Optimal. Leaf size=27 \[ e^{\frac {1}{16} e^x \left (-3+3 e^{-4+x}-x\right ) x (5+x)^2} \]
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Rubi [F] time = 6.89, 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 {1}{16} \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) \left (-75-185 x-94 x^2-17 x^3-x^4+e^{-4+x} \left (75+210 x+69 x^2+6 x^3\right )\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) \left (-75-185 x-94 x^2-17 x^3-x^4+e^{-4+x} \left (75+210 x+69 x^2+6 x^3\right )\right ) \, dx\\ &=\frac {1}{16} \int \left (-75 \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right )-185 \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x-94 \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^2-17 \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^3-\exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^4+3 \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) \left (25+70 x+23 x^2+2 x^3\right )\right ) \, dx\\ &=-\left (\frac {1}{16} \int \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^4 \, dx\right )+\frac {3}{16} \int \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) \left (25+70 x+23 x^2+2 x^3\right ) \, dx-\frac {17}{16} \int \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^3 \, dx-\frac {75}{16} \int \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) \, dx-\frac {47}{8} \int \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^2 \, dx-\frac {185}{16} \int \exp \left (x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x \, dx\\ &=-\left (\frac {1}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x^4 \, dx\right )+\frac {3}{16} \int \left (25 \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right )+70 \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x+23 \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^2+2 \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^3\right ) \, dx-\frac {17}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x^3 \, dx-\frac {75}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) \, dx-\frac {47}{8} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x^2 \, dx-\frac {185}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x \, dx\\ &=-\left (\frac {1}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x^4 \, dx\right )+\frac {3}{8} \int \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^3 \, dx-\frac {17}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x^3 \, dx+\frac {69}{16} \int \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x^2 \, dx-\frac {75}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) \, dx+\frac {75}{16} \int \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) \, dx-\frac {47}{8} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x^2 \, dx-\frac {185}{16} \int \exp \left (\frac {1}{16} x \left (16+3 e^{-4+2 x} (5+x)^2-e^x (3+x) (5+x)^2\right )\right ) x \, dx+\frac {105}{8} \int \exp \left (-4+2 x+\frac {1}{16} e^x \left (-75 x-55 x^2-13 x^3-x^4+e^{-4+x} \left (75 x+30 x^2+3 x^3\right )\right )\right ) x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.22, size = 30, normalized size = 1.11 \begin {gather*} e^{-\frac {1}{16} e^{-4+x} x (5+x)^2 \left (-3 e^x+e^4 (3+x)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.85, size = 56, normalized size = 2.07 \begin {gather*} e^{\left (\frac {1}{16} \, {\left (16 \, x e^{4} + 3 \, {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} - {\left (x^{4} + 13 \, x^{3} + 55 \, x^{2} + 75 \, x\right )} e^{\left (x + 4\right )}\right )} e^{\left (-4\right )} - 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 -\frac {1}{16} \, {\left (x^{4} + 17 \, x^{3} + 94 \, x^{2} - 3 \, {\left (2 \, x^{3} + 23 \, x^{2} + 70 \, x + 25\right )} e^{\left (x - 4\right )} + 185 \, x + 75\right )} e^{\left (-\frac {1}{16} \, {\left (x^{4} + 13 \, x^{3} + 55 \, x^{2} - 3 \, {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{\left (x - 4\right )} + 75 \, x\right )} e^{x} + x\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 21, normalized size = 0.78
method | result | size |
risch | \({\mathrm e}^{-\frac {x \left (5+x \right )^{2} \left (-3 \,{\mathrm e}^{x -4}+3+x \right ) {\mathrm e}^{x}}{16}}\) | \(21\) |
norman | \({\mathrm e}^{\frac {\left (\left (3 x^{3}+30 x^{2}+75 x \right ) {\mathrm e}^{x} {\mathrm e}^{-4}-x^{4}-13 x^{3}-55 x^{2}-75 x \right ) {\mathrm e}^{x}}{16}}\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.08, size = 59, normalized size = 2.19 \begin {gather*} e^{\left (-\frac {1}{16} \, x^{4} e^{x} + \frac {3}{16} \, x^{3} e^{\left (2 \, x - 4\right )} - \frac {13}{16} \, x^{3} e^{x} + \frac {15}{8} \, x^{2} e^{\left (2 \, x - 4\right )} - \frac {55}{16} \, x^{2} e^{x} + \frac {75}{16} \, x e^{\left (2 \, x - 4\right )} - \frac {75}{16} \, x e^{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.89, size = 65, normalized size = 2.41 \begin {gather*} {\mathrm {e}}^{-\frac {75\,x\,{\mathrm {e}}^x}{16}}\,{\mathrm {e}}^{\frac {3\,x^3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-4}}{16}}\,{\mathrm {e}}^{\frac {15\,x^2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-4}}{8}}\,{\mathrm {e}}^{-\frac {x^4\,{\mathrm {e}}^x}{16}}\,{\mathrm {e}}^{-\frac {13\,x^3\,{\mathrm {e}}^x}{16}}\,{\mathrm {e}}^{-\frac {55\,x^2\,{\mathrm {e}}^x}{16}}\,{\mathrm {e}}^{\frac {75\,x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-4}}{16}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.52, size = 49, normalized size = 1.81 \begin {gather*} e^{\left (- \frac {x^{4}}{16} - \frac {13 x^{3}}{16} - \frac {55 x^{2}}{16} - \frac {75 x}{16} + \frac {\left (3 x^{3} + 30 x^{2} + 75 x\right ) e^{x}}{16 e^{4}}\right ) e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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