Optimal. Leaf size=59 \[ -\frac {1}{2} i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+\cos ^{-1}\left (\frac {x}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right ) \]
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Rubi [A] time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5264, 4626, 3719, 2190, 2279, 2391} \[ -\frac {1}{2} i \text {PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+\cos ^{-1}\left (\frac {x}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right ) \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 3719
Rule 4626
Rule 5264
Rubi steps
\begin {align*} \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x} \, dx &=\int \frac {\cos ^{-1}\left (\frac {x}{a}\right )}{x} \, dx\\ &=-\operatorname {Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (\frac {x}{a}\right )\right )\\ &=-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+2 i \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (\frac {x}{a}\right )\right )\\ &=-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+\cos ^{-1}\left (\frac {x}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )-\operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (\frac {x}{a}\right )\right )\\ &=-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+\cos ^{-1}\left (\frac {x}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )+\frac {1}{2} i \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )\\ &=-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+\cos ^{-1}\left (\frac {x}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )-\frac {1}{2} i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )\\ \end {align*}
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Mathematica [A] time = 0.02, size = 59, normalized size = 1.00 \[ -\frac {1}{2} i \text {Li}_2\left (-e^{2 i \sec ^{-1}\left (\frac {a}{x}\right )}\right )-\frac {1}{2} i \sec ^{-1}\left (\frac {a}{x}\right )^2+\sec ^{-1}\left (\frac {a}{x}\right ) \log \left (1+e^{2 i \sec ^{-1}\left (\frac {a}{x}\right )}\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{x}, x\right ) \]
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 \[ \int \frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 76, normalized size = 1.29 \[ -\frac {i \mathrm {arcsec}\left (\frac {a}{x}\right )^{2}}{2}+\mathrm {arcsec}\left (\frac {a}{x}\right ) \ln \left (1+\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )^{2}\right )-\frac {i \polylog \left (2, -\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )^{2}\right )}{2} \]
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 \[ \int \frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acos}\left (\frac {x}{a}\right )}{x} \,d x \]
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 \[ \int \frac {\operatorname {asec}{\left (\frac {a}{x} \right )}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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