Optimal. Leaf size=28 \[ \left (5-5 e^{\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}}+2 x\right )^2 \]
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Rubi [F] time = 1.31, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{16-8 x+x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {320-32 x-44 x^2+8 x^3+e^{20+\frac {2 e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-40-80 x+10 x^2\right )+e^{\frac {e^{20} \left (4+x^2\right )}{-20+5 x}} \left (-320+160 x-20 x^2+e^{20} \left (40+96 x+22 x^2-4 x^3\right )\right )}{(-4+x)^2} \, dx\\ &=\int \left (\frac {320}{(-4+x)^2}-\frac {32 x}{(-4+x)^2}-\frac {44 x^2}{(-4+x)^2}+\frac {8 x^3}{(-4+x)^2}+\frac {10 e^{20+\frac {2 e^{20} \left (4+x^2\right )}{5 (-4+x)}} \left (-4-8 x+x^2\right )}{(-4+x)^2}+\frac {2 e^{\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}} \left (-20 \left (8-e^{20}\right )+16 \left (5+3 e^{20}\right ) x-\left (10-11 e^{20}\right ) x^2-2 e^{20} x^3\right )}{(4-x)^2}\right ) \, dx\\ &=\frac {320}{4-x}+2 \int \frac {e^{\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}} \left (-20 \left (8-e^{20}\right )+16 \left (5+3 e^{20}\right ) x-\left (10-11 e^{20}\right ) x^2-2 e^{20} x^3\right )}{(4-x)^2} \, dx+8 \int \frac {x^3}{(-4+x)^2} \, dx+10 \int \frac {e^{20+\frac {2 e^{20} \left (4+x^2\right )}{5 (-4+x)}} \left (-4-8 x+x^2\right )}{(-4+x)^2} \, dx-32 \int \frac {x}{(-4+x)^2} \, dx-44 \int \frac {x^2}{(-4+x)^2} \, dx\\ &=\frac {320}{4-x}+2 \int \left (-5 e^{\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}} \left (2+e^{20}\right )+\frac {260 e^{20+\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}}}{(-4+x)^2}+\frac {40 e^{20+\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}}}{-4+x}-2 e^{20+\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}} x\right ) \, dx+8 \int \left (8+\frac {64}{(-4+x)^2}+\frac {48}{-4+x}+x\right ) \, dx+10 \int \left (e^{20+\frac {2 e^{20} \left (4+x^2\right )}{5 (-4+x)}}-\frac {20 e^{20+\frac {2 e^{20} \left (4+x^2\right )}{5 (-4+x)}}}{(-4+x)^2}\right ) \, dx-32 \int \left (\frac {4}{(-4+x)^2}+\frac {1}{-4+x}\right ) \, dx-44 \int \left (1+\frac {16}{(-4+x)^2}+\frac {8}{-4+x}\right ) \, dx\\ &=20 x+4 x^2-4 \int e^{20+\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}} x \, dx+10 \int e^{20+\frac {2 e^{20} \left (4+x^2\right )}{5 (-4+x)}} \, dx+80 \int \frac {e^{20+\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}}}{-4+x} \, dx-200 \int \frac {e^{20+\frac {2 e^{20} \left (4+x^2\right )}{5 (-4+x)}}}{(-4+x)^2} \, dx+520 \int \frac {e^{20+\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}}}{(-4+x)^2} \, dx-\left (10 \left (2+e^{20}\right )\right ) \int e^{\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.10, size = 58, normalized size = 2.07 \begin {gather*} 25 e^{\frac {2 e^{20} \left (4+x^2\right )}{5 (-4+x)}}-10 e^{\frac {e^{20} \left (4+x^2\right )}{5 (-4+x)}} (5+2 x)+4 \left (-36+5 x+x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 48, normalized size = 1.71 \begin {gather*} 4 \, x^{2} - 10 \, {\left (2 \, x + 5\right )} e^{\left (\frac {{\left (x^{2} + 4\right )} e^{20}}{5 \, {\left (x - 4\right )}}\right )} + 20 \, x + 25 \, e^{\left (\frac {2 \, {\left (x^{2} + 4\right )} e^{20}}{5 \, {\left (x - 4\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.32, size = 98, normalized size = 3.50 \begin {gather*} {\left (4 \, x^{2} e^{20} + 20 \, x e^{20} - 20 \, x e^{\left (\frac {x^{2} e^{20} + x e^{20}}{5 \, {\left (x - 4\right )}} - \frac {1}{5} \, e^{20} + 20\right )} + 25 \, e^{\left (\frac {2 \, {\left (x^{2} e^{20} + x e^{20}\right )}}{5 \, {\left (x - 4\right )}} - \frac {2}{5} \, e^{20} + 20\right )} - 50 \, e^{\left (\frac {x^{2} e^{20} + x e^{20}}{5 \, {\left (x - 4\right )}} - \frac {1}{5} \, e^{20} + 20\right )}\right )} e^{\left (-20\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 48, normalized size = 1.71
method | result | size |
risch | \(4 x^{2}+25 \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 \left (x -4\right )}}+20 x +\left (-50-20 x \right ) {\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}\) | \(48\) |
norman | \(\frac {4 x^{2}+4 x^{3}-100 \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+30 x \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+25 x \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}-20 x^{2} {\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}+200 \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{20}}{5 x -20}}-320}{x -4}\) | \(118\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.47, size = 65, normalized size = 2.32 \begin {gather*} 4 \, x^{2} - 10 \, {\left (2 \, x e^{\left (\frac {4}{5} \, e^{20}\right )} + 5 \, e^{\left (\frac {4}{5} \, e^{20}\right )}\right )} e^{\left (\frac {1}{5} \, x e^{20} + \frac {4 \, e^{20}}{x - 4}\right )} + 20 \, x + 25 \, e^{\left (\frac {2}{5} \, x e^{20} + \frac {8 \, e^{20}}{x - 4} + \frac {8}{5} \, e^{20}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.31, size = 96, normalized size = 3.43 \begin {gather*} 25\,{\mathrm {e}}^{\frac {8\,{\mathrm {e}}^{20}}{5\,x-20}+\frac {2\,x^2\,{\mathrm {e}}^{20}}{5\,x-20}}-50\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{20}}{5\,x-20}+\frac {x^2\,{\mathrm {e}}^{20}}{5\,x-20}}+4\,x^2-x\,\left (20\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{20}}{5\,x-20}+\frac {x^2\,{\mathrm {e}}^{20}}{5\,x-20}}-20\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.45, size = 48, normalized size = 1.71 \begin {gather*} 4 x^{2} + 20 x + \left (- 20 x - 50\right ) e^{\frac {\left (x^{2} + 4\right ) e^{20}}{5 x - 20}} + 25 e^{\frac {2 \left (x^{2} + 4\right ) e^{20}}{5 x - 20}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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