Integrand size = 11, antiderivative size = 19 \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=\frac {\log (\cosh (x))}{a}+\frac {\log (a+b \text {sech}(x))}{a} \]
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Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3970, 36, 29, 31} \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=\frac {\log (a+b \text {sech}(x))}{a}+\frac {\log (\cosh (x))}{a} \]
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Rule 29
Rule 31
Rule 36
Rule 3970
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{x (a+x)} \, dx,x,b \text {sech}(x)\right ) \\ & = -\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,b \text {sech}(x)\right )}{a}+\frac {\text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \text {sech}(x)\right )}{a} \\ & = \frac {\log (\cosh (x))}{a}+\frac {\log (a+b \text {sech}(x))}{a} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=\frac {\log (b+a \cosh (x))}{a} \]
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Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(-\frac {\ln \left (\operatorname {sech}\left (x \right )\right )}{a}+\frac {\ln \left (a +b \,\operatorname {sech}\left (x \right )\right )}{a}\) | \(21\) |
default | \(-\frac {\ln \left (\operatorname {sech}\left (x \right )\right )}{a}+\frac {\ln \left (a +b \,\operatorname {sech}\left (x \right )\right )}{a}\) | \(21\) |
risch | \(-\frac {x}{a}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 b \,{\mathrm e}^{x}}{a}+1\right )}{a}\) | \(27\) |
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none
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=-\frac {x - \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (15) = 30\).
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16 \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=\begin {cases} \frac {\tilde {\infty }}{\operatorname {sech}{\left (x \right )}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {1}{b \operatorname {sech}{\left (x \right )}} & \text {for}\: a = 0 \\\frac {x - \log {\left (\tanh {\left (x \right )} + 1 \right )}}{a} & \text {for}\: b = 0 \\\frac {x}{a} + \frac {\log {\left (\frac {a}{b} + \operatorname {sech}{\left (x \right )} \right )}}{a} - \frac {\log {\left (\tanh {\left (x \right )} + 1 \right )}}{a} & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=\frac {x}{a} + \frac {\log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a} \]
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Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=\frac {\log \left ({\left | a {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a} \]
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Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx=-\frac {x-\ln \left (a+2\,b\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}\right )}{a} \]
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