Optimal. Leaf size=28 \[ 5 x \left (\frac {3 x}{4}+\frac {1}{4} e^{-x} x^2 \left (e^5+2 x\right )\right ) \]
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Rubi [B] time = 0.53, antiderivative size = 163, normalized size of antiderivative = 5.82, number of steps used = 20, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 6741, 6688, 6742, 2196, 2176, 2194} \begin {gather*} \frac {5}{2} e^{-x} x^4+10 e^{-x} x^3-\frac {5}{4} \left (8-e^5\right ) e^{-x} x^3-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} \left (8-e^5\right ) e^{-x} x^2+\frac {15 x^2}{4}-\frac {15}{2} e^{5-x} x+60 e^{-x} x-\frac {15}{2} \left (8-e^5\right ) e^{-x} x-\frac {15 e^{5-x}}{2}+60 e^{-x}-\frac {15}{2} \left (8-e^5\right ) e^{-x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2194
Rule 2196
Rule 6688
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx\\ &=\frac {1}{4} \int 5 e^{-x} x \left (6 e^x+3 e^5 x+8 \left (1-\frac {e^5}{8}\right ) x^2-2 x^3\right ) \, dx\\ &=\frac {5}{4} \int e^{-x} x \left (6 e^x+3 e^5 x+8 \left (1-\frac {e^5}{8}\right ) x^2-2 x^3\right ) \, dx\\ &=\frac {5}{4} \int e^{-x} x \left (6 e^x-e^5 (-3+x) x-2 (-4+x) x^2\right ) \, dx\\ &=\frac {5}{4} \int \left (6 x+e^{-x} x^2 \left (3 e^5+\left (8-e^5\right ) x-2 x^2\right )\right ) \, dx\\ &=\frac {15 x^2}{4}+\frac {5}{4} \int e^{-x} x^2 \left (3 e^5+\left (8-e^5\right ) x-2 x^2\right ) \, dx\\ &=\frac {15 x^2}{4}+\frac {5}{4} \int \left (3 e^{5-x} x^2-e^{-x} \left (-8+e^5\right ) x^3-2 e^{-x} x^4\right ) \, dx\\ &=\frac {15 x^2}{4}-\frac {5}{2} \int e^{-x} x^4 \, dx+\frac {15}{4} \int e^{5-x} x^2 \, dx+\frac {1}{4} \left (5 \left (8-e^5\right )\right ) \int e^{-x} x^3 \, dx\\ &=\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4+\frac {15}{2} \int e^{5-x} x \, dx-10 \int e^{-x} x^3 \, dx+\frac {1}{4} \left (15 \left (8-e^5\right )\right ) \int e^{-x} x^2 \, dx\\ &=-\frac {15}{2} e^{5-x} x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4+\frac {15}{2} \int e^{5-x} \, dx-30 \int e^{-x} x^2 \, dx+\frac {1}{2} \left (15 \left (8-e^5\right )\right ) \int e^{-x} x \, dx\\ &=-\frac {15 e^{5-x}}{2}-\frac {15}{2} e^{5-x} x-\frac {15}{2} e^{-x} \left (8-e^5\right ) x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4-60 \int e^{-x} x \, dx+\frac {1}{2} \left (15 \left (8-e^5\right )\right ) \int e^{-x} \, dx\\ &=-\frac {15 e^{5-x}}{2}-\frac {15}{2} e^{-x} \left (8-e^5\right )-\frac {15}{2} e^{5-x} x+60 e^{-x} x-\frac {15}{2} e^{-x} \left (8-e^5\right ) x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4-60 \int e^{-x} \, dx\\ &=-\frac {15 e^{5-x}}{2}+60 e^{-x}-\frac {15}{2} e^{-x} \left (8-e^5\right )-\frac {15}{2} e^{5-x} x+60 e^{-x} x-\frac {15}{2} e^{-x} \left (8-e^5\right ) x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 29, normalized size = 1.04 \begin {gather*} \frac {5}{4} \left (3 x^2+e^{-x} \left (e^5 x^3+2 x^4\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 25, normalized size = 0.89 \begin {gather*} \frac {5}{4} \, {\left (2 \, x^{4} + x^{3} e^{5} + 3 \, x^{2} e^{x}\right )} e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 26, normalized size = 0.93 \begin {gather*} \frac {5}{2} \, x^{4} e^{\left (-x\right )} + \frac {5}{4} \, x^{3} e^{\left (-x + 5\right )} + \frac {15}{4} \, x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 26, normalized size = 0.93
method | result | size |
norman | \(\left (\frac {5 x^{4}}{2}+\frac {5 x^{3} {\mathrm e}^{5}}{4}+\frac {15 \,{\mathrm e}^{x} x^{2}}{4}\right ) {\mathrm e}^{-x}\) | \(26\) |
risch | \(\frac {15 x^{2}}{4}+\frac {\left (5 x^{3} {\mathrm e}^{5}+10 x^{4}\right ) {\mathrm e}^{-x}}{4}\) | \(26\) |
default | \(\frac {15 x^{2}}{4}+\frac {5 x^{4} {\mathrm e}^{-x}}{2}+\frac {15 \,{\mathrm e}^{5} \left (-x^{2} {\mathrm e}^{-x}-2 x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}\right )}{4}-\frac {5 \,{\mathrm e}^{5} \left (-x^{3} {\mathrm e}^{-x}-3 x^{2} {\mathrm e}^{-x}-6 x \,{\mathrm e}^{-x}-6 \,{\mathrm e}^{-x}\right )}{4}\) | \(79\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.38, size = 100, normalized size = 3.57 \begin {gather*} \frac {15}{4} \, x^{2} + \frac {5}{2} \, {\left (x^{4} + 4 \, x^{3} + 12 \, x^{2} + 24 \, x + 24\right )} e^{\left (-x\right )} + \frac {5}{4} \, {\left (x^{3} e^{5} + 3 \, x^{2} e^{5} + 6 \, x e^{5} + 6 \, e^{5}\right )} e^{\left (-x\right )} - 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x + 6\right )} e^{\left (-x\right )} - \frac {15}{4} \, {\left (x^{2} e^{5} + 2 \, x e^{5} + 2 \, e^{5}\right )} e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.13, size = 26, normalized size = 0.93 \begin {gather*} \frac {5\,x^4\,{\mathrm {e}}^{-x}}{2}+\frac {15\,x^2}{4}+\frac {5\,x^3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 24, normalized size = 0.86 \begin {gather*} \frac {15 x^{2}}{4} + \frac {\left (10 x^{4} + 5 x^{3} e^{5}\right ) e^{- x}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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