Integrand size = 58, antiderivative size = 18 \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=\left (398-\frac {e^{5+x}}{4}+x-\log (x)\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(18)=36\).
Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.72, number of steps used = 15, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.155, Rules used = {12, 14, 2225, 6818, 6874, 2230, 2209, 2207, 2634} \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=-199 e^{x+5}+\frac {1}{16} e^{2 x+10}-\frac {1}{2} e^{x+5} x+(x-\log (x)+398)^2+\frac {1}{2} e^{x+5} \log (x) \]
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Rule 12
Rule 14
Rule 2207
Rule 2209
Rule 2225
Rule 2230
Rule 2634
Rule 6818
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{x} \, dx \\ & = \frac {1}{8} \int \left (e^{10+2 x}+\frac {16 (-1+x) (398+x-\log (x))}{x}-\frac {4 e^{5+x} \left (-1+399 x+x^2-x \log (x)\right )}{x}\right ) \, dx \\ & = \frac {1}{8} \int e^{10+2 x} \, dx-\frac {1}{2} \int \frac {e^{5+x} \left (-1+399 x+x^2-x \log (x)\right )}{x} \, dx+2 \int \frac {(-1+x) (398+x-\log (x))}{x} \, dx \\ & = \frac {1}{16} e^{10+2 x}+(398+x-\log (x))^2-\frac {1}{2} \int \left (\frac {e^{5+x} \left (-1+399 x+x^2\right )}{x}-e^{5+x} \log (x)\right ) \, dx \\ & = \frac {1}{16} e^{10+2 x}+(398+x-\log (x))^2-\frac {1}{2} \int \frac {e^{5+x} \left (-1+399 x+x^2\right )}{x} \, dx+\frac {1}{2} \int e^{5+x} \log (x) \, dx \\ & = \frac {1}{16} e^{10+2 x}+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x)-\frac {1}{2} \int \frac {e^{5+x}}{x} \, dx-\frac {1}{2} \int \left (399 e^{5+x}-\frac {e^{5+x}}{x}+e^{5+x} x\right ) \, dx \\ & = \frac {1}{16} e^{10+2 x}-\frac {e^5 \operatorname {ExpIntegralEi}(x)}{2}+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x)+\frac {1}{2} \int \frac {e^{5+x}}{x} \, dx-\frac {1}{2} \int e^{5+x} x \, dx-\frac {399}{2} \int e^{5+x} \, dx \\ & = -\frac {399 e^{5+x}}{2}+\frac {1}{16} e^{10+2 x}-\frac {1}{2} e^{5+x} x+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x)+\frac {1}{2} \int e^{5+x} \, dx \\ & = -199 e^{5+x}+\frac {1}{16} e^{10+2 x}-\frac {1}{2} e^{5+x} x+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=\frac {1}{16} \left (-e^{5+x}+4 (398+x)-4 \log (x)\right )^2 \]
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Leaf count of result is larger than twice the leaf count of optimal. \(49\) vs. \(2(15)=30\).
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.78
method | result | size |
default | \(\ln \left (x \right )^{2}+\frac {\ln \left (x \right ) {\mathrm e}^{5+x}}{2}-2 x \ln \left (x \right )+x^{2}+796 x -796 \ln \left (x \right )+\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}\) | \(50\) |
norman | \(\ln \left (x \right )^{2}+\frac {\ln \left (x \right ) {\mathrm e}^{5+x}}{2}-2 x \ln \left (x \right )+x^{2}+796 x -796 \ln \left (x \right )+\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}\) | \(50\) |
parallelrisch | \(\ln \left (x \right )^{2}+\frac {\ln \left (x \right ) {\mathrm e}^{5+x}}{2}-2 x \ln \left (x \right )+x^{2}+796 x -796 \ln \left (x \right )+\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}\) | \(50\) |
parts | \(\ln \left (x \right )^{2}+\frac {\ln \left (x \right ) {\mathrm e}^{5+x}}{2}-2 x \ln \left (x \right )+x^{2}+796 x -796 \ln \left (x \right )+\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}\) | \(50\) |
risch | \(\ln \left (x \right )^{2}+\frac {\left (-16 x +4 \,{\mathrm e}^{5+x}\right ) \ln \left (x \right )}{8}+x^{2}+796 x -796 \ln \left (x \right )+\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}\) | \(51\) |
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (19) = 38\).
Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.39 \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=x^{2} - \frac {1}{2} \, {\left (x + 398\right )} e^{\left (x + 5\right )} - \frac {1}{2} \, {\left (4 \, x - e^{\left (x + 5\right )} + 1592\right )} \log \left (x\right ) + \log \left (x\right )^{2} + 796 \, x + \frac {1}{16} \, e^{\left (2 \, x + 10\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (14) = 28\).
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.67 \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=x^{2} - 2 x \log {\left (x \right )} + 796 x + \frac {\left (- 16 x + 16 \log {\left (x \right )} - 6368\right ) e^{x + 5}}{32} + \frac {e^{2 x + 10}}{16} + \log {\left (x \right )}^{2} - 796 \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (19) = 38\).
Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 3.06 \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=x^{2} - \frac {1}{2} \, {\left (x e^{5} - e^{5}\right )} e^{x} - 2 \, x \log \left (x\right ) + \frac {1}{2} \, e^{\left (x + 5\right )} \log \left (x\right ) + \log \left (x\right )^{2} + 796 \, x + \frac {1}{16} \, e^{\left (2 \, x + 10\right )} - \frac {399}{2} \, e^{\left (x + 5\right )} - 796 \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.72 \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=x^{2} - \frac {1}{2} \, x e^{\left (x + 5\right )} - 2 \, x \log \left (x\right ) + \frac {1}{2} \, e^{\left (x + 5\right )} \log \left (x\right ) + \log \left (x\right )^{2} + 796 \, x + \frac {1}{16} \, e^{\left (2 \, x + 10\right )} - 199 \, e^{\left (x + 5\right )} - 796 \, \log \left (x\right ) \]
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Time = 9.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.72 \[ \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{8 x} \, dx=796\,x-199\,{\mathrm {e}}^{x+5}+\frac {{\mathrm {e}}^{2\,x+10}}{16}-796\,\ln \left (x\right )-\frac {x\,{\mathrm {e}}^{x+5}}{2}+{\ln \left (x\right )}^2+\frac {{\mathrm {e}}^{x+5}\,\ln \left (x\right )}{2}-2\,x\,\ln \left (x\right )+x^2 \]
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