Optimal. Leaf size=31 \[ e^{2 x} \left (4-x-\frac {1}{4} \left (-3+e^5-x\right ) x\right ) \left (5-x^2\right ) \]
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Rubi [B] time = 0.29, antiderivative size = 80, normalized size of antiderivative = 2.58, number of steps used = 30, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2196, 2194, 2176} \begin {gather*} -\frac {1}{4} e^{2 x} x^4+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{2 x+5} x^3-\frac {11}{4} e^{2 x} x^2-\frac {5}{4} e^{2 x} x-\frac {5}{4} e^{2 x+5} x+20 e^{2 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2194
Rule 2196
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx\\ &=\frac {1}{4} \int \left (155 e^{2 x}-32 e^{2 x} x-19 e^{2 x} x^2-2 e^{2 x} x^3-2 e^{2 x} x^4+e^{5+2 x} \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx\\ &=\frac {1}{4} \int e^{5+2 x} \left (-5-10 x+3 x^2+2 x^3\right ) \, dx-\frac {1}{2} \int e^{2 x} x^3 \, dx-\frac {1}{2} \int e^{2 x} x^4 \, dx-\frac {19}{4} \int e^{2 x} x^2 \, dx-8 \int e^{2 x} x \, dx+\frac {155}{4} \int e^{2 x} \, dx\\ &=\frac {155 e^{2 x}}{8}-4 e^{2 x} x-\frac {19}{8} e^{2 x} x^2-\frac {1}{4} e^{2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {1}{4} \int \left (-5 e^{5+2 x}-10 e^{5+2 x} x+3 e^{5+2 x} x^2+2 e^{5+2 x} x^3\right ) \, dx+\frac {3}{4} \int e^{2 x} x^2 \, dx+4 \int e^{2 x} \, dx+\frac {19}{4} \int e^{2 x} x \, dx+\int e^{2 x} x^3 \, dx\\ &=\frac {171 e^{2 x}}{8}-\frac {13}{8} e^{2 x} x-2 e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {1}{2} \int e^{5+2 x} x^3 \, dx-\frac {3}{4} \int e^{2 x} x \, dx+\frac {3}{4} \int e^{5+2 x} x^2 \, dx-\frac {5}{4} \int e^{5+2 x} \, dx-\frac {3}{2} \int e^{2 x} x^2 \, dx-\frac {19}{8} \int e^{2 x} \, dx-\frac {5}{2} \int e^{5+2 x} x \, dx\\ &=\frac {323 e^{2 x}}{16}-\frac {5}{8} e^{5+2 x}-2 e^{2 x} x-\frac {5}{4} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {3}{8} e^{5+2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {3}{8} \int e^{2 x} \, dx-\frac {3}{4} \int e^{5+2 x} x \, dx-\frac {3}{4} \int e^{5+2 x} x^2 \, dx+\frac {5}{4} \int e^{5+2 x} \, dx+\frac {3}{2} \int e^{2 x} x \, dx\\ &=\frac {163 e^{2 x}}{8}-\frac {5}{4} e^{2 x} x-\frac {13}{8} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {3}{8} \int e^{5+2 x} \, dx-\frac {3}{4} \int e^{2 x} \, dx+\frac {3}{4} \int e^{5+2 x} x \, dx\\ &=20 e^{2 x}+\frac {3}{16} e^{5+2 x}-\frac {5}{4} e^{2 x} x-\frac {5}{4} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4-\frac {3}{8} \int e^{5+2 x} \, dx\\ &=20 e^{2 x}-\frac {5}{4} e^{2 x} x-\frac {5}{4} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 38, normalized size = 1.23 \begin {gather*} \frac {1}{4} e^{2 x} \left (80-5 \left (1+e^5\right ) x-11 x^2+\left (1+e^5\right ) x^3-x^4\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 35, normalized size = 1.13 \begin {gather*} -\frac {1}{4} \, {\left (x^{4} - x^{3} + 11 \, x^{2} - {\left (x^{3} - 5 \, x\right )} e^{5} + 5 \, x - 80\right )} e^{\left (2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 40, normalized size = 1.29 \begin {gather*} -\frac {1}{4} \, {\left (x^{4} - x^{3} + 11 \, x^{2} + 5 \, x - 80\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (x^{3} - 5 \, x\right )} e^{\left (2 \, x + 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 36, normalized size = 1.16
method | result | size |
gosper | \(\frac {{\mathrm e}^{2 x} \left (x^{3} {\mathrm e}^{5}-x^{4}+x^{3}-5 x \,{\mathrm e}^{5}-11 x^{2}-5 x +80\right )}{4}\) | \(36\) |
risch | \(\frac {{\mathrm e}^{2 x} \left (x^{3} {\mathrm e}^{5}-x^{4}+x^{3}-5 x \,{\mathrm e}^{5}-11 x^{2}-5 x +80\right )}{4}\) | \(36\) |
norman | \(\left (-\frac {5}{4}-\frac {5 \,{\mathrm e}^{5}}{4}\right ) x \,{\mathrm e}^{2 x}+\left (\frac {{\mathrm e}^{5}}{4}+\frac {1}{4}\right ) x^{3} {\mathrm e}^{2 x}+20 \,{\mathrm e}^{2 x}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}\) | \(52\) |
meijerg | \(-19+\frac {155 \,{\mathrm e}^{2 x}}{8}+\frac {\left (2 \,{\mathrm e}^{5}-2\right ) \left (6-\frac {\left (-32 x^{3}+48 x^{2}-48 x +24\right ) {\mathrm e}^{2 x}}{4}\right )}{64}-\frac {\left (3 \,{\mathrm e}^{5}-19\right ) \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )}{32}+\frac {\left (-10 \,{\mathrm e}^{5}-32\right ) \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{16}-\frac {\left (80 x^{4}-160 x^{3}+240 x^{2}-240 x +120\right ) {\mathrm e}^{2 x}}{320}+\frac {5 \,{\mathrm e}^{5} \left (1-{\mathrm e}^{2 x}\right )}{8}\) | \(125\) |
default | \(20 \,{\mathrm e}^{2 x}-\frac {5 x \,{\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{8}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {{\mathrm e}^{2 x} x^{3}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}-\frac {5 \,{\mathrm e}^{5} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )}{2}+\frac {3 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}-\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {{\mathrm e}^{2 x}}{4}\right )}{4}+\frac {{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x} x^{3}}{2}-\frac {3 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {3 x \,{\mathrm e}^{2 x}}{4}-\frac {3 \,{\mathrm e}^{2 x}}{8}\right )}{2}\) | \(131\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.35, size = 156, normalized size = 5.03 \begin {gather*} -\frac {1}{8} \, {\left (2 \, x^{4} - 4 \, x^{3} + 6 \, x^{2} - 6 \, x + 3\right )} e^{\left (2 \, x\right )} + \frac {1}{16} \, {\left (4 \, x^{3} e^{5} - 6 \, x^{2} e^{5} + 6 \, x e^{5} - 3 \, e^{5}\right )} e^{\left (2 \, x\right )} - \frac {1}{16} \, {\left (4 \, x^{3} - 6 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {3}{16} \, {\left (2 \, x^{2} e^{5} - 2 \, x e^{5} + e^{5}\right )} e^{\left (2 \, x\right )} - \frac {19}{16} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - \frac {5}{8} \, {\left (2 \, x e^{5} - e^{5}\right )} e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {155}{8} \, e^{\left (2 \, x\right )} - \frac {5}{8} \, e^{\left (2 \, x + 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.26, size = 23, normalized size = 0.74 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}\,\left (x^2-5\right )\,\left (x+x\,{\mathrm {e}}^5-x^2-16\right )}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.18, size = 36, normalized size = 1.16 \begin {gather*} \frac {\left (- x^{4} + x^{3} + x^{3} e^{5} - 11 x^{2} - 5 x e^{5} - 5 x + 80\right ) e^{2 x}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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