Optimal. Leaf size=51 \[ \frac{e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.0349531, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2282, 390, 288, 203} \[ \frac{e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 390
Rule 288
Rule 203
Rubi steps
\begin{align*} \int e^{a+b x} \tanh ^2(a+b x) \, dx &=\frac{\operatorname{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{\operatorname{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 \operatorname{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 \operatorname{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 \tan ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.0875275, size = 40, normalized size = 0.78 \[ \frac{e^{a+b x} \left (\frac{2}{e^{2 (a+b x)}+1}+1\right )-2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.007, size = 56, normalized size = 1.1 \begin{align*} -{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{b\cosh \left ( bx+a \right ) }}+2\,{\frac{\cosh \left ( bx+a \right ) }{b}}+{\frac{\sinh \left ( bx+a \right ) }{b}}-2\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.59225, size = 63, normalized size = 1.24 \begin{align*} -\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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.98645, size = 433, normalized size = 8.49 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \tanh ^{2}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.20872, size = 55, normalized size = 1.08 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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