Optimal. Leaf size=310 \[ -2 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+2 \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
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Rubi [A] time = 0.485217, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5258, 4551, 4530, 3719, 2190, 2531, 2282, 6589, 4520} \[ -2 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+2 \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
Antiderivative was successfully verified.
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Rule 5258
Rule 4551
Rule 4530
Rule 3719
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rule 4520
Rubi steps
\begin{align*} \int \frac{\sec ^{-1}(a+b x)^2}{x} \, dx &=\operatorname{Subst}\left (\int \frac{x^2 \sec (x) \tan (x)}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\operatorname{Subst}\left (\int \frac{x^2 \tan (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=a \operatorname{Subst}\left (\int \frac{x^2 \sin (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )+\operatorname{Subst}\left (\int x^2 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\left (2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} x^2}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\right )-(i a) \operatorname{Subst}\left (\int \frac{e^{i x} x^2}{1-\sqrt{1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )-(i a) \operatorname{Subst}\left (\int \frac{e^{i x} x^2}{1+\sqrt{1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 \operatorname{Subst}\left (\int x \log \left (1-\frac{a e^{i x}}{1-\sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )-2 \operatorname{Subst}\left (\int x \log \left (1-\frac{a e^{i x}}{1+\sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+2 \operatorname{Subst}\left (\int x \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )+i \sec ^{-1}(a+b x) \text{Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-i \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+2 i \operatorname{Subst}\left (\int \text{Li}_2\left (\frac{a e^{i x}}{1-\sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+2 i \operatorname{Subst}\left (\int \text{Li}_2\left (\frac{a e^{i x}}{1+\sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )+i \sec ^{-1}(a+b x) \text{Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )+2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{a x}{1-\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )+2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{a x}{1+\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )+i \sec ^{-1}(a+b x) \text{Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+2 \text{Li}_3\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{Li}_3\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\frac{1}{2} \text{Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right )\\ \end{align*}
Mathematica [B] time = 2.23018, size = 813, normalized size = 2.62 \[ \log \left (\frac{e^{i \sec ^{-1}(a+b x)} a}{\sqrt{1-a^2}-1}+1\right ) \sec ^{-1}(a+b x)^2+\log \left (\frac{e^{i \sec ^{-1}(a+b x)} \left (\sqrt{1-a^2}-1\right )}{a}+1\right ) \sec ^{-1}(a+b x)^2+\log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right ) \sec ^{-1}(a+b x)^2+\log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \sec ^{-1}(a+b x)^2-2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \sec ^{-1}(a+b x)^2+\log \left (\frac{2 \left (i \sqrt{1-\frac{1}{(a+b x)^2}}+\frac{1}{a+b x}\right )}{a+b x}\right ) \sec ^{-1}(a+b x)^2-\log \left (\frac{\left (\sqrt{1-a^2}-1\right ) \left (i \sqrt{1-\frac{1}{(a+b x)^2}}+\frac{1}{a+b x}\right )}{a}+1\right ) \sec ^{-1}(a+b x)^2-\log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) \left (i \sqrt{1-\frac{1}{(a+b x)^2}}+\frac{1}{a+b x}\right )}{a}\right ) \sec ^{-1}(a+b x)^2-4 \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (\frac{e^{i \sec ^{-1}(a+b x)} \left (\sqrt{1-a^2}-1\right )}{a}+1\right ) \sec ^{-1}(a+b x)+4 \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \sec ^{-1}(a+b x)+4 \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (\frac{\left (\sqrt{1-a^2}-1\right ) \left (i \sqrt{1-\frac{1}{(a+b x)^2}}+\frac{1}{a+b x}\right )}{a}+1\right ) \sec ^{-1}(a+b x)-4 \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) \left (i \sqrt{1-\frac{1}{(a+b x)^2}}+\frac{1}{a+b x}\right )}{a}\right ) \sec ^{-1}(a+b x)-2 i \text{PolyLog}\left (2,-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}-1}\right ) \sec ^{-1}(a+b x)-2 i \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right ) \sec ^{-1}(a+b x)+i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right ) \sec ^{-1}(a+b x)+2 \text{PolyLog}\left (3,-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}-1}\right )+2 \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right ) \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.627, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}}{x}}\, dx \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{arcsec}\left (b x + a\right )^{2}}{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{arcsec}\left (b x + a\right )^{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{\operatorname{asec}^{2}{\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{arcsec}\left (b x + a\right )^{2}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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