Optimal. Leaf size=102 \[ -\frac{i d \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac{i d \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}+\frac{d \text{sech}(a+b x)}{2 b^2}+\frac{(c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{(c+d x) \tanh (a+b x) \text{sech}(a+b x)}{2 b} \]
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Rubi [A] time = 0.0640534, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4185, 4180, 2279, 2391} \[ -\frac{i d \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac{i d \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}+\frac{d \text{sech}(a+b x)}{2 b^2}+\frac{(c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{(c+d x) \tanh (a+b x) \text{sech}(a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 4185
Rule 4180
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int (c+d x) \text{sech}^3(a+b x) \, dx &=\frac{d \text{sech}(a+b x)}{2 b^2}+\frac{(c+d x) \text{sech}(a+b x) \tanh (a+b x)}{2 b}+\frac{1}{2} \int (c+d x) \text{sech}(a+b x) \, dx\\ &=\frac{(c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{d \text{sech}(a+b x)}{2 b^2}+\frac{(c+d x) \text{sech}(a+b x) \tanh (a+b x)}{2 b}-\frac{(i d) \int \log \left (1-i e^{a+b x}\right ) \, dx}{2 b}+\frac{(i d) \int \log \left (1+i e^{a+b x}\right ) \, dx}{2 b}\\ &=\frac{(c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{d \text{sech}(a+b x)}{2 b^2}+\frac{(c+d x) \text{sech}(a+b x) \tanh (a+b x)}{2 b}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}+\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}\\ &=\frac{(c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{i d \text{Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac{i d \text{Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac{d \text{sech}(a+b x)}{2 b^2}+\frac{(c+d x) \text{sech}(a+b x) \tanh (a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 2.81574, size = 180, normalized size = 1.76 \[ \frac{-i d \left (\text{PolyLog}\left (2,-i e^{a+b x}\right )-\text{PolyLog}\left (2,i e^{a+b x}\right )\right )+b c \tan ^{-1}(\sinh (a+b x))+b c \tanh (a+b x) \text{sech}(a+b x)-\frac{1}{2} d (-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )+\frac{1}{2} (\pi -2 i a) d \log \left (\cot \left (\frac{1}{4} (2 i a+2 i b x+\pi )\right )\right )+b d x \text{sech}(a) \sinh (b x) \text{sech}^2(a+b x)+d (b x \tanh (a)+1) \text{sech}(a+b x)}{2 b^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.023, size = 216, normalized size = 2.1 \begin{align*}{\frac{{{\rm e}^{bx+a}} \left ( bdx{{\rm e}^{2\,bx+2\,a}}+bc{{\rm e}^{2\,bx+2\,a}}-bdx+d{{\rm e}^{2\,bx+2\,a}}-cb+d \right ) }{{b}^{2} \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}+{\frac{c\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}}-{\frac{{\frac{i}{2}}d\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{{\frac{i}{2}}d\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\frac{i}{2}}d\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{{\frac{i}{2}}d\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-{\frac{{\frac{i}{2}}d{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{{\frac{i}{2}}d{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{da\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} d{\left (\frac{{\left (b x e^{\left (3 \, a\right )} + e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} -{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + 8 \, \int \frac{x e^{\left (b x + a\right )}}{8 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x}\right )} - c{\left (\frac{\arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac{e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.54573, size = 3492, normalized size = 34.24 \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*} \int \left (c + d x\right ) \operatorname{sech}^{3}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} \operatorname{sech}\left (b x + a\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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