Optimal. Leaf size=64 \[ \frac{1}{2} i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+\frac{i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right ) \]
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Rubi [A] time = 0.0843917, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5218, 4626, 3719, 2190, 2279, 2391} \[ \frac{1}{2} i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+\frac{i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right ) \]
Antiderivative was successfully verified.
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Rule 5218
Rule 4626
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{a+b \sec ^{-1}(c x)}{x} \, dx &=-\operatorname{Subst}\left (\int \frac{a+b \cos ^{-1}\left (\frac{x}{c}\right )}{x} \, dx,x,\frac{1}{x}\right )\\ &=\operatorname{Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+b \operatorname{Subst}\left (\int \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 )^2}{2 b}-\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )-\frac{1}{2} (i b) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac{1}{2} i b \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )\\ \end{align*}
Mathematica [A] time = 0.0164604, size = 59, normalized size = 0.92 \[ \frac{1}{2} i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+a \log (x)+\frac{1}{2} i b \sec ^{-1}(c x)^2-b \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.281, size = 86, normalized size = 1.3 \begin{align*} a\ln \left ( cx \right ) +{\frac{i}{2}}b \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}-b{\rm arcsec} \left (cx\right )\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) +{\frac{i}{2}}b{\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*} -{\left (c^{2} \int \frac{\sqrt{c x + 1} \sqrt{c x - 1} \log \left (x\right )}{c^{4} x^{3} - c^{2} x}\,{d x} - \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) \log \left (x\right )\right )} b + a \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 \operatorname{arcsec}\left (c x\right ) + a}{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{a + b \operatorname{asec}{\left (c 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{b \operatorname{arcsec}\left (c x\right ) + a}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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