Optimal. Leaf size=95 \[ \frac{e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac{8 e^{3 a+3 b x}}{3 b \left (e^{2 a+2 b x}+1\right )^3}+\frac{\tan ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.0467283, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {2282, 12, 288, 199, 203} \[ \frac{e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac{8 e^{3 a+3 b x}}{3 b \left (e^{2 a+2 b x}+1\right )^3}+\frac{\tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 288
Rule 199
Rule 203
Rubi steps
\begin{align*} \int e^{a+b x} \text{sech}^4(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{16 x^4}{\left (1+x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{16 \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{8 e^{3 a+3 b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}+\frac{8 \operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{8 e^{3 a+3 b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac{2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{8 e^{3 a+3 b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac{2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}+\frac{e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{8 e^{3 a+3 b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac{2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}+\frac{e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}+\frac{\tan ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.0828558, size = 64, normalized size = 0.67 \[ \frac{e^{a+b x} \left (-8 e^{2 (a+b x)}+3 e^{4 (a+b x)}-3\right )}{3 b \left (e^{2 (a+b x)}+1\right )^3}+\frac{\tan ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.04, size = 83, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,b\cosh \left ( bx+a \right ) }}-{\frac{\cosh \left ( bx+a \right ) }{3\,b}}+{\frac{{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{2\,b}}+{\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 = 1.56712, size = 112, normalized size = 1.18 \begin{align*} \frac{\arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac{3 \, e^{\left (5 \, b x + 5 \, a\right )} - 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{3 \, b{\left (e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, 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 = 1.8413, size = 1424, normalized size = 14.99 \begin{align*} \text{result too large to display} \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} \operatorname{sech}^{4}{\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 = 1.25156, size = 82, normalized size = 0.86 \begin{align*} \frac{\arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac{3 \, e^{\left (5 \, b x + 5 \, a\right )} - 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{3 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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