Integrand size = 18, antiderivative size = 117 \[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=\frac {e^{c (a+b x)}}{b c}-\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}+\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c} \]
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Time = 0.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5592, 2225, 2283} \[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=-\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}+\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}+\frac {e^{c (a+b x)}}{b c} \]
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Rule 2225
Rule 2283
Rule 5592
Rubi steps \begin{align*} \text {integral}& = \int \left (e^{c (a+b x)}+\frac {4 e^{c (a+b x)}}{\left (1+e^{2 (d+e x)}\right )^2}-\frac {4 e^{c (a+b x)}}{1+e^{2 (d+e x)}}\right ) \, dx \\ & = 4 \int \frac {e^{c (a+b x)}}{\left (1+e^{2 (d+e x)}\right )^2} \, dx-4 \int \frac {e^{c (a+b x)}}{1+e^{2 (d+e x)}} \, dx+\int e^{c (a+b x)} \, dx \\ & = \frac {e^{c (a+b x)}}{b c}-\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}+\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c} \\ \end{align*}
Time = 1.31 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.44 \[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=\frac {e^{c (a+b x)} \left (2 b^2 c^2 e^{2 (d+e x)} \operatorname {Hypergeometric2F1}\left (1,1+\frac {b c}{2 e},2+\frac {b c}{2 e},-e^{2 (d+e x)}\right )-(b c+2 e) \left (2 b c e^{2 d} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )-\left (1+e^{2 d}\right ) (e-b c \text {sech}(d) \text {sech}(d+e x) \sinh (e x))\right )\right )}{b c e (b c+2 e) \left (1+e^{2 d}\right )} \]
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\[\int {\mathrm e}^{c \left (b x +a \right )} \tanh \left (e x +d \right )^{2}d x\]
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\[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{2} \,d x } \]
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\[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=e^{a c} \int e^{b c x} \tanh ^{2}{\left (d + e x \right )}\, dx \]
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\[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{2} \,d x } \]
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\[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{2} \,d x } \]
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Timed out. \[ \int e^{c (a+b x)} \tanh ^2(d+e x) \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\mathrm {tanh}\left (d+e\,x\right )}^2 \,d x \]
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