Integrand size = 50, antiderivative size = 86 \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=\frac {5828}{9075 (2-5 x)}-\frac {313+502 x}{1452 \left (1+2 x^2\right )}-\frac {251 \arctan \left (\sqrt {2} x\right )}{726 \sqrt {2}}+\frac {272 \sqrt {2} \arctan \left (\sqrt {2} x\right )}{1331}-\frac {59096 \log (2-5 x)}{99825}+\frac {2843 \log \left (1+2 x^2\right )}{7986} \]
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Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2099, 653, 209, 649, 266} \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=\frac {272 \sqrt {2} \arctan \left (\sqrt {2} x\right )}{1331}-\frac {251 \arctan \left (\sqrt {2} x\right )}{726 \sqrt {2}}-\frac {502 x+313}{1452 \left (2 x^2+1\right )}+\frac {2843 \log \left (2 x^2+1\right )}{7986}+\frac {5828}{9075 (2-5 x)}-\frac {59096 \log (2-5 x)}{99825} \]
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Rule 209
Rule 266
Rule 649
Rule 653
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5828}{1815 (-2+5 x)^2}-\frac {59096}{19965 (-2+5 x)}+\frac {-251+313 x}{363 \left (1+2 x^2\right )^2}+\frac {2 (816+2843 x)}{3993 \left (1+2 x^2\right )}\right ) \, dx \\ & = \frac {5828}{9075 (2-5 x)}-\frac {59096 \log (2-5 x)}{99825}+\frac {2 \int \frac {816+2843 x}{1+2 x^2} \, dx}{3993}+\frac {1}{363} \int \frac {-251+313 x}{\left (1+2 x^2\right )^2} \, dx \\ & = \frac {5828}{9075 (2-5 x)}-\frac {313+502 x}{1452 \left (1+2 x^2\right )}-\frac {59096 \log (2-5 x)}{99825}-\frac {251}{726} \int \frac {1}{1+2 x^2} \, dx+\frac {544 \int \frac {1}{1+2 x^2} \, dx}{1331}+\frac {5686 \int \frac {x}{1+2 x^2} \, dx}{3993} \\ & = \frac {5828}{9075 (2-5 x)}-\frac {313+502 x}{1452 \left (1+2 x^2\right )}-\frac {251 \arctan \left (\sqrt {2} x\right )}{726 \sqrt {2}}+\frac {272 \sqrt {2} \arctan \left (\sqrt {2} x\right )}{1331}-\frac {59096 \log (2-5 x)}{99825}+\frac {2843 \log \left (1+2 x^2\right )}{7986} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.78 \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=\frac {-\frac {33 \left (2554+4675 x+36458 x^2\right )}{-2+5 x-4 x^2+10 x^3}+12575 \sqrt {2} \arctan \left (\sqrt {2} x\right )-236384 \log (2-5 x)+142150 \log \left (1+2 x^2\right )}{399300} \]
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Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.63
method | result | size |
default | \(\frac {-\frac {2761 x}{4}-\frac {3443}{8}}{3993 x^{2}+\frac {3993}{2}}+\frac {2843 \ln \left (2 x^{2}+1\right )}{7986}+\frac {503 \arctan \left (x \sqrt {2}\right ) \sqrt {2}}{15972}-\frac {5828}{9075 \left (5 x -2\right )}-\frac {59096 \ln \left (5 x -2\right )}{99825}\) | \(54\) |
risch | \(\frac {-\frac {18229}{60500} x^{2}-\frac {17}{440} x -\frac {1277}{60500}}{x^{3}-\frac {2}{5} x^{2}+\frac {1}{2} x -\frac {1}{5}}+\frac {2843 \ln \left (4 x^{2}+2\right )}{7986}+\frac {503 \arctan \left (x \sqrt {2}\right ) \sqrt {2}}{15972}-\frac {59096 \ln \left (5 x -2\right )}{99825}\) | \(57\) |
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Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.20 \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=\frac {12575 \, \sqrt {2} {\left (10 \, x^{3} - 4 \, x^{2} + 5 \, x - 2\right )} \arctan \left (\sqrt {2} x\right ) - 1203114 \, x^{2} + 142150 \, {\left (10 \, x^{3} - 4 \, x^{2} + 5 \, x - 2\right )} \log \left (2 \, x^{2} + 1\right ) - 236384 \, {\left (10 \, x^{3} - 4 \, x^{2} + 5 \, x - 2\right )} \log \left (5 \, x - 2\right ) - 154275 \, x - 84282}{399300 \, {\left (10 \, x^{3} - 4 \, x^{2} + 5 \, x - 2\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=\frac {- 36458 x^{2} - 4675 x - 2554}{121000 x^{3} - 48400 x^{2} + 60500 x - 24200} - \frac {59096 \log {\left (x - \frac {2}{5} \right )}}{99825} + \frac {2843 \log {\left (x^{2} + \frac {1}{2} \right )}}{7986} + \frac {503 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x \right )}}{15972} \]
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Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=\frac {503}{15972} \, \sqrt {2} \arctan \left (\sqrt {2} x\right ) - \frac {36458 \, x^{2} + 4675 \, x + 2554}{12100 \, {\left (10 \, x^{3} - 4 \, x^{2} + 5 \, x - 2\right )}} + \frac {2843}{7986} \, \log \left (2 \, x^{2} + 1\right ) - \frac {59096}{99825} \, \log \left (5 \, x - 2\right ) \]
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Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=\frac {503}{15972} \, \sqrt {2} \arctan \left (\sqrt {2} x\right ) - \frac {36458 \, x^{2} + 4675 \, x + 2554}{12100 \, {\left (2 \, x^{2} + 1\right )} {\left (5 \, x - 2\right )}} + \frac {2843}{7986} \, \log \left (2 \, x^{2} + 1\right ) - \frac {59096}{99825} \, \log \left ({\left | 5 \, x - 2 \right |}\right ) \]
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Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int \frac {8-13 x^2-7 x^3+12 x^5}{4-20 x+41 x^2-80 x^3+116 x^4-80 x^5+100 x^6} \, dx=-\frac {59096\,\ln \left (x-\frac {2}{5}\right )}{99825}-\frac {\frac {18229\,x^2}{60500}+\frac {17\,x}{440}+\frac {1277}{60500}}{x^3-\frac {2\,x^2}{5}+\frac {x}{2}-\frac {1}{5}}-\ln \left (x-\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {2843}{7986}+\frac {\sqrt {2}\,503{}\mathrm {i}}{31944}\right )+\ln \left (x+\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {2843}{7986}+\frac {\sqrt {2}\,503{}\mathrm {i}}{31944}\right ) \]
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