Integrand size = 114, antiderivative size = 27 \[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=\left (3+e^5+x\right ) \left (-9+\frac {3}{4+\frac {5 x}{4}+x^2-\log (x)}\right ) \]
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\[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=\int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {3 \left (48-748 x-576 x^2-475 x^3-120 x^4-48 x^5-4 e^5 \left (-4+5 x+8 x^2\right )+8 x \left (46+15 x+12 x^2\right ) \log (x)-48 x \log ^2(x)\right )}{x \left (16+5 x+4 x^2-4 \log (x)\right )^2} \, dx \\ & = 3 \int \frac {48-748 x-576 x^2-475 x^3-120 x^4-48 x^5-4 e^5 \left (-4+5 x+8 x^2\right )+8 x \left (46+15 x+12 x^2\right ) \log (x)-48 x \log ^2(x)}{x \left (16+5 x+4 x^2-4 \log (x)\right )^2} \, dx \\ & = 3 \int \left (-3-\frac {4 \left (3+e^5+x\right ) \left (-4+5 x+8 x^2\right )}{x \left (16+5 x+4 x^2-4 \log (x)\right )^2}+\frac {4}{16+5 x+4 x^2-4 \log (x)}\right ) \, dx \\ & = -9 x-12 \int \frac {\left (3+e^5+x\right ) \left (-4+5 x+8 x^2\right )}{x \left (16+5 x+4 x^2-4 \log (x)\right )^2} \, dx+12 \int \frac {1}{16+5 x+4 x^2-4 \log (x)} \, dx \\ & = -9 x-12 \int \left (\frac {11 \left (1+\frac {5 e^5}{11}\right )}{\left (16+5 x+4 x^2-4 \log (x)\right )^2}-\frac {4 \left (3+e^5\right )}{x \left (16+5 x+4 x^2-4 \log (x)\right )^2}+\frac {\left (29+8 e^5\right ) x}{\left (16+5 x+4 x^2-4 \log (x)\right )^2}+\frac {8 x^2}{\left (16+5 x+4 x^2-4 \log (x)\right )^2}\right ) \, dx+12 \int \frac {1}{16+5 x+4 x^2-4 \log (x)} \, dx \\ & = -9 x+12 \int \frac {1}{16+5 x+4 x^2-4 \log (x)} \, dx-96 \int \frac {x^2}{\left (16+5 x+4 x^2-4 \log (x)\right )^2} \, dx+\left (48 \left (3+e^5\right )\right ) \int \frac {1}{x \left (16+5 x+4 x^2-4 \log (x)\right )^2} \, dx-\left (12 \left (11+5 e^5\right )\right ) \int \frac {1}{\left (16+5 x+4 x^2-4 \log (x)\right )^2} \, dx-\left (12 \left (29+8 e^5\right )\right ) \int \frac {x}{\left (16+5 x+4 x^2-4 \log (x)\right )^2} \, dx \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=-3 \left (3 x+\frac {4 \left (3+e^5+x\right )}{-16-5 x-4 x^2+4 \log (x)}\right ) \]
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Time = 0.82 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-9 x +\frac {12 \,{\mathrm e}^{5}+36+12 x}{4 x^{2}+5 x -4 \ln \left (x \right )+16}\) | \(28\) |
norman | \(\frac {-\frac {303 x}{4}-45 \ln \left (x \right )+36 x \ln \left (x \right )-36 x^{3}+216+12 \,{\mathrm e}^{5}}{4 x^{2}+5 x -4 \ln \left (x \right )+16}\) | \(41\) |
default | \(-\frac {3 \left (-44 x -15 x^{2}-12 x^{3}+12 x \ln \left (x \right )+12+4 \,{\mathrm e}^{5}\right )}{-4 x^{2}+4 \ln \left (x \right )-5 x -16}\) | \(43\) |
parallelrisch | \(\frac {-144 x^{3}+144-180 x^{2}+144 x \ln \left (x \right )+48 \,{\mathrm e}^{5}-528 x}{16 x^{2}+20 x -16 \ln \left (x \right )+64}\) | \(43\) |
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Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=-\frac {3 \, {\left (12 \, x^{3} + 15 \, x^{2} - 12 \, x \log \left (x\right ) + 44 \, x - 4 \, e^{5} - 12\right )}}{4 \, x^{2} + 5 \, x - 4 \, \log \left (x\right ) + 16} \]
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Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=- 9 x + \frac {- 12 x - 12 e^{5} - 36}{- 4 x^{2} - 5 x + 4 \log {\left (x \right )} - 16} \]
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Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=-\frac {3 \, {\left (12 \, x^{3} + 15 \, x^{2} - 12 \, x \log \left (x\right ) + 44 \, x - 4 \, e^{5} - 12\right )}}{4 \, x^{2} + 5 \, x - 4 \, \log \left (x\right ) + 16} \]
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Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=-\frac {3 \, {\left (12 \, x^{3} + 15 \, x^{2} - 12 \, x \log \left (x\right ) + 44 \, x - 4 \, e^{5} - 12\right )}}{4 \, x^{2} + 5 \, x - 4 \, \log \left (x\right ) + 16} \]
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Time = 15.50 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {144-2244 x-1728 x^2-1425 x^3-360 x^4-144 x^5+e^5 \left (48-60 x-96 x^2\right )+\left (1104 x+360 x^2+288 x^3\right ) \log (x)-144 x \log ^2(x)}{256 x+160 x^2+153 x^3+40 x^4+16 x^5+\left (-128 x-40 x^2-32 x^3\right ) \log (x)+16 x \log ^2(x)} \, dx=-\frac {3\,\left (44\,x-4\,{\mathrm {e}}^5-12\,x\,\ln \left (x\right )+15\,x^2+12\,x^3-12\right )}{5\,x-4\,\ln \left (x\right )+4\,x^2+16} \]
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