Optimal. Leaf size=77 \[ \frac{e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac{3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.0498214, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2282, 390, 1158, 12, 288, 203} \[ \frac{e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac{3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 390
Rule 1158
Rule 12
Rule 288
Rule 203
Rubi steps
\begin{align*} \int e^{a+b x} \tanh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{\left (1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1-\frac{2 \left (1+3 x^4\right )}{\left (1+x^2\right )^3}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{e^{a+b x}}{b}-\frac{2 \operatorname{Subst}\left (\int \frac{1+3 x^4}{\left (1+x^2\right )^3} \, 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 )^2}+\frac{\operatorname{Subst}\left (\int -\frac{12 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac{e^{a+b x}}{b}-\frac{2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}-\frac{6 \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 )^2}+\frac{3 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac{3 \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 )^2}+\frac{3 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac{3 \tan ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.0887836, size = 60, normalized size = 0.78 \[ \frac{e^{a+b x} \left (5 e^{2 (a+b x)}+e^{4 (a+b x)}+2\right )}{b \left (e^{2 (a+b x)}+1\right )^2}-\frac{3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.012, size = 102, normalized size = 1.3 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}{b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+3\,{\frac{\sinh \left ( bx+a \right ) }{b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}-{\frac{3\,{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{2\,b}}-3\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}}-{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{b\cosh \left ( bx+a \right ) }}+2\,{\frac{\cosh \left ( bx+a \right ) }{b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.58349, size = 93, normalized size = 1.21 \begin{align*} -\frac{3 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac{e^{\left (b x + a\right )}}{b} + \frac{3 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}}{b{\left (e^{\left (4 \, b x + 4 \, a\right )} + 2 \, 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 = 2.11078, size = 956, normalized size = 12.42 \begin{align*} \frac{\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{5} + 5 \,{\left (2 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 5 \, \cosh \left (b x + a\right )^{3} + 5 \,{\left (2 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 3 \,{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) +{\left (5 \, \cosh \left (b x + a\right )^{4} + 15 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right ) + 2 \, \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 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 ^{3}{\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 = 2.13441, size = 70, normalized size = 0.91 \begin{align*} \frac{\frac{3 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} - 3 \, \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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