Integrand size = 36, antiderivative size = 49 \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=-\frac {15033 \arctan (5-x)}{1025}-\frac {4607 \arctan \left (\frac {1}{4} (-1+x)\right )}{4100}+\frac {1003 \log \left (26-10 x+x^2\right )}{1025}+\frac {22 \log \left (17-2 x+x^2\right )}{1025} \]
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Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {6860, 648, 632, 210, 642} \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=-\frac {15033 \arctan (5-x)}{1025}-\frac {4607 \arctan \left (\frac {x-1}{4}\right )}{4100}+\frac {1003 \log \left (x^2-10 x+26\right )}{1025}+\frac {22 \log \left (x^2-2 x+17\right )}{1025} \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5003+2006 x}{1025 \left (26-10 x+x^2\right )}+\frac {-4651+44 x}{1025 \left (17-2 x+x^2\right )}\right ) \, dx \\ & = \frac {\int \frac {5003+2006 x}{26-10 x+x^2} \, dx}{1025}+\frac {\int \frac {-4651+44 x}{17-2 x+x^2} \, dx}{1025} \\ & = \frac {22 \int \frac {-2+2 x}{17-2 x+x^2} \, dx}{1025}+\frac {1003 \int \frac {-10+2 x}{26-10 x+x^2} \, dx}{1025}-\frac {4607 \int \frac {1}{17-2 x+x^2} \, dx}{1025}+\frac {15033 \int \frac {1}{26-10 x+x^2} \, dx}{1025} \\ & = \frac {1003 \log \left (26-10 x+x^2\right )}{1025}+\frac {22 \log \left (17-2 x+x^2\right )}{1025}+\frac {9214 \text {Subst}\left (\int \frac {1}{-64-x^2} \, dx,x,-2+2 x\right )}{1025}-\frac {30066 \text {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,-10+2 x\right )}{1025} \\ & = -\frac {15033 \tan ^{-1}(5-x)}{1025}-\frac {4607 \tan ^{-1}\left (\frac {1}{4} (-1+x)\right )}{4100}+\frac {1003 \log \left (26-10 x+x^2\right )}{1025}+\frac {22 \log \left (17-2 x+x^2\right )}{1025} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=-\frac {15033 \arctan (5-x)}{1025}-\frac {4607 \arctan \left (\frac {1}{4} (-1+x)\right )}{4100}+\frac {1003 \log \left (26-10 x+x^2\right )}{1025}+\frac {22 \log \left (17-2 x+x^2\right )}{1025} \]
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Time = 1.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {15033 \arctan \left (-5+x \right )}{1025}-\frac {4607 \arctan \left (-\frac {1}{4}+\frac {x}{4}\right )}{4100}+\frac {1003 \ln \left (x^{2}-10 x +26\right )}{1025}+\frac {22 \ln \left (x^{2}-2 x +17\right )}{1025}\) | \(38\) |
risch | \(\frac {15033 \arctan \left (-5+x \right )}{1025}-\frac {4607 \arctan \left (-\frac {1}{4}+\frac {x}{4}\right )}{4100}+\frac {1003 \ln \left (x^{2}-10 x +26\right )}{1025}+\frac {22 \ln \left (x^{2}-2 x +17\right )}{1025}\) | \(38\) |
parallelrisch | \(\frac {1003 \ln \left (x -5-i\right )}{1025}-\frac {15033 i \ln \left (x -5-i\right )}{2050}+\frac {1003 \ln \left (x -5+i\right )}{1025}+\frac {15033 i \ln \left (x -5+i\right )}{2050}+\frac {22 \ln \left (x -1-4 i\right )}{1025}+\frac {4607 i \ln \left (x -1-4 i\right )}{8200}+\frac {22 \ln \left (x -1+4 i\right )}{1025}-\frac {4607 i \ln \left (x -1+4 i\right )}{8200}\) | \(70\) |
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76 \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=\frac {15033}{1025} \, \arctan \left (x - 5\right ) - \frac {4607}{4100} \, \arctan \left (\frac {1}{4} \, x - \frac {1}{4}\right ) + \frac {22}{1025} \, \log \left (x^{2} - 2 \, x + 17\right ) + \frac {1003}{1025} \, \log \left (x^{2} - 10 \, x + 26\right ) \]
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Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=\frac {1003 \log {\left (x^{2} - 10 x + 26 \right )}}{1025} + \frac {22 \log {\left (x^{2} - 2 x + 17 \right )}}{1025} - \frac {4607 \operatorname {atan}{\left (\frac {x}{4} - \frac {1}{4} \right )}}{4100} + \frac {15033 \operatorname {atan}{\left (x - 5 \right )}}{1025} \]
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76 \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=\frac {15033}{1025} \, \arctan \left (x - 5\right ) - \frac {4607}{4100} \, \arctan \left (\frac {1}{4} \, x - \frac {1}{4}\right ) + \frac {22}{1025} \, \log \left (x^{2} - 2 \, x + 17\right ) + \frac {1003}{1025} \, \log \left (x^{2} - 10 \, x + 26\right ) \]
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Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76 \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=\frac {15033}{1025} \, \arctan \left (x - 5\right ) - \frac {4607}{4100} \, \arctan \left (\frac {1}{4} \, x - \frac {1}{4}\right ) + \frac {22}{1025} \, \log \left (x^{2} - 2 \, x + 17\right ) + \frac {1003}{1025} \, \log \left (x^{2} - 10 \, x + 26\right ) \]
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Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx=\ln \left (x-1-4{}\mathrm {i}\right )\,\left (\frac {22}{1025}+\frac {4607}{8200}{}\mathrm {i}\right )+\ln \left (x-1+4{}\mathrm {i}\right )\,\left (\frac {22}{1025}-\frac {4607}{8200}{}\mathrm {i}\right )+\ln \left (x-5-\mathrm {i}\right )\,\left (\frac {1003}{1025}-\frac {15033}{2050}{}\mathrm {i}\right )+\ln \left (x-5+1{}\mathrm {i}\right )\,\left (\frac {1003}{1025}+\frac {15033}{2050}{}\mathrm {i}\right ) \]
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