Optimal. Leaf size=80 \[ \frac{2 i \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{2 i \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \text{sech}^{-1}(a+b x)^2}{b}-\frac{4 \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b} \]
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Rubi [A] time = 0.067922, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6315, 6279, 5418, 4180, 2279, 2391} \[ \frac{2 i \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{2 i \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \text{sech}^{-1}(a+b x)^2}{b}-\frac{4 \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 6315
Rule 6279
Rule 5418
Rule 4180
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \text{sech}^{-1}(a+b x)^2 \, dx &=\frac{\operatorname{Subst}\left (\int \text{sech}^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int x^2 \text{sech}(x) \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^2}{b}-\frac{2 \operatorname{Subst}\left (\int x \text{sech}(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^2}{b}-\frac{4 \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(2 i) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}-\frac{(2 i) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^2}{b}-\frac{4 \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^2}{b}-\frac{4 \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{2 i \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{2 i \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.21023, size = 105, normalized size = 1.31 \[ \frac{i \left (2 \text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(a+b x)}\right )-2 \text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x) \left (-i (a+b x) \text{sech}^{-1}(a+b x)+2 \log \left (1-i e^{-\text{sech}^{-1}(a+b x)}\right )-2 \log \left (1+i e^{-\text{sech}^{-1}(a+b x)}\right )\right )\right )}{b} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.285, size = 209, normalized size = 2.6 \begin{align*} x \left ({\rm arcsech} \left (bx+a\right ) \right ) ^{2}+{\frac{ \left ({\rm arcsech} \left (bx+a\right ) \right ) ^{2}a}{b}}+{\frac{2\,i{\rm arcsech} \left (bx+a\right )}{b}\ln \left ( 1+i \left ( \left ( bx+a \right ) ^{-1}+\sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) \right ) }-{\frac{2\,i{\rm arcsech} \left (bx+a\right )}{b}\ln \left ( 1-i \left ( \left ( bx+a \right ) ^{-1}+\sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) \right ) }-{\frac{2\,i}{b}{\it dilog} \left ( 1-i \left ( \left ( bx+a \right ) ^{-1}+\sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) \right ) }+{\frac{2\,i}{b}{\it dilog} \left ( 1+i \left ( \left ( bx+a \right ) ^{-1}+\sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) \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*} x \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )^{2} - \int -\frac{2 \,{\left (2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} \log \left (b x + a\right )^{2} + 2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )^{2} -{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} +{\left (a^{2} b - b\right )} x + 2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right ) +{\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \log \left (b x + a\right ) +{\left (2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} +{\left (2 \, a^{2} b - b\right )} x +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )\right )} \sqrt{b x + a + 1}\right )} \sqrt{-b x - a + 1}\right )} \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )\right )}}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} +{\left (3 \, a^{2} b - b\right )} x - a}\,{d x} \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 (\operatorname{arsech}\left (b x + a\right )^{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 \operatorname{asech}^{2}{\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 \operatorname{arsech}\left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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