Integrand size = 78, antiderivative size = 23 \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=8 \left (x+\frac {e^x+2 (1+x)}{x}\right ) (x+\log (4))^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(23)=46\).
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.91, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 2230, 2207, 2225, 2208, 2209} \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=8 x^3+16 x^2 (1+\log (4))+8 e^x x-8 e^x+8 x \left (2+\log ^2(4)+\log (256)\right )+\frac {8 e^x \log ^2(4)}{x}+\frac {16 \log ^2(4)}{x}+8 e^x (1+\log (16)) \]
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Rule 14
Rule 2207
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {8 e^x (x+\log (4)) \left (x^2-\log (4)+x (1+\log (4))\right )}{x^2}+\frac {8 \left (3 x^4-2 \log ^2(4)+4 x^3 (1+\log (4))+x^2 \left (2+\log ^2(4)+\log (256)\right )\right )}{x^2}\right ) \, dx \\ & = 8 \int \frac {e^x (x+\log (4)) \left (x^2-\log (4)+x (1+\log (4))\right )}{x^2} \, dx+8 \int \frac {3 x^4-2 \log ^2(4)+4 x^3 (1+\log (4))+x^2 \left (2+\log ^2(4)+\log (256)\right )}{x^2} \, dx \\ & = 8 \int \left (e^x x-\frac {e^x \log ^2(4)}{x^2}+\frac {e^x \log ^2(4)}{x}+e^x (1+\log (16))\right ) \, dx+8 \int \left (3 x^2-\frac {2 \log ^2(4)}{x^2}+4 x (1+\log (4))+2 \left (1+\frac {\log ^2(4)}{2}+\log (16)\right )\right ) \, dx \\ & = 8 x^3+\frac {16 \log ^2(4)}{x}+16 x^2 (1+\log (4))+8 x \left (2+\log ^2(4)+\log (256)\right )+8 \int e^x x \, dx-\left (8 \log ^2(4)\right ) \int \frac {e^x}{x^2} \, dx+\left (8 \log ^2(4)\right ) \int \frac {e^x}{x} \, dx+(8 (1+\log (16))) \int e^x \, dx \\ & = 8 e^x x+8 x^3+\frac {16 \log ^2(4)}{x}+\frac {8 e^x \log ^2(4)}{x}+8 \operatorname {ExpIntegralEi}(x) \log ^2(4)+16 x^2 (1+\log (4))+8 e^x (1+\log (16))+8 x \left (2+\log ^2(4)+\log (256)\right )-8 \int e^x \, dx-\left (8 \log ^2(4)\right ) \int \frac {e^x}{x} \, dx \\ & = -8 e^x+8 e^x x+8 x^3+\frac {16 \log ^2(4)}{x}+\frac {8 e^x \log ^2(4)}{x}+16 x^2 (1+\log (4))+8 e^x (1+\log (16))+8 x \left (2+\log ^2(4)+\log (256)\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(47\) vs. \(2(23)=46\).
Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=8 \left (x^3+\frac {\left (2+e^x\right ) \log ^2(4)}{x}+2 x^2 (1+\log (4))+e^x \log (16)+x \left (2+e^x+\log ^2(4)+\log (256)\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(64\) vs. \(2(23)=46\).
Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.83
method | result | size |
norman | \(\frac {\left (32 \ln \left (2\right )+16\right ) x^{3}+\left (32 \ln \left (2\right )^{2}+64 \ln \left (2\right )+16\right ) x^{2}+8 x^{4}+64 \ln \left (2\right )^{2}+8 \,{\mathrm e}^{x} x^{2}+32 \ln \left (2\right )^{2} {\mathrm e}^{x}+32 x \ln \left (2\right ) {\mathrm e}^{x}}{x}\) | \(65\) |
risch | \(32 x \ln \left (2\right )^{2}+32 x^{2} \ln \left (2\right )+8 x^{3}+64 x \ln \left (2\right )+16 x^{2}+16 x +\frac {64 \ln \left (2\right )^{2}}{x}+\frac {8 \left (4 \ln \left (2\right )^{2}+4 x \ln \left (2\right )+x^{2}\right ) {\mathrm e}^{x}}{x}\) | \(65\) |
parts | \(32 x \ln \left (2\right )^{2}+32 x^{2} \ln \left (2\right )+8 x^{3}+64 x \ln \left (2\right )+16 x^{2}+16 x +\frac {64 \ln \left (2\right )^{2}}{x}+8 \,{\mathrm e}^{x} x +32 \,{\mathrm e}^{x} \ln \left (2\right )+\frac {32 \ln \left (2\right )^{2} {\mathrm e}^{x}}{x}\) | \(65\) |
parallelrisch | \(\frac {32 x^{2} \ln \left (2\right )^{2}+32 x^{3} \ln \left (2\right )+8 x^{4}+32 \ln \left (2\right )^{2} {\mathrm e}^{x}+64 x^{2} \ln \left (2\right )+32 x \ln \left (2\right ) {\mathrm e}^{x}+16 x^{3}+8 \,{\mathrm e}^{x} x^{2}+64 \ln \left (2\right )^{2}+16 x^{2}}{x}\) | \(72\) |
default | \(8 x^{3}+16 x^{2}+16 x +32 x \ln \left (2\right )^{2}+32 x^{2} \ln \left (2\right )+8 \,{\mathrm e}^{x} x +32 \,{\mathrm e}^{x} \ln \left (2\right )+\frac {64 \ln \left (2\right )^{2}}{x}-32 \ln \left (2\right )^{2} \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )-32 \ln \left (2\right )^{2} \operatorname {Ei}_{1}\left (-x \right )+64 x \ln \left (2\right )\) | \(86\) |
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Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.65 \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=\frac {8 \, {\left (x^{4} + 2 \, x^{3} + 4 \, {\left (x^{2} + 2\right )} \log \left (2\right )^{2} + 2 \, x^{2} + {\left (x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}\right )} e^{x} + 4 \, {\left (x^{3} + 2 \, x^{2}\right )} \log \left (2\right )\right )}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (20) = 40\).
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.65 \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=8 x^{3} + x^{2} \cdot \left (16 + 32 \log {\left (2 \right )}\right ) + x \left (32 \log {\left (2 \right )}^{2} + 16 + 64 \log {\left (2 \right )}\right ) + \frac {\left (8 x^{2} + 32 x \log {\left (2 \right )} + 32 \log {\left (2 \right )}^{2}\right ) e^{x}}{x} + \frac {64 \log {\left (2 \right )}^{2}}{x} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.39 \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=8 \, x^{3} + 32 \, x^{2} \log \left (2\right ) + 32 \, x \log \left (2\right )^{2} + 32 \, {\rm Ei}\left (x\right ) \log \left (2\right )^{2} - 32 \, \Gamma \left (-1, -x\right ) \log \left (2\right )^{2} + 16 \, x^{2} + 8 \, {\left (x - 1\right )} e^{x} + 64 \, x \log \left (2\right ) + 32 \, e^{x} \log \left (2\right ) + 16 \, x + \frac {64 \, \log \left (2\right )^{2}}{x} + 8 \, e^{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (23) = 46\).
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.00 \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=\frac {8 \, {\left (x^{4} + 4 \, x^{3} \log \left (2\right ) + 4 \, x^{2} \log \left (2\right )^{2} + 2 \, x^{3} + x^{2} e^{x} + 8 \, x^{2} \log \left (2\right ) + 4 \, x e^{x} \log \left (2\right ) + 4 \, e^{x} \log \left (2\right )^{2} + 2 \, x^{2} + 8 \, \log \left (2\right )^{2}\right )}}{x} \]
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Time = 13.82 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.57 \[ \int \frac {16 x^2+32 x^3+24 x^4+\left (32 x^2+32 x^3\right ) \log (4)+\left (-16+8 x^2\right ) \log ^2(4)+e^x \left (8 x^2+8 x^3+16 x^2 \log (4)+(-8+8 x) \log ^2(4)\right )}{x^2} \, dx=x^2\,\left (32\,\ln \left (2\right )+16\right )+x\,\left (64\,\ln \left (2\right )+8\,{\mathrm {e}}^x+32\,{\ln \left (2\right )}^2+16\right )+32\,{\mathrm {e}}^x\,\ln \left (2\right )+\frac {32\,{\mathrm {e}}^x\,{\ln \left (2\right )}^2+64\,{\ln \left (2\right )}^2}{x}+8\,x^3 \]
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