Optimal. Leaf size=25 \[ e^{\left (3-e^{-x+x \left (26-x^2\right )}\right )^4 x} \]
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Rubi [F] time = 7.43, 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 \exp \left (81 x+e^{100 x-4 x^3} x-12 e^{75 x-3 x^3} x+54 e^{50 x-2 x^3} x-108 e^{25 x-x^3} x\right ) \left (81+e^{50 x-2 x^3} \left (54+2700 x-324 x^3\right )+e^{100 x-4 x^3} \left (1+100 x-12 x^3\right )+e^{75 x-3 x^3} \left (-12-900 x+108 x^3\right )+e^{25 x-x^3} \left (-108-2700 x+324 x^3\right )\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) \left (e^{25 x}-3 e^{x^3}\right )^3 \left (-3 e^{x^3}-e^{25 x} \left (-1-100 x+12 x^3\right )\right ) \, dx\\ &=\int \left (81 e^{e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x}+108 \exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right ) \left (-1-25 x+3 x^3\right )-54 \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right ) \left (-1-50 x+6 x^3\right )+12 \exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right ) \left (-1-75 x+9 x^3\right )-\exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) \left (-1-100 x+12 x^3\right )\right ) \, dx\\ &=12 \int \exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right ) \left (-1-75 x+9 x^3\right ) \, dx-54 \int \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right ) \left (-1-50 x+6 x^3\right ) \, dx+81 \int e^{e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x} \, dx+108 \int \exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right ) \left (-1-25 x+3 x^3\right ) \, dx-\int \exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) \left (-1-100 x+12 x^3\right ) \, dx\\ &=12 \int \left (-\exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right )-75 \exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right ) x+9 \exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right ) x^3\right ) \, dx-54 \int \left (-\exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right )-50 \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right ) x+6 \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right ) x^3\right ) \, dx+81 \int e^{e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x} \, dx+108 \int \left (-\exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right )-25 \exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right ) x+3 \exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right ) x^3\right ) \, dx-\int \left (-\exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right )-100 \exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) x+12 \exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) x^3\right ) \, dx\\ &=-\left (12 \int \exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right ) \, dx\right )-12 \int \exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) x^3 \, dx+54 \int \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right ) \, dx+81 \int e^{e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x} \, dx+100 \int \exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) x \, dx-108 \int \exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right ) \, dx+108 \int \exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right ) x^3 \, dx+324 \int \exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right ) x^3 \, dx-324 \int \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right ) x^3 \, dx-900 \int \exp \left (75 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-3 x^3\right ) x \, dx-2700 \int \exp \left (25 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-x^3\right ) x \, dx+2700 \int \exp \left (e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3+2 x \left (25+x^2\right )\right ) x \, dx+\int \exp \left (100 x+e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x-4 x^3\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 2.62, size = 26, normalized size = 1.04 \begin {gather*} e^{e^{-4 x^3} \left (e^{25 x}-3 e^{x^3}\right )^4 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.60, size = 56, normalized size = 2.24 \begin {gather*} e^{\left (-108 \, x e^{\left (-x^{3} + 25 \, x\right )} + 54 \, x e^{\left (-2 \, x^{3} + 50 \, x\right )} - 12 \, x e^{\left (-3 \, x^{3} + 75 \, x\right )} + x e^{\left (-4 \, x^{3} + 100 \, x\right )} + 81 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (108 \, {\left (3 \, x^{3} - 25 \, x - 1\right )} e^{\left (-x^{3} + 25 \, x\right )} - 54 \, {\left (6 \, x^{3} - 50 \, x - 1\right )} e^{\left (-2 \, x^{3} + 50 \, x\right )} + 12 \, {\left (9 \, x^{3} - 75 \, x - 1\right )} e^{\left (-3 \, x^{3} + 75 \, x\right )} - {\left (12 \, x^{3} - 100 \, x - 1\right )} e^{\left (-4 \, x^{3} + 100 \, x\right )} + 81\right )} e^{\left (-108 \, x e^{\left (-x^{3} + 25 \, x\right )} + 54 \, x e^{\left (-2 \, x^{3} + 50 \, x\right )} - 12 \, x e^{\left (-3 \, x^{3} + 75 \, x\right )} + x e^{\left (-4 \, x^{3} + 100 \, x\right )} + 81 \, x\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 52, normalized size = 2.08
method | result | size |
risch | \({\mathrm e}^{x \left ({\mathrm e}^{-4 x \left (x -5\right ) \left (5+x \right )}-12 \,{\mathrm e}^{-3 x \left (x -5\right ) \left (5+x \right )}+54 \,{\mathrm e}^{-2 x \left (x -5\right ) \left (5+x \right )}-108 \,{\mathrm e}^{-x \left (x -5\right ) \left (5+x \right )}+81\right )}\) | \(52\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (108 \, {\left (3 \, x^{3} - 25 \, x - 1\right )} e^{\left (-x^{3} + 25 \, x\right )} - 54 \, {\left (6 \, x^{3} - 50 \, x - 1\right )} e^{\left (-2 \, x^{3} + 50 \, x\right )} + 12 \, {\left (9 \, x^{3} - 75 \, x - 1\right )} e^{\left (-3 \, x^{3} + 75 \, x\right )} - {\left (12 \, x^{3} - 100 \, x - 1\right )} e^{\left (-4 \, x^{3} + 100 \, x\right )} + 81\right )} e^{\left (-108 \, x e^{\left (-x^{3} + 25 \, x\right )} + 54 \, x e^{\left (-2 \, x^{3} + 50 \, x\right )} - 12 \, x e^{\left (-3 \, x^{3} + 75 \, x\right )} + x e^{\left (-4 \, x^{3} + 100 \, x\right )} + 81 \, x\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.88, size = 60, normalized size = 2.40 \begin {gather*} {\mathrm {e}}^{-12\,x\,{\mathrm {e}}^{75\,x}\,{\mathrm {e}}^{-3\,x^3}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{100\,x}\,{\mathrm {e}}^{-4\,x^3}}\,{\mathrm {e}}^{54\,x\,{\mathrm {e}}^{50\,x}\,{\mathrm {e}}^{-2\,x^3}}\,{\mathrm {e}}^{-108\,x\,{\mathrm {e}}^{25\,x}\,{\mathrm {e}}^{-x^3}}\,{\mathrm {e}}^{81\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.71, size = 54, normalized size = 2.16 \begin {gather*} e^{x e^{- 4 x^{3} + 100 x} - 12 x e^{- 3 x^{3} + 75 x} + 54 x e^{- 2 x^{3} + 50 x} - 108 x e^{- x^{3} + 25 x} + 81 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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