Optimal. Leaf size=29 \[ x-\frac {4 x^2}{\left (\frac {e^{\frac {x}{5-e}} (-4+x)}{x}+x\right )^2} \]
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Rubi [F] time = 4.08, 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 {-5 x^6+e x^6+e^{-\frac {3 x}{-5+e}} \left (320-240 x+60 x^2-5 x^3+e \left (-64+48 x-12 x^2+x^3\right )\right )+e^{-\frac {2 x}{-5+e}} \left (-240 x^2+120 x^3-15 x^4+e \left (48 x^2-24 x^3+3 x^4\right )\right )+e^{-\frac {x}{-5+e}} \left (-320 x^3+132 x^4-23 x^5+e \left (64 x^3-20 x^4+3 x^5\right )\right )}{-5 x^6+e x^6+e^{-\frac {3 x}{-5+e}} \left (320-240 x+60 x^2-5 x^3+e \left (-64+48 x-12 x^2+x^3\right )\right )+e^{-\frac {2 x}{-5+e}} \left (-240 x^2+120 x^3-15 x^4+e \left (48 x^2-24 x^3+3 x^4\right )\right )+e^{-\frac {x}{-5+e}} \left (60 x^4-15 x^5+e \left (-12 x^4+3 x^5\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(-5+e) x^6+e^{-\frac {3 x}{-5+e}} \left (320-240 x+60 x^2-5 x^3+e \left (-64+48 x-12 x^2+x^3\right )\right )+e^{-\frac {2 x}{-5+e}} \left (-240 x^2+120 x^3-15 x^4+e \left (48 x^2-24 x^3+3 x^4\right )\right )+e^{-\frac {x}{-5+e}} \left (-320 x^3+132 x^4-23 x^5+e \left (64 x^3-20 x^4+3 x^5\right )\right )}{-5 x^6+e x^6+e^{-\frac {3 x}{-5+e}} \left (320-240 x+60 x^2-5 x^3+e \left (-64+48 x-12 x^2+x^3\right )\right )+e^{-\frac {2 x}{-5+e}} \left (-240 x^2+120 x^3-15 x^4+e \left (48 x^2-24 x^3+3 x^4\right )\right )+e^{-\frac {x}{-5+e}} \left (60 x^4-15 x^5+e \left (-12 x^4+3 x^5\right )\right )} \, dx\\ &=\int \frac {(-5+e) x^6+e^{-\frac {3 x}{-5+e}} \left (320-240 x+60 x^2-5 x^3+e \left (-64+48 x-12 x^2+x^3\right )\right )+e^{-\frac {2 x}{-5+e}} \left (-240 x^2+120 x^3-15 x^4+e \left (48 x^2-24 x^3+3 x^4\right )\right )+e^{-\frac {x}{-5+e}} \left (-320 x^3+132 x^4-23 x^5+e \left (64 x^3-20 x^4+3 x^5\right )\right )}{(-5+e) x^6+e^{-\frac {3 x}{-5+e}} \left (320-240 x+60 x^2-5 x^3+e \left (-64+48 x-12 x^2+x^3\right )\right )+e^{-\frac {2 x}{-5+e}} \left (-240 x^2+120 x^3-15 x^4+e \left (48 x^2-24 x^3+3 x^4\right )\right )+e^{-\frac {x}{-5+e}} \left (60 x^4-15 x^5+e \left (-12 x^4+3 x^5\right )\right )} \, dx\\ &=\int \frac {-5 \left (1-\frac {e}{5}\right ) (-4+x)^3-15 \left (1-\frac {e}{5}\right ) e^{\frac {x}{-5+e}} (-4+x)^2 x^2-5 \left (1-\frac {e}{5}\right ) e^{\frac {3 x}{-5+e}} x^6+e^{1+\frac {2 x}{-5+e}} x^3 \left (64-20 x+3 x^2\right )-e^{\frac {2 x}{-5+e}} x^3 \left (320-132 x+23 x^2\right )}{(5-e) \left (4-x-e^{\frac {x}{-5+e}} x^2\right )^3} \, dx\\ &=\frac {\int \frac {-5 \left (1-\frac {e}{5}\right ) (-4+x)^3-15 \left (1-\frac {e}{5}\right ) e^{\frac {x}{-5+e}} (-4+x)^2 x^2-5 \left (1-\frac {e}{5}\right ) e^{\frac {3 x}{-5+e}} x^6+e^{1+\frac {2 x}{-5+e}} x^3 \left (64-20 x+3 x^2\right )-e^{\frac {2 x}{-5+e}} x^3 \left (320-132 x+23 x^2\right )}{\left (4-x-e^{\frac {x}{-5+e}} x^2\right )^3} \, dx}{5-e}\\ &=\frac {\int \left (5 \left (1-\frac {e}{5}\right )+\frac {8 (4-x)^2 \left (-8 (5-e)+(9-e) x-x^2\right )}{x \left (4-x-e^{\frac {x}{-5+e}} x^2\right )^3}+\frac {16 (4-x) \left (8 (5-e)-(9-e) x+x^2\right )}{x \left (4-x-e^{\frac {x}{-5+e}} x^2\right )^2}+\frac {8 \left (-8 (5-e)+(9-e) x-x^2\right )}{x \left (4-x-e^{\frac {x}{-5+e}} x^2\right )}\right ) \, dx}{5-e}\\ &=x+\frac {8 \int \frac {(4-x)^2 \left (-8 (5-e)+(9-e) x-x^2\right )}{x \left (4-x-e^{\frac {x}{-5+e}} x^2\right )^3} \, dx}{5-e}+\frac {8 \int \frac {-8 (5-e)+(9-e) x-x^2}{x \left (4-x-e^{\frac {x}{-5+e}} x^2\right )} \, dx}{5-e}+\frac {16 \int \frac {(4-x) \left (8 (5-e)-(9-e) x+x^2\right )}{x \left (4-x-e^{\frac {x}{-5+e}} x^2\right )^2} \, dx}{5-e}\\ &=x+\frac {8 \int \left (\frac {16 (-29+5 e)}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3}-\frac {128 (-5+e)}{x \left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3}-\frac {16 (-8+e) x}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3}+\frac {(-17+e) x^2}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3}+\frac {x^3}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3}\right ) \, dx}{5-e}+\frac {8 \int \left (-\frac {9 \left (1-\frac {e}{9}\right )}{-4+x+e^{\frac {x}{-5+e}} x^2}-\frac {8 (-5+e)}{x \left (-4+x+e^{\frac {x}{-5+e}} x^2\right )}+\frac {x}{-4+x+e^{\frac {x}{-5+e}} x^2}\right ) \, dx}{5-e}+\frac {16 \int \left (\frac {4 (-19+3 e)}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2}-\frac {32 (-5+e)}{x \left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2}-\frac {(-13+e) x}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2}-\frac {x^2}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2}\right ) \, dx}{5-e}\\ &=x+64 \int \frac {1}{x \left (-4+x+e^{\frac {x}{-5+e}} x^2\right )} \, dx+512 \int \frac {1}{x \left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2} \, dx+1024 \int \frac {1}{x \left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3} \, dx+\frac {8 \int \frac {x^3}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3} \, dx}{5-e}+\frac {8 \int \frac {x}{-4+x+e^{\frac {x}{-5+e}} x^2} \, dx}{5-e}-\frac {16 \int \frac {x^2}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2} \, dx}{5-e}-\frac {(128 (29-5 e)) \int \frac {1}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3} \, dx}{5-e}-\frac {(64 (19-3 e)) \int \frac {1}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2} \, dx}{5-e}+\frac {(128 (8-e)) \int \frac {x}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3} \, dx}{5-e}-\frac {(8 (9-e)) \int \frac {1}{-4+x+e^{\frac {x}{-5+e}} x^2} \, dx}{5-e}+\frac {(16 (13-e)) \int \frac {x}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2} \, dx}{5-e}-\frac {(8 (17-e)) \int \frac {x^2}{\left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^3} \, dx}{5-e}\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 2.15, size = 576, normalized size = 19.86 \begin {gather*} \frac {e^{-\frac {20}{-5+e}} (-4+x) x \left (2 e^{\frac {4 e}{-5+e}} (-8+x)^3 (-4+x)+4 e^{\frac {4 e+x}{-5+e}} (-8+x)^3 x^2+2 e^{\frac {2 (2 e+x)}{-5+e}} (-8+x)^3 x^3+4 e^{\frac {15+e+x}{-5+e}} x^2 \left (160-24 x+x^2\right ) \left (40-9 x+x^2\right )^2+2 e^{\frac {15+e+2 x}{-5+e}} x^3 \left (160-24 x+x^2\right ) \left (40-9 x+x^2\right )^2-10 e^{\frac {20}{-5+e}} (-4+x) \left (40-9 x+x^2\right )^3-20 e^{\frac {20+x}{-5+e}} x^2 \left (40-9 x+x^2\right )^3-10 e^{\frac {2 (10+x)}{-5+e}} x^3 \left (40-9 x+x^2\right )^3+4 e^{\frac {5+3 e+x}{-5+e}} (-8+x)^2 x^2 \left (160-32 x+3 x^2\right )+2 e^{\frac {5+3 e+2 x}{-5+e}} (-8+x)^2 x^3 \left (160-32 x+3 x^2\right )+2 e^{\frac {15+e}{-5+e}} \left (40-9 x+x^2\right )^2 \left (-640+256 x-28 x^2+x^3\right )-8 e^{\frac {5+3 e}{-5+e}} \left (4416-3104 x+864 x^2-111 x^3+5 x^4\right )+12 e^{\frac {10+2 e+x}{-5+e}} x^2 \left (-25600+13440 x-3248 x^2+430 x^3-31 x^4+x^5\right )+6 e^{\frac {2 (5+e+x)}{-5+e}} x^3 \left (-25600+13440 x-3248 x^2+430 x^3-31 x^4+x^5\right )+2 e^{3+\frac {20}{-5+e}} \left (-23296+16256 x-4608 x^2+740 x^3-72 x^4+3 x^5\right )+6 e^{\frac {2 (5+e)}{-5+e}} \left (102400-79360 x+26432 x^2-4968 x^3+554 x^4-35 x^5+x^6\right )\right )}{2 (-5+e) \left (40+e (-8+x)-9 x+x^2\right )^3 \left (-4+x+e^{\frac {x}{-5+e}} x^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.53, size = 100, normalized size = 3.45 \begin {gather*} \frac {x^{5} - 4 \, x^{4} + 2 \, {\left (x^{4} - 4 \, x^{3}\right )} e^{\left (-\frac {x}{e - 5}\right )} + {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} e^{\left (-\frac {2 \, x}{e - 5}\right )}}{x^{4} + 2 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{\left (-\frac {x}{e - 5}\right )} + {\left (x^{2} - 8 \, x + 16\right )} e^{\left (-\frac {2 \, x}{e - 5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 5.01, size = 108, normalized size = 3.72 \begin {gather*} \frac {x^{5} e^{\left (\frac {2 \, x}{e - 5}\right )} + 2 \, x^{4} e^{\left (\frac {x}{e - 5}\right )} + x^{3} - 32 \, x^{2} e^{\left (\frac {x}{e - 5}\right )} - 4 \, x^{2} - 16 \, x + 64}{x^{4} e^{\left (\frac {2 \, x}{e - 5}\right )} + 2 \, x^{3} e^{\left (\frac {x}{e - 5}\right )} - 8 \, x^{2} e^{\left (\frac {x}{e - 5}\right )} + x^{2} - 8 \, x + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 5.74, size = 38, normalized size = 1.31
method | result | size |
risch | \(x -\frac {4 x^{4}}{\left (x^{2}+x \,{\mathrm e}^{-\frac {x}{{\mathrm e}-5}}-4 \,{\mathrm e}^{-\frac {x}{{\mathrm e}-5}}\right )^{2}}\) | \(38\) |
norman | \(\frac {x^{5}+64 \,{\mathrm e}^{-\frac {2 x}{{\mathrm e}-5}}+{\mathrm e}^{-\frac {2 x}{{\mathrm e}-5}} x^{3}-32 x^{2} {\mathrm e}^{-\frac {x}{{\mathrm e}-5}}-16 x \,{\mathrm e}^{-\frac {2 x}{{\mathrm e}-5}}-4 x^{2} {\mathrm e}^{-\frac {2 x}{{\mathrm e}-5}}+2 x^{4} {\mathrm e}^{-\frac {x}{{\mathrm e}-5}}}{\left (x^{2}+x \,{\mathrm e}^{-\frac {x}{{\mathrm e}-5}}-4 \,{\mathrm e}^{-\frac {x}{{\mathrm e}-5}}\right )^{2}}\) | \(128\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.67, size = 92, normalized size = 3.17 \begin {gather*} \frac {x^{5} e^{\left (\frac {2 \, x}{e - 5}\right )} + x^{3} - 4 \, x^{2} + 2 \, {\left (x^{4} - 16 \, x^{2}\right )} e^{\left (\frac {x}{e - 5}\right )} - 16 \, x + 64}{x^{4} e^{\left (\frac {2 \, x}{e - 5}\right )} + x^{2} + 2 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{\left (\frac {x}{e - 5}\right )} - 8 \, x + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.93, size = 59, normalized size = 2.03 \begin {gather*} \frac {x\,\left (x-4\right )\,\left (x+x^3\,{\mathrm {e}}^{\frac {2\,x}{\mathrm {e}-5}}-4\right )+2\,x^3\,{\mathrm {e}}^{\frac {x}{\mathrm {e}-5}}\,\left (x-4\right )}{{\left (x+x^2\,{\mathrm {e}}^{\frac {x}{\mathrm {e}-5}}-4\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.58, size = 46, normalized size = 1.59 \begin {gather*} - \frac {4 x^{4}}{x^{4} + \left (2 x^{3} - 8 x^{2}\right ) e^{- \frac {x}{-5 + e}} + \left (x^{2} - 8 x + 16\right ) e^{- \frac {2 x}{-5 + e}}} + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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