Optimal. Leaf size=21 \[ 25-\frac {5 e^{2 x} (4+x)}{-1-\frac {14 x}{5}} \]
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Rubi [A] time = 0.12, antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 2199, 2194, 2177, 2178} \begin {gather*} \frac {25 e^{2 x}}{14}+\frac {1275 e^{2 x}}{14 (14 x+5)} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 2177
Rule 2178
Rule 2194
Rule 2199
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{(5+14 x)^2} \, dx\\ &=\int \left (\frac {25 e^{2 x}}{7}-\frac {1275 e^{2 x}}{(5+14 x)^2}+\frac {1275 e^{2 x}}{7 (5+14 x)}\right ) \, dx\\ &=\frac {25}{7} \int e^{2 x} \, dx+\frac {1275}{7} \int \frac {e^{2 x}}{5+14 x} \, dx-1275 \int \frac {e^{2 x}}{(5+14 x)^2} \, dx\\ &=\frac {25 e^{2 x}}{14}+\frac {1275 e^{2 x}}{14 (5+14 x)}+\frac {1275 \text {Ei}\left (\frac {1}{7} (5+14 x)\right )}{98 e^{5/7}}-\frac {1275}{7} \int \frac {e^{2 x}}{5+14 x} \, dx\\ &=\frac {25 e^{2 x}}{14}+\frac {1275 e^{2 x}}{14 (5+14 x)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 17, normalized size = 0.81 \begin {gather*} \frac {25 e^{2 x} (4+x)}{5+14 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.93, size = 16, normalized size = 0.76 \begin {gather*} \frac {25 \, {\left (x + 4\right )} e^{\left (2 \, x\right )}}{14 \, x + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 22, normalized size = 1.05 \begin {gather*} \frac {25 \, {\left (x e^{\left (2 \, x\right )} + 4 \, e^{\left (2 \, x\right )}\right )}}{14 \, x + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 17, normalized size = 0.81
method | result | size |
gosper | \(\frac {25 \left (4+x \right ) {\mathrm e}^{2 x}}{14 x +5}\) | \(17\) |
risch | \(\frac {25 \left (4+x \right ) {\mathrm e}^{2 x}}{14 x +5}\) | \(17\) |
default | \(\frac {1275 \,{\mathrm e}^{2 x}}{196 \left (x +\frac {5}{14}\right )}+\frac {25 \,{\mathrm e}^{2 x}}{14}\) | \(19\) |
norman | \(\frac {100 \,{\mathrm e}^{2 x}+25 x \,{\mathrm e}^{2 x}}{14 x +5}\) | \(23\) |
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*} \frac {25 \, {\left (14 \, x^{2} + 61 \, x\right )} e^{\left (2 \, x\right )}}{196 \, x^{2} + 140 \, x + 25} + \frac {275 \, e^{\left (-\frac {5}{7}\right )} E_{2}\left (-2 \, x - \frac {5}{7}\right )}{14 \, {\left (14 \, x + 5\right )}} + 25 \, \int \frac {{\left (714 \, x - 305\right )} e^{\left (2 \, x\right )}}{2744 \, x^{3} + 2940 \, x^{2} + 1050 \, x + 125}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.09, size = 16, normalized size = 0.76 \begin {gather*} \frac {25\,{\mathrm {e}}^{2\,x}\,\left (x+4\right )}{14\,x+5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.10, size = 14, normalized size = 0.67 \begin {gather*} \frac {\left (25 x + 100\right ) e^{2 x}}{14 x + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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