Optimal. Leaf size=61 \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )}{2 d}+\frac{\text{sech}^{-1}(a+b x)^2}{2 d}-\frac{\text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right )}{d} \]
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Rubi [A] time = 0.0977936, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {6319, 12, 6281, 5660, 3718, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )}{2 d}+\frac{\text{sech}^{-1}(a+b x)^2}{2 d}-\frac{\text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right )}{d} \]
Antiderivative was successfully verified.
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Rule 6319
Rule 12
Rule 6281
Rule 5660
Rule 3718
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\text{sech}^{-1}(a+b x)}{\frac{a d}{b}+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b \text{sech}^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\text{sech}^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\cosh ^{-1}(x)}{x} \, dx,x,\frac{1}{a+b x}\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int x \tanh (x) \, dx,x,\cosh ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{\cosh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1+e^{2 x}} \, dx,x,\cosh ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{\cosh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\cosh ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}+\frac{\operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{\cosh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\cosh ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}+\frac{\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{2 d}\\ &=\frac{\cosh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\cosh ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}-\frac{\text{Li}_2\left (-e^{2 \cosh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{2 d}\\ \end{align*}
Mathematica [A] time = 0.0552964, size = 52, normalized size = 0.85 \[ \frac{\text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(a+b x)}\right )-\text{sech}^{-1}(a+b x) \left (\text{sech}^{-1}(a+b x)+2 \log \left (e^{-2 \text{sech}^{-1}(a+b x)}+1\right )\right )}{2 d} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.283, size = 104, normalized size = 1.7 \begin{align*}{\frac{ \left ({\rm arcsech} \left (bx+a\right ) \right ) ^{2}}{2\,d}}-{\frac{{\rm arcsech} \left (bx+a\right )}{d}\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{1}{2\,d}{\it polylog} \left ( 2,- \left ( \left ( bx+a \right ) ^{-1}+\sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) ^{2} \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{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 ) \log \left (b x + a\right ) - 3 \, \log \left (b x + a\right )^{2}}{2 \, d} - \frac{\log \left (b x + a + 1\right ) \log \left (b x + a\right ) +{\rm Li}_2\left (-b x - a\right )}{2 \, d} - \frac{\log \left (b x + a\right ) \log \left (-b x - a + 1\right ) +{\rm Li}_2\left (b x + a\right )}{2 \, d} + \int \frac{{\left (b^{2} x + a b\right )} \log \left (b x + a\right )}{b^{2} d x^{2} + 2 \, a b d x + a^{2} d +{\left (b^{2} d x^{2} + 2 \, a b d x + a^{2} d - d\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} - d}\,{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{b \operatorname{arsech}\left (b x + a\right )}{b d x + a d}, 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*} \frac{b \int \frac{\operatorname{asech}{\left (a + b x \right )}}{a + b x}\, dx}{d} \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 )}{d x + \frac{a d}{b}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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