Optimal. Leaf size=78 \[ \frac{2 i b^2 \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{2 i b^2 \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(c x)}\right )}{c}+x \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{4 b \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c} \]
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Rubi [A] time = 0.0702562, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6279, 5451, 4180, 2279, 2391} \[ \frac{2 i b^2 \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{2 i b^2 \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(c x)}\right )}{c}+x \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{4 b \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c} \]
Antiderivative was successfully verified.
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Rule 6279
Rule 5451
Rule 4180
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \left (a+b \text{sech}^{-1}(c x)\right )^2 \, dx &=-\frac{\operatorname{Subst}\left (\int (a+b x)^2 \text{sech}(x) \tanh (x) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{4 b \left (a+b \text{sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}-\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{4 b \left (a+b \text{sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\text{sech}^{-1}(c x)}\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{4 b \left (a+b \text{sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{2 i b^2 \text{Li}_2\left (-i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{2 i b^2 \text{Li}_2\left (i e^{\text{sech}^{-1}(c x)}\right )}{c}\\ \end{align*}
Mathematica [A] time = 0.224079, size = 126, normalized size = 1.62 \[ \frac{i b^2 \left (2 \text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(c x)}\right )-2 \text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(c x)}\right )+\text{sech}^{-1}(c x) \left (-i c x \text{sech}^{-1}(c x)+2 \log \left (1-i e^{-\text{sech}^{-1}(c x)}\right )-2 \log \left (1+i e^{-\text{sech}^{-1}(c x)}\right )\right )\right )}{c}+a^2 x+\frac{2 a b \left (c x \text{sech}^{-1}(c x)-2 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \text{sech}^{-1}(c x)\right )\right )\right )}{c} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.221, size = 250, normalized size = 3.2 \begin{align*} x{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}+{\frac{2\,i{\rm arcsech} \left (cx\right ){b}^{2}}{c}\ln \left ( 1+i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }-{\frac{2\,i{\rm arcsech} \left (cx\right ){b}^{2}}{c}\ln \left ( 1-i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }+2\,xab{\rm arcsech} \left (cx\right )+{\frac{2\,i{b}^{2}}{c}{\it dilog} \left ( 1+i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }-{\frac{2\,i{b}^{2}}{c}{\it dilog} \left ( 1-i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }+{a}^{2}x-2\,{\frac{ab}{c}\arctan \left ( \sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) } \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*}{\left (x \log \left (\sqrt{c x + 1} \sqrt{-c x + 1} + 1\right )^{2} - \int -\frac{c^{2} x^{2} \log \left (c\right )^{2} +{\left (c^{2} x^{2} - 1\right )} \log \left (x\right )^{2} +{\left (c^{2} x^{2} \log \left (c\right )^{2} +{\left (c^{2} x^{2} - 1\right )} \log \left (x\right )^{2} - \log \left (c\right )^{2} + 2 \,{\left (c^{2} x^{2} \log \left (c\right ) - \log \left (c\right )\right )} \log \left (x\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1} - 2 \,{\left (c^{2} x^{2} \log \left (c\right ) +{\left (c^{2} x^{2}{\left (\log \left (c\right ) + 1\right )} +{\left (c^{2} x^{2} - 1\right )} \log \left (x\right ) - \log \left (c\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1} +{\left (c^{2} x^{2} - 1\right )} \log \left (x\right ) - \log \left (c\right )\right )} \log \left (\sqrt{c x + 1} \sqrt{-c x + 1} + 1\right ) - \log \left (c\right )^{2} + 2 \,{\left (c^{2} x^{2} \log \left (c\right ) - \log \left (c\right )\right )} \log \left (x\right )}{c^{2} x^{2} +{\left (c^{2} x^{2} - 1\right )} \sqrt{c x + 1} \sqrt{-c x + 1} - 1}\,{d x}\right )} b^{2} + a^{2} x + \frac{2 \,{\left (c x \operatorname{arsech}\left (c x\right ) - \arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )\right )} a b}{c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{2} \operatorname{arsech}\left (c x\right )^{2} + 2 \, a b \operatorname{arsech}\left (c x\right ) + a^{2}, x\right ) \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 (a + b \operatorname{asech}{\left (c x \right )}\right )^{2}\, 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 (b \operatorname{arsech}\left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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