Integrand size = 13, antiderivative size = 4 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^x\right ) \]
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Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2281, 209} \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^x\right ) \]
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Rule 209
Rule 2281
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right ) \\ & = \arctan \left (e^x\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^x\right ) \]
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Time = 0.03 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00
method | result | size |
default | \(\arctan \left ({\mathrm e}^{x}\right )\) | \(4\) |
risch | \(\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2}\) | \(20\) |
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none
Time = 0.24 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^{x}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (3) = 6\).
Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 3.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\operatorname {RootSum} {\left (4 z^{2} + 1, \left ( i \mapsto i \log {\left (2 i + e^{x} \right )} \right )\right )} \]
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none
Time = 0.27 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^{x}\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^{x}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\mathrm {atan}\left ({\mathrm {e}}^x\right ) \]
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