5\right )^{3}}-\frac {10 x^{2}}{\ln \left (5\right )^{2}}+\frac {30 x}{\ln \left (5\right )}+60\right )}{15 \ln \left (5\right ) \left (1+\frac {x}{\ln \left (5\right )}\right )}-4 \ln \left (1+\frac {x}{\ln \left (5\right )}\right )\right )-8 \ln \left (5\right ) \left (-\frac {x}{\ln \left (5\right ) \left (1+\frac {x}{\ln \left (5\right )}\right )}+\ln \left (1+\frac {x}{\ln \left (5\right )}\right )\right )-125 \ln \left (5\right )^{5} \left (\frac {x \left (\frac {14 x^{5}}{\ln \left (5\right )^{5}}-\frac {21 x^{4}}{\ln \left (5\right )^{4}}+\frac {35 x^{3}}{\ln \left (5\right )^{3}}-\frac {70 x^{2}}{\ln \left (5\right )^{2}}+\frac {210 x}{\ln \left (5\right )}+420\right )}{70 \ln \left (5\right ) \left (1+\frac {x}{\ln \left (5\right )}\right )}-6 \ln \left (1+\frac {x}{\ln \left (5\right )}\right )\right )-4 \ln \left (5\right ) \left (\frac {x \left (6+\frac {3 x}{\ln \left (5\right )}\right )}{3 \ln \left (5\right ) \left (1+\frac {x}{\ln \left (5\right )}\right )}-2 \ln \left (1+\frac {x}{\ln \left (5\right )}\right )\right )+\frac {196 x}{\ln \left (5\right ) \left (1+\frac {x}{\ln \left (5\right )}\right )}\) | \(310\) |
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Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (21) = 42\).
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.68 \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=-\frac {25 \, x^{6} + 25 \, {\left (x + 49\right )} \log \left (5\right )^{5} + 25 \, \log \left (5\right )^{6} - 1225 \, x^{5} + 1225 \, x \log \left (5\right )^{4} + 4 \, x^{2} + 4 \, {\left (x + 49\right )} \log \left (5\right ) + 4 \, \log \left (5\right )^{2}}{x + \log \left (5\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (17) = 34\).
Time = 0.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 4.50 \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=- 25 x^{5} - x^{4} \left (-1225 - 25 \log {\left (5 \right )}\right ) - x^{3} \cdot \left (25 \log {\left (5 \right )}^{2} + 1225 \log {\left (5 \right )}\right ) - x^{2} \left (- 1225 \log {\left (5 \right )}^{2} - 25 \log {\left (5 \right )}^{3}\right ) - x \left (4 + 25 \log {\left (5 \right )}^{4} + 1225 \log {\left (5 \right )}^{3}\right ) - \frac {4 \log {\left (5 \right )}^{2} + 196 \log {\left (5 \right )} + 25 \log {\left (5 \right )}^{6} + 1225 \log {\left (5 \right )}^{5}}{x + \log {\left (5 \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (21) = 42\).
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 4.23 \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=-25 \, x^{5} + 25 \, x^{4} {\left (\log \left (5\right ) + 49\right )} - 25 \, {\left (\log \left (5\right )^{2} + 49 \, \log \left (5\right )\right )} x^{3} + 25 \, {\left (\log \left (5\right )^{3} + 49 \, \log \left (5\right )^{2}\right )} x^{2} - {\left (25 \, \log \left (5\right )^{4} + 1225 \, \log \left (5\right )^{3} + 4\right )} x - \frac {25 \, \log \left (5\right )^{6} + 1225 \, \log \left (5\right )^{5} + 4 \, \log \left (5\right )^{2} + 196 \, \log \left (5\right )}{x + \log \left (5\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (21) = 42\).
Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.55 \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=-25 \, x^{5} + 25 \, x^{4} \log \left (5\right ) - 25 \, x^{3} \log \left (5\right )^{2} + 25 \, x^{2} \log \left (5\right )^{3} - 25 \, x \log \left (5\right )^{4} + 1225 \, x^{4} - 1225 \, x^{3} \log \left (5\right ) + 1225 \, x^{2} \log \left (5\right )^{2} - 1225 \, x \log \left (5\right )^{3} - 4 \, x - \frac {25 \, \log \left (5\right )^{6} + 1225 \, \log \left (5\right )^{5} + 4 \, \log \left (5\right )^{2} + 196 \, \log \left (5\right )}{x + \log \left (5\right )} \]
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Time = 14.44 (sec) , antiderivative size = 185, normalized size of antiderivative = 8.41 \[ \int \frac {-4 x^2+4900 x^5-125 x^6+\left (196-8 x+6125 x^4-150 x^5\right ) \log (5)}{x^2+2 x \log (5)+\log ^2(5)} \, dx=x^4\,\left (25\,\ln \left (5\right )+1225\right )-x\,\left ({\ln \left (5\right )}^2\,\left (6125\,\ln \left (5\right )+125\,{\ln \left (5\right )}^2-2\,\ln \left (5\right )\,\left (100\,\ln \left (5\right )+4900\right )\right )-2\,\ln \left (5\right )\,\left (2\,\ln \left (5\right )\,\left (6125\,\ln \left (5\right )+125\,{\ln \left (5\right )}^2-2\,\ln \left (5\right )\,\left (100\,\ln \left (5\right )+4900\right )\right )+{\ln \left (5\right )}^2\,\left (100\,\ln \left (5\right )+4900\right )\right )+4\right )-x^2\,\left (\ln \left (5\right )\,\left (6125\,\ln \left (5\right )+125\,{\ln \left (5\right )}^2-2\,\ln \left (5\right )\,\left (100\,\ln \left (5\right )+4900\right )\right )+\frac {{\ln \left (5\right )}^2\,\left (100\,\ln \left (5\right )+4900\right )}{2}\right )+x^3\,\left (\frac {6125\,\ln \left (5\right )}{3}+\frac {125\,{\ln \left (5\right )}^2}{3}-\frac {2\,\ln \left (5\right )\,\left (100\,\ln \left (5\right )+4900\right )}{3}\right )-\frac {196\,\ln \left (5\right )+\ln \left (5\right )\,\ln \left (625\right )+1225\,{\ln \left (5\right )}^5+25\,{\ln \left (5\right )}^6}{x+\ln \left (5\right )}-25\,x^5 \]
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