Optimal. Leaf size=83 \[ -b \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )+\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{3 b}-\log \left (e^{2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )^2+\frac {1}{2} b^2 \text {Li}_3\left (-e^{2 \text {sech}^{-1}(c x)}\right ) \]
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Rubi [A] time = 0.13, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6285, 3718, 2190, 2531, 2282, 6589} \[ -b \text {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{2} b^2 \text {PolyLog}\left (3,-e^{2 \text {sech}^{-1}(c x)}\right )+\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{3 b}-\log \left (e^{2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )^2 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 6285
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{x} \, dx &=-\operatorname {Subst}\left (\int (a+b x)^2 \tanh (x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{3 b}-2 \operatorname {Subst}\left (\int \frac {e^{2 x} (a+b x)^2}{1+e^{2 x}} \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{3 b}-\left (a+b \text {sech}^{-1}(c x)\right )^2 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )+(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{3 b}-\left (a+b \text {sech}^{-1}(c x)\right )^2 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )-b \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )+b^2 \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{3 b}-\left (a+b \text {sech}^{-1}(c x)\right )^2 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )-b \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )+\frac {1}{2} b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 \text {sech}^{-1}(c x)}\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{3 b}-\left (a+b \text {sech}^{-1}(c x)\right )^2 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )-b \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )+\frac {1}{2} b^2 \text {Li}_3\left (-e^{2 \text {sech}^{-1}(c x)}\right )\\ \end {align*}
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Mathematica [A] time = 0.17, size = 116, normalized size = 1.40 \[ a^2 \log (c x)+a b \left (\text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right )-\text {sech}^{-1}(c x) \left (\text {sech}^{-1}(c x)+2 \log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right )\right )\right )+b^2 \left (\text {sech}^{-1}(c x) \text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \text {sech}^{-1}(c x)}\right )-\frac {1}{3} \text {sech}^{-1}(c x)^3-\text {sech}^{-1}(c x)^2 \log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right )\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsech}\left (c x\right )^{2} + 2 \, a b \operatorname {arsech}\left (c x\right ) + a^{2}}{x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )}^{2}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 250, normalized size = 3.01 \[ a^{2} \ln \left (c x \right )+\frac {b^{2} \mathrm {arcsech}\left (c x \right )^{3}}{3}-b^{2} \mathrm {arcsech}\left (c x \right )^{2} \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-b^{2} \mathrm {arcsech}\left (c x \right ) \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )+\frac {b^{2} \polylog \left (3, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}+a b \mathrm {arcsech}\left (c x \right )^{2}-2 a b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-a b \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \log \relax (x) + \int \frac {b^{2} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{2}}{x} + \frac {2 \, a b \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^2}{x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right )^{2}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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