Integrand size = 40, antiderivative size = 35 \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=1-e^x+5 \left (3-e-e^5+x+5 \left (-x+\frac {x^2}{9}\right )\right )^2 \]
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Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6, 12, 2225} \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=\frac {125 x^4}{81}-\frac {200 x^3}{9}+\frac {290 x^2}{3}-120 x-e^x-\frac {2}{9} e \left (1+e^4\right ) (18-5 x)^2 \]
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Rule 6
Rule 12
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{81} \left (-9720-81 e^x+\left (e+e^5\right ) (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx \\ & = \frac {1}{81} \int \left (-9720-81 e^x+\left (e+e^5\right ) (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx \\ & = -\frac {2}{9} e \left (1+e^4\right ) (18-5 x)^2-120 x+\frac {290 x^2}{3}-\frac {200 x^3}{9}+\frac {125 x^4}{81}-\int e^x \, dx \\ & = -e^x-\frac {2}{9} e \left (1+e^4\right ) (18-5 x)^2-120 x+\frac {290 x^2}{3}-\frac {200 x^3}{9}+\frac {125 x^4}{81} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.66 \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=-e^x-120 x+40 e x+40 e^5 x+\frac {290 x^2}{3}-\frac {50 e x^2}{9}-\frac {50 e^5 x^2}{9}-\frac {200 x^3}{9}+\frac {125 x^4}{81} \]
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Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20
method | result | size |
norman | \(\left (40 \,{\mathrm e}^{5}+40 \,{\mathrm e}-120\right ) x +\left (-\frac {50 \,{\mathrm e}^{5}}{9}-\frac {50 \,{\mathrm e}}{9}+\frac {290}{3}\right ) x^{2}-\frac {200 x^{3}}{9}+\frac {125 x^{4}}{81}-{\mathrm e}^{x}\) | \(42\) |
risch | \(-{\mathrm e}^{x}-\frac {50 x^{2} {\mathrm e}^{5}}{9}+40 x \,{\mathrm e}^{5}-\frac {50 x^{2} {\mathrm e}}{9}+40 x \,{\mathrm e}+\frac {125 x^{4}}{81}-\frac {200 x^{3}}{9}+\frac {290 x^{2}}{3}-120 x\) | \(48\) |
parallelrisch | \(-{\mathrm e}^{x}-\frac {50 x^{2} {\mathrm e}^{5}}{9}+40 x \,{\mathrm e}^{5}-\frac {50 x^{2} {\mathrm e}}{9}+40 x \,{\mathrm e}+\frac {125 x^{4}}{81}-\frac {200 x^{3}}{9}+\frac {290 x^{2}}{3}-120 x\) | \(48\) |
parts | \(-{\mathrm e}^{x}-\frac {50 x^{2} {\mathrm e}^{5}}{9}+40 x \,{\mathrm e}^{5}-\frac {50 x^{2} {\mathrm e}}{9}+40 x \,{\mathrm e}+\frac {125 x^{4}}{81}-\frac {200 x^{3}}{9}+\frac {290 x^{2}}{3}-120 x\) | \(48\) |
default | \(-120 x +\frac {20 \,{\mathrm e} \left (-\frac {5}{2} x^{2}+18 x \right )}{9}+\frac {20 \,{\mathrm e}^{5} \left (-\frac {5}{2} x^{2}+18 x \right )}{9}+\frac {290 x^{2}}{3}-\frac {200 x^{3}}{9}+\frac {125 x^{4}}{81}-{\mathrm e}^{x}\) | \(50\) |
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Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=\frac {125}{81} \, x^{4} - \frac {200}{9} \, x^{3} + \frac {290}{3} \, x^{2} - \frac {10}{9} \, {\left (5 \, x^{2} - 36 \, x\right )} e^{5} - \frac {10}{9} \, {\left (5 \, x^{2} - 36 \, x\right )} e - 120 \, x - e^{x} \]
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Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=\frac {125 x^{4}}{81} - \frac {200 x^{3}}{9} + x^{2} \left (- \frac {50 e^{5}}{9} - \frac {50 e}{9} + \frac {290}{3}\right ) + x \left (-120 + 40 e + 40 e^{5}\right ) - e^{x} \]
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Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=\frac {125}{81} \, x^{4} - \frac {200}{9} \, x^{3} + \frac {290}{3} \, x^{2} - \frac {10}{9} \, {\left (5 \, x^{2} - 36 \, x\right )} e^{5} - \frac {10}{9} \, {\left (5 \, x^{2} - 36 \, x\right )} e - 120 \, x - e^{x} \]
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Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=\frac {125}{81} \, x^{4} - \frac {200}{9} \, x^{3} + \frac {290}{3} \, x^{2} - \frac {10}{9} \, {\left (5 \, x^{2} - 36 \, x\right )} e^{5} - \frac {10}{9} \, {\left (5 \, x^{2} - 36 \, x\right )} e - 120 \, x - e^{x} \]
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Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int \frac {1}{81} \left (-9720-81 e^x+e (3240-900 x)+e^5 (3240-900 x)+15660 x-5400 x^2+500 x^3\right ) \, dx=x\,\left (40\,\mathrm {e}+40\,{\mathrm {e}}^5-120\right )-x^2\,\left (\frac {50\,\mathrm {e}}{9}+\frac {50\,{\mathrm {e}}^5}{9}-\frac {290}{3}\right )-{\mathrm {e}}^x-\frac {200\,x^3}{9}+\frac {125\,x^4}{81} \]
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