Optimal. Leaf size=93 \[ i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right )+\frac{i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2 \]
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Rubi [A] time = 0.118987, antiderivative size = 93, normalized size of antiderivative = 1., 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 = {5222, 3719, 2190, 2531, 2282, 6589} \[ i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right )+\frac{i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2 \]
Antiderivative was successfully verified.
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Rule 5222
Rule 3719
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx &=\operatorname{Subst}\left (\int (a+b x)^2 \tan (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^2}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac{1}{2} b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac{1}{2} b^2 \text{Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )\\ \end{align*}
Mathematica [A] time = 0.120485, size = 129, normalized size = 1.39 \[ i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right )+a^2 \log (c x)+i a b \sec ^{-1}(c x)^2-2 a b \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac{1}{3} i b^2 \sec ^{-1}(c x)^3-b^2 \sec ^{-1}(c x)^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.358, size = 215, normalized size = 2.3 \begin{align*}{a}^{2}\ln \left ( cx \right ) +{\frac{i}{3}}{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{3}-{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) +i{b}^{2}{\rm arcsec} \left (cx\right ){\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) -{\frac{{b}^{2}}{2}{\it polylog} \left ( 3,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) }+iab \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}-2\,ab{\rm arcsec} \left (cx\right )\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) +iab{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \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{1}{2} \, b^{2} c^{2}{\left (\frac{\log \left (c x + 1\right )}{c^{2}} + \frac{\log \left (c x - 1\right )}{c^{2}}\right )} \log \left (c\right )^{2} + b^{2} c^{2} \int \frac{x^{2} \log \left (c^{2} x^{2}\right )}{c^{2} x^{3} - x}\,{d x} \log \left (c\right ) - 2 \, b^{2} c^{2} \int \frac{x^{2} \log \left (x\right )}{c^{2} x^{3} - x}\,{d x} \log \left (c\right ) + 2 \, b^{2} c^{2} \int \frac{x^{2} \log \left (c^{2} x^{2}\right ) \log \left (x\right )}{c^{2} x^{3} - x}\,{d x} - b^{2} c^{2} \int \frac{x^{2} \log \left (x\right )^{2}}{c^{2} x^{3} - x}\,{d x} + 2 \, a b c^{2} \int \frac{x^{2} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )}{c^{2} x^{3} - x}\,{d x} + \frac{1}{2} \, b^{2}{\left (\log \left (c x + 1\right ) + \log \left (c x - 1\right ) - 2 \, \log \left (x\right )\right )} \log \left (c\right )^{2} + b^{2} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{2} \log \left (x\right ) - \frac{1}{4} \, b^{2} \log \left (c^{2} x^{2}\right )^{2} \log \left (x\right ) - b^{2} \int \frac{\log \left (c^{2} x^{2}\right )}{c^{2} x^{3} - x}\,{d x} \log \left (c\right ) + 2 \, b^{2} \int \frac{\log \left (x\right )}{c^{2} x^{3} - x}\,{d x} \log \left (c\right ) - 2 \, b^{2} \int \frac{\sqrt{c x + 1} \sqrt{c x - 1} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) \log \left (x\right )}{c^{2} x^{3} - x}\,{d x} - 2 \, b^{2} \int \frac{\log \left (c^{2} x^{2}\right ) \log \left (x\right )}{c^{2} x^{3} - x}\,{d x} + b^{2} \int \frac{\log \left (x\right )^{2}}{c^{2} x^{3} - x}\,{d x} - 2 \, a b \int \frac{\arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )}{c^{2} x^{3} - x}\,{d x} + a^{2} \log \left (x\right ) \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^{2} \operatorname{arcsec}\left (c x\right )^{2} + 2 \, a b \operatorname{arcsec}\left (c x\right ) + a^{2}}{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{\left (a + b \operatorname{asec}{\left (c x \right )}\right )^{2}}{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{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{2}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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