Integrand size = 10, antiderivative size = 200 \[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right ) \]
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Time = 0.23 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5366, 4647, 4626, 3800, 2221, 2317, 2438, 4616} \[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
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Rule 2221
Rule 2317
Rule 2438
Rule 3800
Rule 4616
Rule 4626
Rule 4647
Rule 5366
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x \sec (x) \tan (x)}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right ) \\ & = \text {Subst}\left (\int \frac {x \tan (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right ) \\ & = a \text {Subst}\left (\int \frac {x \sin (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )+\text {Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right ) \\ & = -\left (2 i \text {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\right )-(i a) \text {Subst}\left (\int \frac {e^{i x} x}{1-\sqrt {1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )-(i a) \text {Subst}\left (\int \frac {e^{i x} x}{1+\sqrt {1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right ) \\ & = \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-\text {Subst}\left (\int \log \left (1-\frac {a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )-\text {Subst}\left (\int \log \left (1-\frac {a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+\text {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right ) \\ & = \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {1}{2} i \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )+i \text {Subst}\left (\int \frac {\log \left (1-\frac {a x}{1-\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )+i \text {Subst}\left (\int \frac {\log \left (1-\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) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right ) \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=-4 i \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right ) \arctan \left (\frac {(1+a) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a^2}}\right )+\left (\sec ^{-1}(a+b x)-2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right )\right ) \log \left (1+\frac {\left (-1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+\left (\sec ^{-1}(a+b x)+2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right )\right ) \log \left (1-\frac {\left (1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-i \left (\operatorname {PolyLog}\left (2,-\frac {\left (-1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+\operatorname {PolyLog}\left (2,\frac {\left (1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right ) \]
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Time = 1.64 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.87
method | result | size |
derivativedivides | \(\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )+\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-i \operatorname {dilog}\left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )-i \operatorname {dilog}\left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )\) | \(374\) |
default | \(\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )+\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-i \operatorname {dilog}\left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )-i \operatorname {dilog}\left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )\) | \(374\) |
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\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{x} \,d x } \]
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\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int \frac {\operatorname {asec}{\left (a + b x \right )}}{x}\, dx \]
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\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{x} \,d x } \]
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\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{x} \,d x } \]
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Timed out. \[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int \frac {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}{x} \,d x \]
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