Integrand size = 95, antiderivative size = 19 \[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=\left (-5+10 x \left (5+e^x+\frac {4}{1+x}\right )\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(19)=38\).
Time = 0.34 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.79, number of steps used = 29, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {6820, 12, 6874, 37, 45, 2227, 2207, 2225, 2230, 2208, 2209} \[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=1000 e^x x^2+100 e^{2 x} x^2+\frac {7150 x^2}{(x+1)^2}+2500 x^2+700 e^x x+3500 x-800 e^x+\frac {800 e^x}{x+1}+\frac {15500}{x+1}-\frac {5550}{(x+1)^2} \]
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Rule 12
Rule 37
Rule 45
Rule 2207
Rule 2208
Rule 2209
Rule 2225
Rule 2227
Rule 2230
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {100 \left (-9+143 x+255 x^2+185 x^3+50 x^4+2 e^{2 x} x (1+x)^4+e^x \left (-1+32 x+96 x^2+110 x^3+57 x^4+10 x^5\right )\right )}{(1+x)^3} \, dx \\ & = 100 \int \frac {-9+143 x+255 x^2+185 x^3+50 x^4+2 e^{2 x} x (1+x)^4+e^x \left (-1+32 x+96 x^2+110 x^3+57 x^4+10 x^5\right )}{(1+x)^3} \, dx \\ & = 100 \int \left (-\frac {9}{(1+x)^3}+\frac {143 x}{(1+x)^3}+\frac {255 x^2}{(1+x)^3}+\frac {185 x^3}{(1+x)^3}+\frac {50 x^4}{(1+x)^3}+2 e^{2 x} x (1+x)+\frac {e^x \left (-1+33 x+63 x^2+47 x^3+10 x^4\right )}{(1+x)^2}\right ) \, dx \\ & = \frac {450}{(1+x)^2}+100 \int \frac {e^x \left (-1+33 x+63 x^2+47 x^3+10 x^4\right )}{(1+x)^2} \, dx+200 \int e^{2 x} x (1+x) \, dx+5000 \int \frac {x^4}{(1+x)^3} \, dx+14300 \int \frac {x}{(1+x)^3} \, dx+18500 \int \frac {x^3}{(1+x)^3} \, dx+25500 \int \frac {x^2}{(1+x)^3} \, dx \\ & = \frac {450}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+100 \int \left (-e^x+27 e^x x+10 e^x x^2-\frac {8 e^x}{(1+x)^2}+\frac {8 e^x}{1+x}\right ) \, dx+200 \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx+5000 \int \left (-3+x+\frac {1}{(1+x)^3}-\frac {4}{(1+x)^2}+\frac {6}{1+x}\right ) \, dx+18500 \int \left (1-\frac {1}{(1+x)^3}+\frac {3}{(1+x)^2}-\frac {3}{1+x}\right ) \, dx+25500 \int \left (\frac {1}{(1+x)^3}-\frac {2}{(1+x)^2}+\frac {1}{1+x}\right ) \, dx \\ & = 3500 x+2500 x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}-100 \int e^x \, dx+200 \int e^{2 x} x \, dx+200 \int e^{2 x} x^2 \, dx-800 \int \frac {e^x}{(1+x)^2} \, dx+800 \int \frac {e^x}{1+x} \, dx+1000 \int e^x x^2 \, dx+2700 \int e^x x \, dx \\ & = -100 e^x+3500 x+2700 e^x x+100 e^{2 x} x+2500 x^2+1000 e^x x^2+100 e^{2 x} x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}+\frac {800 e^x}{1+x}+\frac {800 \operatorname {ExpIntegralEi}(1+x)}{e}-100 \int e^{2 x} \, dx-200 \int e^{2 x} x \, dx-800 \int \frac {e^x}{1+x} \, dx-2000 \int e^x x \, dx-2700 \int e^x \, dx \\ & = -2800 e^x-50 e^{2 x}+3500 x+700 e^x x+2500 x^2+1000 e^x x^2+100 e^{2 x} x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}+\frac {800 e^x}{1+x}+100 \int e^{2 x} \, dx+2000 \int e^x \, dx \\ & = -800 e^x+3500 x+700 e^x x+2500 x^2+1000 e^x x^2+100 e^{2 x} x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}+\frac {800 e^x}{1+x} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(19)=38\).
Time = 4.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.74 \[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=100 \left (35 x+25 x^2+e^{2 x} x^2+\frac {4 (7+3 x)}{(1+x)^2}+\frac {e^x x \left (-1+17 x+10 x^2\right )}{1+x}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(54\) vs. \(2(18)=36\).
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.89
method | result | size |
risch | \(2500 x^{2}+3500 x +\frac {1200 x +2800}{x^{2}+2 x +1}+100 \,{\mathrm e}^{2 x} x^{2}+\frac {100 x \left (10 x^{2}+17 x -1\right ) {\mathrm e}^{x}}{1+x}\) | \(55\) |
default | \(\frac {1600}{\left (1+x \right )^{2}}+\frac {1200}{1+x}+3500 x +2500 x^{2}+\frac {800 \,{\mathrm e}^{x}}{1+x}+1000 \,{\mathrm e}^{x} x^{2}+700 \,{\mathrm e}^{x} x -800 \,{\mathrm e}^{x}+100 \,{\mathrm e}^{2 x} x^{2}\) | \(58\) |
parts | \(\frac {1600}{\left (1+x \right )^{2}}+\frac {1200}{1+x}+3500 x +2500 x^{2}+\frac {800 \,{\mathrm e}^{x}}{1+x}+1000 \,{\mathrm e}^{x} x^{2}+700 \,{\mathrm e}^{x} x -800 \,{\mathrm e}^{x}+100 \,{\mathrm e}^{2 x} x^{2}\) | \(58\) |
norman | \(\frac {-14300 x +8500 x^{3}+2500 x^{4}-100 \,{\mathrm e}^{x} x +1600 \,{\mathrm e}^{x} x^{2}+2700 \,{\mathrm e}^{x} x^{3}+1000 \,{\mathrm e}^{x} x^{4}+100 \,{\mathrm e}^{2 x} x^{2}+200 \,{\mathrm e}^{2 x} x^{3}+100 \,{\mathrm e}^{2 x} x^{4}-6700}{\left (1+x \right )^{2}}\) | \(75\) |
parallelrisch | \(\frac {-14300 x +8500 x^{3}+2500 x^{4}-100 \,{\mathrm e}^{x} x +1600 \,{\mathrm e}^{x} x^{2}+2700 \,{\mathrm e}^{x} x^{3}+1000 \,{\mathrm e}^{x} x^{4}+100 \,{\mathrm e}^{2 x} x^{2}+200 \,{\mathrm e}^{2 x} x^{3}+100 \,{\mathrm e}^{2 x} x^{4}-6700}{x^{2}+2 x +1}\) | \(80\) |
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.74 \[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=\frac {100 \, {\left (25 \, x^{4} + 85 \, x^{3} + 95 \, x^{2} + {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + {\left (10 \, x^{4} + 27 \, x^{3} + 16 \, x^{2} - x\right )} e^{x} + 47 \, x + 28\right )}}{x^{2} + 2 \, x + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (15) = 30\).
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.95 \[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=2500 x^{2} + 3500 x + \frac {1200 x + 2800}{x^{2} + 2 x + 1} + \frac {\left (100 x^{3} + 100 x^{2}\right ) e^{2 x} + \left (1000 x^{3} + 1700 x^{2} - 100 x\right ) e^{x}}{x + 1} \]
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\[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=\int { \frac {100 \, {\left (50 \, x^{4} + 185 \, x^{3} + 255 \, x^{2} + 2 \, {\left (x^{5} + 4 \, x^{4} + 6 \, x^{3} + 4 \, x^{2} + x\right )} e^{\left (2 \, x\right )} + {\left (10 \, x^{5} + 57 \, x^{4} + 110 \, x^{3} + 96 \, x^{2} + 32 \, x - 1\right )} e^{x} + 143 \, x - 9\right )}}{x^{3} + 3 \, x^{2} + 3 \, x + 1} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 4.37 \[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=\frac {100 \, {\left (x^{4} e^{\left (2 \, x\right )} + 10 \, x^{4} e^{x} + 25 \, x^{4} + 2 \, x^{3} e^{\left (2 \, x\right )} + 27 \, x^{3} e^{x} + 85 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} e^{x} + 95 \, x^{2} - x e^{x} + 47 \, x + 28\right )}}{x^{2} + 2 \, x + 1} \]
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Time = 8.00 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.58 \[ \int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} \left (200 x+800 x^2+1200 x^3+800 x^4+200 x^5\right )+e^x \left (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5\right )}{1+3 x+3 x^2+x^3} \, dx=x^2\,\left (100\,{\mathrm {e}}^{2\,x}+1000\,{\mathrm {e}}^x+2500\right )-800\,{\mathrm {e}}^x+x\,\left (700\,{\mathrm {e}}^x+3500\right )+\frac {800\,{\mathrm {e}}^x+x\,\left (800\,{\mathrm {e}}^x+1200\right )+2800}{{\left (x+1\right )}^2} \]
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