Integrand size = 185, antiderivative size = 36 \[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=\frac {3 \log \left (\frac {25 \left (4 e^{-x}+\frac {4}{(4+2 x)^2}\right )^2}{x^2}\right )}{5+\frac {5}{x}} \]
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\[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=\int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {3 \left (-64-224 x-304 x^2-200 x^3-64 x^4-8 x^5-2 e^x \left (2+5 x+3 x^2\right )+(2+x) \left (e^x+4 (2+x)^2\right ) \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )\right )}{5 (1+x)^2 (2+x) \left (e^x+4 (2+x)^2\right )} \, dx \\ & = \frac {3}{5} \int \frac {-64-224 x-304 x^2-200 x^3-64 x^4-8 x^5-2 e^x \left (2+5 x+3 x^2\right )+(2+x) \left (e^x+4 (2+x)^2\right ) \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x) \left (e^x+4 (2+x)^2\right )} \, dx \\ & = \frac {3}{5} \int \left (-\frac {8 x^2 (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )}+\frac {-4-10 x-6 x^2+2 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )+x \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x)}\right ) \, dx \\ & = \frac {3}{5} \int \frac {-4-10 x-6 x^2+2 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )+x \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x)} \, dx-\frac {24}{5} \int \frac {x^2 (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx \\ & = \frac {3}{5} \int \frac {-2 \left (2+5 x+3 x^2\right )+(2+x) \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x)} \, dx-\frac {24}{5} \int \left (-\frac {1}{16+e^x+16 x+4 x^2}+\frac {x}{16+e^x+16 x+4 x^2}+\frac {x^2}{16+e^x+16 x+4 x^2}+\frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )}\right ) \, dx \\ & = \frac {3}{5} \int \left (-\frac {2 (2+3 x)}{2+3 x+x^2}+\frac {\log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2}\right ) \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx \\ & = \frac {3}{5} \int \frac {\log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2} \, dx-\frac {6}{5} \int \frac {2+3 x}{2+3 x+x^2} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx \\ & = -\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {3}{5} \int \frac {2 \left (-4 (1+x) (2+x)^3-e^x (2+3 x)\right )}{x (1+x) (2+x) \left (e^x+4 (2+x)^2\right )} \, dx+\frac {6}{5} \int \frac {1}{1+x} \, dx-\frac {24}{5} \int \frac {1}{2+x} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx \\ & = \frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \frac {-4 (1+x) (2+x)^3-e^x (2+3 x)}{x (1+x) (2+x) \left (e^x+4 (2+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx \\ & = \frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \left (\frac {-2-3 x}{x \left (2+3 x+x^2\right )}-\frac {4 x (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )}\right ) \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx \\ & = \frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \frac {-2-3 x}{x \left (2+3 x+x^2\right )} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx-\frac {24}{5} \int \frac {x (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx \\ & = \frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \left (\frac {1}{-1-x}-\frac {1}{x}+\frac {2}{2+x}\right ) \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx-\frac {24}{5} \int \left (\frac {1}{16+e^x+16 x+4 x^2}+\frac {x}{16+e^x+16 x+4 x^2}-\frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )}\right ) \, dx \\ & = -\frac {6 \log (x)}{5}-\frac {12}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}-2 \left (\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx\right )-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(93\) vs. \(2(36)=72\).
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.58 \[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=-\frac {3 \left (2+4 x+2 x^2+2 (1+x) \log (x)+4 (1+x) \log (2+x)-2 \log \left (e^x+4 (2+x)^2\right )-2 x \log \left (e^x+4 (2+x)^2\right )+\log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )\right )}{5 (1+x)} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(76\) vs. \(2(35)=70\).
Time = 1.64 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.14
method | result | size |
parallelrisch | \(\frac {3 \ln \left (\frac {\left (25 \,{\mathrm e}^{2 x}+\left (200 x^{2}+800 x +800\right ) {\mathrm e}^{x}+400 x^{4}+3200 x^{3}+9600 x^{2}+12800 x +6400\right ) {\mathrm e}^{-2 x}}{x^{2} \left (x^{4}+8 x^{3}+24 x^{2}+32 x +16\right )}\right ) x}{5 \left (1+x \right )}\) | \(77\) |
risch | \(\text {Expression too large to display}\) | \(1110\) |
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Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (29) = 58\).
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.14 \[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=\frac {3 \, x \log \left (\frac {25 \, {\left (16 \, x^{4} + 128 \, x^{3} + 384 \, x^{2} + 8 \, {\left (x^{2} + 4 \, x + 4\right )} e^{x} + 512 \, x + e^{\left (2 \, x\right )} + 256\right )} e^{\left (-2 \, x\right )}}{x^{6} + 8 \, x^{5} + 24 \, x^{4} + 32 \, x^{3} + 16 \, x^{2}}\right )}{5 \, {\left (x + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (27) = 54\).
Time = 0.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 3.22 \[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=- \frac {6 x}{5} - \frac {6 \log {\left (x \right )}}{5} - \frac {12 \log {\left (x + 2 \right )}}{5} + \frac {6 \log {\left (4 x^{2} + 16 x + e^{x} + 16 \right )}}{5} - \frac {3 \log {\left (\frac {\left (400 x^{4} + 3200 x^{3} + 9600 x^{2} + 12800 x + \left (200 x^{2} + 800 x + 800\right ) e^{x} + 25 e^{2 x} + 6400\right ) e^{- 2 x}}{x^{6} + 8 x^{5} + 24 x^{4} + 32 x^{3} + 16 x^{2}} \right )}}{5 x + 5} \]
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Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=-\frac {6 \, {\left (x^{2} - x \log \left (4 \, x^{2} + 16 \, x + e^{x} + 16\right ) + 2 \, x \log \left (x + 2\right ) + x \log \left (x\right ) + x + \log \left (5\right ) + 1\right )}}{5 \, {\left (x + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (29) = 58\).
Time = 0.47 (sec) , antiderivative size = 163, normalized size of antiderivative = 4.53 \[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=-\frac {3 \, {\left (2 \, x^{2} - 2 \, x \log \left (-4 \, x^{2} - 16 \, x - e^{x} - 16\right ) + 4 \, x \log \left (x + 2\right ) + 2 \, x \log \left (x\right ) + 2 \, x - 2 \, \log \left (-4 \, x^{2} - 16 \, x - e^{x} - 16\right ) + 4 \, \log \left (x + 2\right ) + 2 \, \log \left (x\right ) + \log \left (\frac {25 \, {\left (16 \, x^{4} + 128 \, x^{3} + 8 \, x^{2} e^{x} + 384 \, x^{2} + 32 \, x e^{x} + 512 \, x + e^{\left (2 \, x\right )} + 32 \, e^{x} + 256\right )}}{x^{6} e^{\left (2 \, x\right )} + 8 \, x^{5} e^{\left (2 \, x\right )} + 24 \, x^{4} e^{\left (2 \, x\right )} + 32 \, x^{3} e^{\left (2 \, x\right )} + 16 \, x^{2} e^{\left (2 \, x\right )}}\right )\right )}}{5 \, {\left (x + 1\right )}} \]
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Timed out. \[ \int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx=\int -\frac {672\,x-\ln \left (\frac {{\mathrm {e}}^{-2\,x}\,\left (12800\,x+25\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (200\,x^2+800\,x+800\right )+9600\,x^2+3200\,x^3+400\,x^4+6400\right )}{x^6+8\,x^5+24\,x^4+32\,x^3+16\,x^2}\right )\,\left (144\,x+{\mathrm {e}}^x\,\left (3\,x+6\right )+72\,x^2+12\,x^3+96\right )+{\mathrm {e}}^x\,\left (18\,x^2+30\,x+12\right )+912\,x^2+600\,x^3+192\,x^4+24\,x^5+192}{560\,x+760\,x^2+500\,x^3+160\,x^4+20\,x^5+{\mathrm {e}}^x\,\left (5\,x^3+20\,x^2+25\,x+10\right )+160} \,d x \]
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