Integrand size = 69, antiderivative size = 30 \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=5 \left (-5+e^4-e^x+x\right ) \left (5-(4-x)^2+\log \left (-\frac {4}{x}\right )\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(30)=60\).
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.70, number of steps used = 20, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {14, 6874, 2230, 2225, 2209, 2207, 2634, 2332} \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=-5 x^3+5 e^x x^2+5 \left (13-e^4\right ) x^2-40 e^x x-20 \left (13-2 e^4\right ) x+5 x+55 e^x+5 x \log \left (-\frac {4}{x}\right )-5 e^x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x) \]
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Rule 14
Rule 2207
Rule 2209
Rule 2225
Rule 2230
Rule 2332
Rule 2634
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5 e^x \left (1+3 x-6 x^2+x^3-x \log \left (-\frac {4}{x}\right )\right )}{x}+\frac {5 \left (5 \left (1-\frac {e^4}{5}\right )-52 \left (1-\frac {2 e^4}{13}\right ) x+26 \left (1-\frac {e^4}{13}\right ) x^2-3 x^3+x \log \left (-\frac {4}{x}\right )\right )}{x}\right ) \, dx \\ & = 5 \int \frac {e^x \left (1+3 x-6 x^2+x^3-x \log \left (-\frac {4}{x}\right )\right )}{x} \, dx+5 \int \frac {5 \left (1-\frac {e^4}{5}\right )-52 \left (1-\frac {2 e^4}{13}\right ) x+26 \left (1-\frac {e^4}{13}\right ) x^2-3 x^3+x \log \left (-\frac {4}{x}\right )}{x} \, dx \\ & = 5 \int \left (\frac {5-e^4-4 \left (13-2 e^4\right ) x+2 \left (13-e^4\right ) x^2-3 x^3}{x}+\log \left (-\frac {4}{x}\right )\right ) \, dx+5 \int \left (\frac {e^x \left (1+3 x-6 x^2+x^3\right )}{x}-e^x \log \left (-\frac {4}{x}\right )\right ) \, dx \\ & = 5 \int \frac {5-e^4-4 \left (13-2 e^4\right ) x+2 \left (13-e^4\right ) x^2-3 x^3}{x} \, dx+5 \int \frac {e^x \left (1+3 x-6 x^2+x^3\right )}{x} \, dx+5 \int \log \left (-\frac {4}{x}\right ) \, dx-5 \int e^x \log \left (-\frac {4}{x}\right ) \, dx \\ & = 5 x-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )-5 \int \frac {e^x}{x} \, dx+5 \int \left (-4 \left (13-2 e^4\right )+\frac {5-e^4}{x}+2 \left (13-e^4\right ) x-3 x^2\right ) \, dx+5 \int \left (3 e^x+\frac {e^x}{x}-6 e^x x+e^x x^2\right ) \, dx \\ & = 5 x-20 \left (13-2 e^4\right ) x+5 \left (13-e^4\right ) x^2-5 x^3-5 \text {Ei}(x)-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x)+5 \int \frac {e^x}{x} \, dx+5 \int e^x x^2 \, dx+15 \int e^x \, dx-30 \int e^x x \, dx \\ & = 15 e^x+5 x-30 e^x x-20 \left (13-2 e^4\right ) x+5 e^x x^2+5 \left (13-e^4\right ) x^2-5 x^3-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x)-10 \int e^x x \, dx+30 \int e^x \, dx \\ & = 45 e^x+5 x-40 e^x x-20 \left (13-2 e^4\right ) x+5 e^x x^2+5 \left (13-e^4\right ) x^2-5 x^3-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x)+10 \int e^x \, dx \\ & = 55 e^x+5 x-40 e^x x-20 \left (13-2 e^4\right ) x+5 e^x x^2+5 \left (13-e^4\right ) x^2-5 x^3-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(30)=60\).
Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.13 \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=5 \left (\left (-51+8 e^4\right ) x+\left (13-e^4\right ) x^2-x^3+e^x \left (11-8 x+x^2\right )+\left (-e^x+x\right ) \log \left (-\frac {4}{x}\right )+\left (5-e^4\right ) \log (x)\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(72\) vs. \(2(28)=56\).
Time = 2.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.43
method | result | size |
norman | \(\left (-255+40 \,{\mathrm e}^{4}\right ) x +\left (-5 \,{\mathrm e}^{4}+65\right ) x^{2}+\left (5 \,{\mathrm e}^{4}-25\right ) \ln \left (-\frac {4}{x}\right )-5 x^{3}-40 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{2}-5 \,{\mathrm e}^{x} \ln \left (-\frac {4}{x}\right )+5 \ln \left (-\frac {4}{x}\right ) x +55 \,{\mathrm e}^{x}\) | \(73\) |
default | \(5 \left (11-\ln \left (-\frac {4}{x}\right )-\ln \left (x \right )\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} \ln \left (x \right )-40 \,{\mathrm e}^{x} x -5 x^{2} {\mathrm e}^{4}-5 x^{3}+40 x \,{\mathrm e}^{4}+65 x^{2}-255 x -5 \left ({\mathrm e}^{4}-5\right ) \ln \left (x \right )+5 \ln \left (-\frac {4}{x}\right ) x\) | \(80\) |
parallelrisch | \(-5 x^{2} {\mathrm e}^{4}-5 x^{3}+5 \,{\mathrm e}^{x} x^{2}+40 x \,{\mathrm e}^{4}+5 \ln \left (-\frac {4}{x}\right ) {\mathrm e}^{4}+65 x^{2}-40 \,{\mathrm e}^{x} x +5 \ln \left (-\frac {4}{x}\right ) x -5 \,{\mathrm e}^{x} \ln \left (-\frac {4}{x}\right )-255 x +55 \,{\mathrm e}^{x}-25 \ln \left (-\frac {4}{x}\right )\) | \(80\) |
parts | \(5 \left (11-\ln \left (-\frac {4}{x}\right )-\ln \left (x \right )\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} \ln \left (x \right )-40 \,{\mathrm e}^{x} x -5 x^{2} {\mathrm e}^{4}-5 x^{3}+40 x \,{\mathrm e}^{4}+65 x^{2}-255 x -5 \left ({\mathrm e}^{4}-5\right ) \ln \left (x \right )+5 \ln \left (-\frac {4}{x}\right ) x\) | \(80\) |
risch | \(\left (5 \,{\mathrm e}^{x}-5 x \right ) \ln \left (x \right )-5 i \pi x \operatorname {csgn}\left (\frac {i}{x}\right )^{2}+5 i \pi x \operatorname {csgn}\left (\frac {i}{x}\right )^{3}-5 i {\mathrm e}^{x} \pi \operatorname {csgn}\left (\frac {i}{x}\right )^{3}+5 i {\mathrm e}^{x} \pi \operatorname {csgn}\left (\frac {i}{x}\right )^{2}-5 x^{2} {\mathrm e}^{4}-5 x^{3}+40 x \,{\mathrm e}^{4}+10 x \ln \left (2\right )+65 x^{2}+5 i \pi x -255 x -5 \,{\mathrm e}^{4} \ln \left (x \right )+25 \ln \left (x \right )-5 i {\mathrm e}^{x} \pi -10 \,{\mathrm e}^{x} \ln \left (2\right )+55 \,{\mathrm e}^{x}-40 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{2}\) | \(144\) |
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).
Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=-5 \, x^{3} + 65 \, x^{2} - 5 \, {\left (x^{2} - 8 \, x\right )} e^{4} + 5 \, {\left (x^{2} - 8 \, x + 11\right )} e^{x} + 5 \, {\left (x + e^{4} - e^{x} - 5\right )} \log \left (-\frac {4}{x}\right ) - 255 \, x \]
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Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (27) = 54\).
Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=- 5 x^{3} - x^{2} \left (-65 + 5 e^{4}\right ) + 5 x \log {\left (- \frac {4}{x} \right )} - x \left (255 - 40 e^{4}\right ) + \left (5 x^{2} - 40 x - 5 \log {\left (- \frac {4}{x} \right )} + 55\right ) e^{x} - 5 \left (-5 + e^{4}\right ) \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (26) = 52\).
Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.60 \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=-5 \, x^{3} - 5 \, x^{2} e^{4} + 65 \, x^{2} + 40 \, x e^{4} + 5 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 30 \, {\left (x - 1\right )} e^{x} - 5 \, e^{4} \log \left (x\right ) + 5 \, x \log \left (-\frac {4}{x}\right ) - 5 \, e^{x} \log \left (-\frac {4}{x}\right ) - 255 \, x + 15 \, e^{x} + 25 \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (26) = 52\).
Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.37 \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=-5 \, x^{3} - 5 \, x^{2} e^{4} + 5 \, x^{2} e^{x} + 65 \, x^{2} + 40 \, x e^{4} - 40 \, x e^{x} - 5 \, e^{4} \log \left (x\right ) + 5 \, x \log \left (-\frac {4}{x}\right ) - 5 \, e^{x} \log \left (-\frac {4}{x}\right ) - 255 \, x + 55 \, e^{x} + 25 \, \log \left (x\right ) \]
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Time = 10.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int \frac {25-260 x+130 x^2-15 x^3+e^4 \left (-5+40 x-10 x^2\right )+e^x \left (5+15 x-30 x^2+5 x^3\right )+\left (5 x-5 e^x x\right ) \log \left (-\frac {4}{x}\right )}{x} \, dx=55\,{\mathrm {e}}^x+x\,\left (40\,{\mathrm {e}}^4+5\,\ln \left (-\frac {4}{x}\right )-40\,{\mathrm {e}}^x-255\right )+x^2\,\left (5\,{\mathrm {e}}^x-5\,{\mathrm {e}}^4+65\right )+\ln \left (\frac {1}{x}\right )\,\left (5\,{\mathrm {e}}^4-25\right )-5\,x^3-5\,{\mathrm {e}}^x\,\ln \left (-\frac {4}{x}\right ) \]
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