Optimal. Leaf size=29 \[ e^x-e^{5 (2+x)}+x+\frac {3}{1+x}+\frac {4}{-3+\log (2)} \]
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Rubi [A] time = 0.21, antiderivative size = 21, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {27, 6742, 2194, 683} \begin {gather*} x+e^x-e^{5 x+10}+\frac {3}{x+1} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 683
Rule 2194
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2+2 x+x^2+e^{10+5 x} \left (-5-10 x-5 x^2\right )+e^x \left (1+2 x+x^2\right )}{(1+x)^2} \, dx\\ &=\int \left (e^x-5 e^{10+5 x}+\frac {-2+2 x+x^2}{(1+x)^2}\right ) \, dx\\ &=-\left (5 \int e^{10+5 x} \, dx\right )+\int e^x \, dx+\int \frac {-2+2 x+x^2}{(1+x)^2} \, dx\\ &=e^x-e^{10+5 x}+\int \left (1-\frac {3}{(1+x)^2}\right ) \, dx\\ &=e^x-e^{10+5 x}+x+\frac {3}{1+x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 21, normalized size = 0.72 \begin {gather*} e^x-e^{10+5 x}+x+\frac {3}{1+x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.77, size = 29, normalized size = 1.00 \begin {gather*} \frac {x^{2} - {\left (x + 1\right )} e^{\left (5 \, x + 10\right )} + {\left (x + 1\right )} e^{x} + x + 3}{x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.39, size = 35, normalized size = 1.21 \begin {gather*} \frac {x^{2} - x e^{\left (5 \, x + 10\right )} + x e^{x} + x - e^{\left (5 \, x + 10\right )} + e^{x} + 3}{x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.49, size = 20, normalized size = 0.69
method | result | size |
risch | \(x +\frac {3}{x +1}-{\mathrm e}^{5 x +10}+{\mathrm e}^{x}\) | \(20\) |
norman | \(\frac {x^{2}+{\mathrm e}^{x} x -{\mathrm e}^{5 x} {\mathrm e}^{10}-{\mathrm e}^{5 x} {\mathrm e}^{10} x +2+{\mathrm e}^{x}}{x +1}\) | \(35\) |
default | \(x +\frac {3}{x +1}+{\mathrm e}^{x}-5 \,{\mathrm e}^{10} \left (-\frac {{\mathrm e}^{5 x}}{x +1}-5 \,{\mathrm e}^{-5} \expIntegralEi \left (1, -5 x -5\right )\right )-10 \,{\mathrm e}^{10} \left (4 \,{\mathrm e}^{-5} \expIntegralEi \left (1, -5 x -5\right )+\frac {{\mathrm e}^{5 x}}{x +1}\right )-5 \,{\mathrm e}^{10} \left (\frac {{\mathrm e}^{5 x}}{5}-3 \,{\mathrm e}^{-5} \expIntegralEi \left (1, -5 x -5\right )-\frac {{\mathrm e}^{5 x}}{x +1}\right )\) | \(98\) |
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*} x - \frac {e^{\left (-1\right )} E_{2}\left (-x - 1\right )}{x + 1} + \frac {x^{2} e^{x} - {\left (x^{2} e^{10} + 2 \, x e^{10} + e^{10}\right )} e^{\left (5 \, x\right )}}{x^{2} + 2 \, x + 1} + \frac {2 \, e^{x}}{x + 1} + \frac {3}{x + 1} - 2 \, \int \frac {x e^{x}}{x^{3} + 3 \, x^{2} + 3 \, x + 1}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.11, size = 19, normalized size = 0.66 \begin {gather*} x-{\mathrm {e}}^{5\,x+10}+{\mathrm {e}}^x+\frac {3}{x+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.18, size = 17, normalized size = 0.59 \begin {gather*} x - e^{10} e^{5 x} + e^{x} + \frac {3}{x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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