Integrand size = 39, antiderivative size = 29 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {1}{x+\frac {4 \left (3+\frac {16 \log ^2(3)}{(-x+x (1+\log (3)))^2}\right )}{x}} \]
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Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1608, 2100, 652, 632, 210} \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x+4}{2 \left (x^2+2 x+8\right )}-\frac {4-x}{2 \left (x^2-2 x+8\right )} \]
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Rule 210
Rule 632
Rule 652
Rule 1608
Rule 2100
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (192+12 x^2-x^4\right )}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx \\ & = \int \left (\frac {4+3 x}{\left (8-2 x+x^2\right )^2}-\frac {1}{2 \left (8-2 x+x^2\right )}+\frac {4-3 x}{\left (8+2 x+x^2\right )^2}-\frac {1}{2 \left (8+2 x+x^2\right )}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{8-2 x+x^2} \, dx\right )-\frac {1}{2} \int \frac {1}{8+2 x+x^2} \, dx+\int \frac {4+3 x}{\left (8-2 x+x^2\right )^2} \, dx+\int \frac {4-3 x}{\left (8+2 x+x^2\right )^2} \, dx \\ & = -\frac {4-x}{2 \left (8-2 x+x^2\right )}+\frac {4+x}{2 \left (8+2 x+x^2\right )}+\frac {1}{2} \int \frac {1}{8-2 x+x^2} \, dx+\frac {1}{2} \int \frac {1}{8+2 x+x^2} \, dx+\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,-2+2 x\right )+\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,2+2 x\right ) \\ & = -\frac {4-x}{2 \left (8-2 x+x^2\right )}+\frac {4+x}{2 \left (8+2 x+x^2\right )}-\frac {\arctan \left (\frac {-1+x}{\sqrt {7}}\right )}{2 \sqrt {7}}-\frac {\arctan \left (\frac {1+x}{\sqrt {7}}\right )}{2 \sqrt {7}}-\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,-2+2 x\right )-\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,2+2 x\right ) \\ & = -\frac {4-x}{2 \left (8-2 x+x^2\right )}+\frac {4+x}{2 \left (8+2 x+x^2\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^3}{64+12 x^2+x^4} \]
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Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59
method | result | size |
gosper | \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) | \(17\) |
norman | \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) | \(17\) |
risch | \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) | \(17\) |
parallelrisch | \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) | \(17\) |
default | \(-\frac {-x +4}{2 \left (x^{2}-2 x +8\right )}-\frac {-4-x}{2 \left (x^{2}+2 x +8\right )}\) | \(36\) |
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Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \]
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Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.41 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 x^{2} + 64} \]
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Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \]
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Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \]
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Time = 7.46 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^3}{x^4+12\,x^2+64} \]
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