Optimal. Leaf size=170 \[ \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\frac{1}{2} \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right ) \]
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Rubi [A] time = 0.289761, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {6321, 5595, 5570, 3718, 2190, 2279, 2391, 5562} \[ \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\frac{1}{2} \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right ) \]
Antiderivative was successfully verified.
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Rule 6321
Rule 5595
Rule 5570
Rule 3718
Rule 2190
Rule 2279
Rule 2391
Rule 5562
Rubi steps
\begin{align*} \int \frac{\text{sech}^{-1}(a+b x)}{x} \, dx &=-\operatorname{Subst}\left (\int \frac{x \text{sech}(x) \tanh (x)}{-a+\text{sech}(x)} \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=-\operatorname{Subst}\left (\int \frac{x \tanh (x)}{1-a \cosh (x)} \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=-\left (a \operatorname{Subst}\left (\int \frac{x \sinh (x)}{1-a \cosh (x)} \, dx,x,\text{sech}^{-1}(a+b x)\right )\right )-\operatorname{Subst}\left (\int x \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}(a+b x)\right )\right )-a \operatorname{Subst}\left (\int \frac{e^x x}{1-\sqrt{1-a^2}-a e^x} \, dx,x,\text{sech}^{-1}(a+b x)\right )-a \operatorname{Subst}\left (\int \frac{e^x x}{1+\sqrt{1-a^2}-a e^x} \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )-\operatorname{Subst}\left (\int \log \left (1-\frac{a e^x}{1-\sqrt{1-a^2}}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )-\operatorname{Subst}\left (\int \log \left (1-\frac{a e^x}{1+\sqrt{1-a^2}}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )+\operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \text{sech}^{-1}(a+b x)}\right )-\operatorname{Subst}\left (\int \frac{\log \left (1-\frac{a x}{1-\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )-\operatorname{Subst}\left (\int \frac{\log \left (1-\frac{a x}{1+\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )\\ &=\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )+\text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\frac{1}{2} \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a+b x)}\right )\\ \end{align*}
Mathematica [C] time = 0.191604, size = 332, normalized size = 1.95 \[ -\text{PolyLog}\left (2,-\frac{\left (\sqrt{1-a^2}-1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )-\text{PolyLog}\left (2,\frac{\left (\sqrt{1-a^2}+1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )+\frac{1}{2} \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x) \log \left (\frac{\left (\sqrt{1-a^2}-1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}+1\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )+2 i \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (\frac{\left (\sqrt{1-a^2}-1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}+1\right )-2 i \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )-4 i \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \tanh ^{-1}\left (\frac{(a+1) \tanh \left (\frac{1}{2} \text{sech}^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \log \left (e^{-2 \text{sech}^{-1}(a+b x)}+1\right ) \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.352, size = 886, normalized size = 5.2 \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (b x + a\right )}{x}\,{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 (\frac{\operatorname{arsech}\left (b x + a\right )}{x}, 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 \frac{\operatorname{asech}{\left (a + b x \right )}}{x}\, 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 \frac{\operatorname{arsech}\left (b x + a\right )}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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