Integrand size = 16, antiderivative size = 51 \[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=\frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {2 \arctan \left (e^{a+b x}\right )}{b} \]
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Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 398, 294, 209} \[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=-\frac {2 \arctan \left (e^{a+b x}\right )}{b}+\frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )} \]
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Rule 209
Rule 294
Rule 398
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (1-\frac {4 x^2}{\left (1+x^2\right )^2}\right ) \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}-\frac {4 \text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {2 \arctan \left (e^{a+b x}\right )}{b} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=\frac {e^{a+b x} \left (1+\frac {2}{1+e^{2 (a+b x)}}\right )-2 \arctan \left (e^{a+b x}\right )}{b} \]
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Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\sinh \left (b x +a \right )^{2}}{\cosh \left (b x +a \right )}+\frac {2}{\cosh \left (b x +a \right )}+\sinh \left (b x +a \right )-2 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(48\) |
default | \(\frac {\frac {\sinh \left (b x +a \right )^{2}}{\cosh \left (b x +a \right )}+\frac {2}{\cosh \left (b x +a \right )}+\sinh \left (b x +a \right )-2 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(48\) |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {2 \,{\mathrm e}^{b x +a}}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}+\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{b}-\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{b}\) | \(68\) |
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Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (47) = 94\).
Time = 0.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.88 \[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=\frac {\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} - 2 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 3 \, {\left (\cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + 3 \, \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} + b} \]
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\[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=e^{a} \int e^{b x} \tanh ^{2}{\left (a + b x \right )}\, dx \]
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none
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=-\frac {2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {e^{\left (b x + a\right )}}{b} + \frac {2 \, e^{\left (b x + a\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=\frac {\frac {2 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1} - 2 \, \arctan \left (e^{\left (b x + a\right )}\right ) + e^{\left (b x + a\right )}}{b} \]
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Time = 1.68 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int e^{a+b x} \tanh ^2(a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}+\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]
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