Integrand size = 120, antiderivative size = 28 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=\frac {4}{2+e^{16+2 x}}-\left (-x^2+\log (5 x)\right )^2 \]
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Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 17, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.092, Rules used = {6820, 6874, 2320, 46, 36, 29, 31, 14, 2393, 2338, 2341} \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=-x^4+2 x^2 \log (5 x)+\frac {4}{e^{2 x+16}+2}-\log ^2(5 x) \]
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Rule 14
Rule 29
Rule 31
Rule 36
Rule 46
Rule 2320
Rule 2338
Rule 2341
Rule 2393
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \left (-4 x+8 x^3+e^{4 (8+x)} x \left (-1+2 x^2\right )+e^{2 (8+x)} \left (4-4 x+8 x^3\right )\right )}{\left (2+e^{2 (8+x)}\right )^2}+\frac {2 \left (-1+2 x^2\right ) \log (5 x)}{x}\right ) \, dx \\ & = -\left (2 \int \frac {-4 x+8 x^3+e^{4 (8+x)} x \left (-1+2 x^2\right )+e^{2 (8+x)} \left (4-4 x+8 x^3\right )}{\left (2+e^{2 (8+x)}\right )^2} \, dx\right )+2 \int \frac {\left (-1+2 x^2\right ) \log (5 x)}{x} \, dx \\ & = -\left (2 \int \left (-\frac {8}{\left (2+e^{16+2 x}\right )^2}+\frac {4}{2+e^{16+2 x}}+x \left (-1+2 x^2\right )\right ) \, dx\right )+2 \int \left (-\frac {\log (5 x)}{x}+2 x \log (5 x)\right ) \, dx \\ & = -\left (2 \int x \left (-1+2 x^2\right ) \, dx\right )-2 \int \frac {\log (5 x)}{x} \, dx+4 \int x \log (5 x) \, dx-8 \int \frac {1}{2+e^{16+2 x}} \, dx+16 \int \frac {1}{\left (2+e^{16+2 x}\right )^2} \, dx \\ & = -x^2+2 x^2 \log (5 x)-\log ^2(5 x)-2 \int \left (-x+2 x^3\right ) \, dx-4 \text {Subst}\left (\int \frac {1}{x (2+x)} \, dx,x,e^{16+2 x}\right )+8 \text {Subst}\left (\int \frac {1}{x (2+x)^2} \, dx,x,e^{16+2 x}\right ) \\ & = -x^4+2 x^2 \log (5 x)-\log ^2(5 x)-2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^{16+2 x}\right )+2 \text {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^{16+2 x}\right )+8 \text {Subst}\left (\int \left (\frac {1}{4 x}-\frac {1}{2 (2+x)^2}-\frac {1}{4 (2+x)}\right ) \, dx,x,e^{16+2 x}\right ) \\ & = \frac {4}{2+e^{16+2 x}}-x^4+2 x^2 \log (5 x)-\log ^2(5 x) \\ \end{align*}
Time = 1.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=\frac {4}{2+e^{2 (8+x)}}-x^4+2 x^2 \log (5 x)-\log ^2(5 x) \]
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Time = 1.45 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
method | result | size |
risch | \(-\ln \left (5 x \right )^{2}+2 x^{2} \ln \left (5 x \right )-\frac {{\mathrm e}^{2 x +16} x^{4}+2 x^{4}-4}{{\mathrm e}^{2 x +16}+2}\) | \(48\) |
parallelrisch | \(\frac {-2 \,{\mathrm e}^{16} {\mathrm e}^{2 x} x^{4}+4 \,{\mathrm e}^{16} {\mathrm e}^{2 x} \ln \left (5 x \right ) x^{2}-2 \,{\mathrm e}^{16} \ln \left (5 x \right )^{2} {\mathrm e}^{2 x}-4 \,{\mathrm e}^{2 x} {\mathrm e}^{16}-4 x^{4}+8 x^{2} \ln \left (5 x \right )-4 \ln \left (5 x \right )^{2}}{2 \,{\mathrm e}^{2 x} {\mathrm e}^{16}+4}\) | \(95\) |
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=-\frac {x^{4} e^{\left (2 \, x + 16\right )} + 2 \, x^{4} + {\left (e^{\left (2 \, x + 16\right )} + 2\right )} \log \left (5 \, x\right )^{2} - 2 \, {\left (x^{2} e^{\left (2 \, x + 16\right )} + 2 \, x^{2}\right )} \log \left (5 \, x\right ) - 4}{e^{\left (2 \, x + 16\right )} + 2} \]
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Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=- x^{4} + 2 x^{2} \log {\left (5 x \right )} - \log {\left (5 x \right )}^{2} + \frac {4}{e^{16} e^{2 x} + 2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (27) = 54\).
Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.18 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=-\frac {2 \, x^{4} - 4 \, x^{2} \log \left (5\right ) + {\left (x^{4} e^{16} - 2 \, x^{2} e^{16} \log \left (5\right ) + e^{16} \log \left (x\right )^{2} - 2 \, {\left (x^{2} e^{16} - e^{16} \log \left (5\right )\right )} \log \left (x\right )\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{2} - \log \left (5\right )\right )} \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 4}{e^{\left (2 \, x + 16\right )} + 2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (27) = 54\).
Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.71 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=-\frac {x^{4} e^{\left (2 \, x + 16\right )} + 2 \, x^{4} - 2 \, x^{2} e^{\left (2 \, x + 16\right )} \log \left (5\right ) - 2 \, x^{2} e^{\left (2 \, x + 16\right )} \log \left (x\right ) - 4 \, x^{2} \log \left (5\right ) - 4 \, x^{2} \log \left (x\right ) + 2 \, e^{\left (2 \, x + 16\right )} \log \left (5\right ) \log \left (x\right ) + e^{\left (2 \, x + 16\right )} \log \left (x\right )^{2} + 4 \, \log \left (5\right ) \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 4}{e^{\left (2 \, x + 16\right )} + 2} \]
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Time = 12.86 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=2\,x^2\,\ln \left (5\,x\right )+\frac {4\,{\mathrm {e}}^{-16}}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^{-16}}-{\ln \left (5\,x\right )}^2-x^4 \]
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