Optimal. Leaf size=149 \[ -\frac{2 i a \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{2 i a \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{\log (a+b x)}{b^2}-\frac{\sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1) \text{sech}^{-1}(a+b x)}{b^2}+\frac{4 a \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2 \]
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Rubi [A] time = 0.138342, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {6321, 5468, 4190, 4180, 2279, 2391, 4184, 3475} \[ -\frac{2 i a \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{2 i a \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{\log (a+b x)}{b^2}-\frac{\sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1) \text{sech}^{-1}(a+b x)}{b^2}+\frac{4 a \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
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Rule 6321
Rule 5468
Rule 4190
Rule 4180
Rule 2279
Rule 2391
Rule 4184
Rule 3475
Rubi steps
\begin{align*} \int x \text{sech}^{-1}(a+b x)^2 \, dx &=-\frac{\operatorname{Subst}\left (\int x^2 \text{sech}(x) (-a+\text{sech}(x)) \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2-\frac{\operatorname{Subst}\left (\int x (-a+\text{sech}(x))^2 \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2-\frac{\operatorname{Subst}\left (\int \left (a^2 x-2 a x \text{sech}(x)+x \text{sech}^2(x)\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{a^2 \text{sech}^{-1}(a+b x)^2}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2-\frac{\operatorname{Subst}\left (\int x \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}+\frac{(2 a) \operatorname{Subst}\left (\int x \text{sech}(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{\sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)}{b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^2}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2+\frac{4 a \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{\operatorname{Subst}\left (\int \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}-\frac{(2 i a) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}+\frac{(2 i a) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{\sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)}{b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^2}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2+\frac{4 a \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{\log (a+b x)}{b^2}-\frac{(2 i a) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{(2 i a) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}\\ &=-\frac{\sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)}{b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^2}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^2+\frac{4 a \text{sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{\log (a+b x)}{b^2}-\frac{2 i a \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{2 i a \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}\\ \end{align*}
Mathematica [A] time = 0.417642, size = 172, normalized size = 1.15 \[ \frac{-4 i a \left (\text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(a+b x)}\right )-\text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(a+b x)}\right )\right )+2 \log \left (\frac{1}{a+b x}\right )+(a+b x)^2 \text{sech}^{-1}(a+b x)^2-2 a (a+b x) \text{sech}^{-1}(a+b x)^2-2 \sqrt{-\frac{a+b x-1}{a+b x+1}} (a+b x+1) \text{sech}^{-1}(a+b x)-4 i a \text{sech}^{-1}(a+b x) \left (\log \left (1-i e^{-\text{sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text{sech}^{-1}(a+b x)}\right )\right )}{2 b^2} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.392, size = 396, normalized size = 2.7 \begin{align*}{\frac{{x}^{2} \left ({\rm arcsech} \left (bx+a\right ) \right ) ^{2}}{2}}-{\frac{x{\rm arcsech} \left (bx+a\right )}{b}\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}}-{\frac{{a}^{2} \left ({\rm arcsech} \left (bx+a\right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{{\rm arcsech} \left (bx+a\right )a}{{b}^{2}}\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}}+{\frac{{\rm arcsech} \left (bx+a\right )}{{b}^{2}}}-2\,{\frac{\ln \left ( \left ( bx+a \right ) ^{-1}+\sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) }{{b}^{2}}}+{\frac{1}{{b}^{2}}\ln \left ( 1+ \left ( \left ( bx+a \right ) ^{-1}+\sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) ^{2} \right ) }-{\frac{2\,ia{\rm arcsech} \left (bx+a\right )}{{b}^{2}}\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\,ia{\rm arcsech} \left (bx+a\right )}{{b}^{2}}\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\,ia}{{b}^{2}}{\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\,ia}{{b}^{2}}{\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*} \frac{1}{2} \, x^{2} \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{4 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} b - b\right )} x^{2} +{\left (a^{3} - a\right )} x\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} \log \left (b x + a\right )^{2} + 4 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} b - b\right )} x^{2} +{\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} -{\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} +{\left (a^{2} b - b\right )} x^{2} + 4 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} b - b\right )} x^{2} +{\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right ) +{\left (2 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} b - b\right )} x^{2} +{\left (a^{3} - a\right )} x\right )} \sqrt{b x + a + 1} \log \left (b x + a\right ) +{\left (2 \, b^{3} x^{4} + 4 \, a b^{2} x^{3} +{\left (2 \, a^{2} b - b\right )} x^{2} + 2 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} b - b\right )} x^{2} +{\left (a^{3} - a\right )} x\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 )}{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 (x \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 x \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 x \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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