Optimal. Leaf size=49 \[ \frac{\log \left (1-e^{2 c (a+b x)}\right )}{b c}+\frac{e^{a c+b c x} \tanh ^{-1}(\text{sech}(c (a+b x)))}{b c} \]
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Rubi [A] time = 0.0678644, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2194, 6275, 2282, 12, 260} \[ \frac{\log \left (1-e^{2 c (a+b x)}\right )}{b c}+\frac{e^{a c+b c x} \tanh ^{-1}(\text{sech}(c (a+b x)))}{b c} \]
Antiderivative was successfully verified.
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Rule 2194
Rule 6275
Rule 2282
Rule 12
Rule 260
Rubi steps
\begin{align*} \int e^{c (a+b x)} \tanh ^{-1}(\text{sech}(a c+b c x)) \, dx &=\frac{\operatorname{Subst}\left (\int e^x \tanh ^{-1}(\text{sech}(x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int e^x \text{csch}(x) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int \frac{2 x}{-1+x^2} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{2 \operatorname{Subst}\left (\int \frac{x}{-1+x^2} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\log \left (1-e^{2 c (a+b x)}\right )}{b c}\\ \end{align*}
Mathematica [A] time = 0.0825592, size = 59, normalized size = 1.2 \[ \frac{\log \left (1-e^{2 c (a+b x)}\right )+e^{c (a+b x)} \tanh ^{-1}\left (\frac{2 e^{c (a+b x)}}{e^{2 c (a+b x)}+1}\right )}{b c} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.335, size = 872, normalized size = 17.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.999006, size = 86, normalized size = 1.76 \begin{align*} \frac{\operatorname{artanh}\left (\operatorname{sech}\left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac{\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac{\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.94412, size = 236, normalized size = 4.82 \begin{align*} \frac{{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \log \left (\frac{\cosh \left (b c x + a c\right ) + 1}{\cosh \left (b c x + a c\right ) - 1}\right ) + 2 \, \log \left (\frac{2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{2 \, b c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.18145, size = 132, normalized size = 2.69 \begin{align*} \frac{e^{\left ({\left (b x + a\right )} c\right )} \log \left (-\frac{\frac{2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}} + 1}{\frac{2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}} - 1}\right )}{2 \, b c} + \frac{\log \left ({\left | e^{\left (2 \, b c x + 2 \, a c\right )} - 1 \right |}\right )}{b c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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