Integrand size = 7, antiderivative size = 15 \[ \int F^{2+5 x} \, dx=\frac {F^{2+5 x}}{5 \log (F)} \]
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Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2225} \[ \int F^{2+5 x} \, dx=\frac {F^{5 x+2}}{5 \log (F)} \]
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Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {F^{2+5 x}}{5 \log (F)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int F^{2+5 x} \, dx=\frac {F^{2+5 x}}{5 \log (F)} \]
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Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93
method | result | size |
gosper | \(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\) | \(14\) |
derivativedivides | \(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\) | \(14\) |
default | \(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\) | \(14\) |
risch | \(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\) | \(14\) |
parallelrisch | \(\frac {F^{2+5 x}}{5 \ln \left (F \right )}\) | \(14\) |
norman | \(\frac {{\mathrm e}^{\left (2+5 x \right ) \ln \left (F \right )}}{5 \ln \left (F \right )}\) | \(16\) |
meijerg | \(-\frac {F^{2} \left (1-{\mathrm e}^{5 x \ln \left (F \right )}\right )}{5 \ln \left (F \right )}\) | \(20\) |
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none
Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int F^{2+5 x} \, dx=\frac {F^{5 \, x + 2}}{5 \, \log \left (F\right )} \]
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Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int F^{2+5 x} \, dx=\begin {cases} \frac {F^{5 x + 2}}{5 \log {\left (F \right )}} & \text {for}\: \log {\left (F \right )} \neq 0 \\x & \text {otherwise} \end {cases} \]
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none
Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int F^{2+5 x} \, dx=\frac {F^{5 \, x + 2}}{5 \, \log \left (F\right )} \]
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none
Time = 0.33 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int F^{2+5 x} \, dx=\frac {F^{5 \, x + 2}}{5 \, \log \left (F\right )} \]
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Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int F^{2+5 x} \, dx=\frac {F^{5\,x+2}}{5\,\ln \left (F\right )} \]
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