Optimal. Leaf size=92 \[ -\frac{2 i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac{2 i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^2+\frac{4 i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c} \]
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Rubi [A] time = 0.0719971, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5216, 4409, 4181, 2279, 2391} \[ -\frac{2 i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac{2 i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^2+\frac{4 i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c} \]
Antiderivative was successfully verified.
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Rule 5216
Rule 4409
Rule 4181
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \left (a+b \sec ^{-1}(c x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^2-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^2+\frac{4 i b \left (a+b \sec ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^2+\frac{4 i b \left (a+b \sec ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c}+\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^2+\frac{4 i b \left (a+b \sec ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac{2 i b^2 \text{Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac{2 i b^2 \text{Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c}\\ \end{align*}
Mathematica [A] time = 0.166325, size = 163, normalized size = 1.77 \[ \frac{b^2 \left (-2 i \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )+2 i \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )+\sec ^{-1}(c x) \left (c x \sec ^{-1}(c x)-2 \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+2 \log \left (1+i e^{i \sec ^{-1}(c x)}\right )\right )\right )+a^2 c x+2 a b \left (c x \sec ^{-1}(c x)+\log \left (\cos \left (\frac{1}{2} \sec ^{-1}(c x)\right )-\sin \left (\frac{1}{2} \sec ^{-1}(c x)\right )\right )-\log \left (\cos \left (\frac{1}{2} \sec ^{-1}(c x)\right )+\sin \left (\frac{1}{2} \sec ^{-1}(c x)\right )\right )\right )}{c} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.294, size = 212, normalized size = 2.3 \begin{align*} x{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}+2\,xab{\rm arcsec} \left (cx\right )-2\,{\frac{{b}^{2}{\rm arcsec} \left (cx\right )}{c}\ln \left ( 1-i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }+2\,{\frac{{b}^{2}{\rm arcsec} \left (cx\right )}{c}\ln \left ( 1+i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }-{\frac{2\,i{b}^{2}}{c}{\it dilog} \left ( 1+i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }+{\frac{2\,i{b}^{2}}{c}{\it dilog} \left ( 1-i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }+{a}^{2}x-2\,{\frac{ab}{c}\ln \left ( cx+cx\sqrt{1-{\frac{1}{{c}^{2}{x}^{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}{4} \,{\left (2 \, c^{2}{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )} \log \left (c\right )^{2} - 4 \, c^{2} \int \frac{x^{2} \log \left (c^{2} x^{2}\right )}{c^{2} x^{2} - 1}\,{d x} \log \left (c\right ) + 8 \, c^{2} \int \frac{x^{2} \log \left (x\right )}{c^{2} x^{2} - 1}\,{d x} \log \left (c\right ) - 4 \, x \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{2} - 4 \, c^{2} \int \frac{x^{2} \log \left (c^{2} x^{2}\right ) \log \left (x\right )}{c^{2} x^{2} - 1}\,{d x} + 4 \, c^{2} \int \frac{x^{2} \log \left (x\right )^{2}}{c^{2} x^{2} - 1}\,{d x} - 4 \, c^{2} \int \frac{x^{2} \log \left (c^{2} x^{2}\right )}{c^{2} x^{2} - 1}\,{d x} + x \log \left (c^{2} x^{2}\right )^{2} + 2 \,{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} \log \left (c\right )^{2} + 4 \, \int \frac{\log \left (c^{2} x^{2}\right )}{c^{2} x^{2} - 1}\,{d x} \log \left (c\right ) - 8 \, \int \frac{\log \left (x\right )}{c^{2} x^{2} - 1}\,{d x} \log \left (c\right ) + 8 \, \int \frac{\sqrt{c x + 1} \sqrt{c x - 1} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )}{c^{2} x^{2} - 1}\,{d x} + 4 \, \int \frac{\log \left (c^{2} x^{2}\right ) \log \left (x\right )}{c^{2} x^{2} - 1}\,{d x} - 4 \, \int \frac{\log \left (x\right )^{2}}{c^{2} x^{2} - 1}\,{d x} + 4 \, \int \frac{\log \left (c^{2} x^{2}\right )}{c^{2} x^{2} - 1}\,{d x}\right )} b^{2} + a^{2} x + \frac{{\left (2 \, c x \operatorname{arcsec}\left (c x\right ) - \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right ) + \log \left (-\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )\right )} a b}{c} \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 (b^{2} \operatorname{arcsec}\left (c x\right )^{2} + 2 \, a b \operatorname{arcsec}\left (c x\right ) + a^{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 \left (a + b \operatorname{asec}{\left (c x \right )}\right )^{2}\, 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{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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