Optimal. Leaf size=64 \[ \frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac {1}{2} i b \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \]
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Rubi [A]
time = 0.06, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5326, 4722,
3800, 2221, 2317, 2438} \begin {gather*} \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 )+\frac {1}{2} i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2317
Rule 2438
Rule 3800
Rule 4722
Rule 5326
Rubi steps
\begin {align*} \int \frac {a+b \sec ^{-1}(c x)}{x} \, dx &=-\text {Subst}\left (\int \frac {a+b \cos ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,\frac {1}{x}\right )\\ &=\text {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 \text {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 \text {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) \text {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*}
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Mathematica [A]
time = 0.02, size = 59, normalized size = 0.92 \begin {gather*} \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 )+a \log (x)+\frac {1}{2} i b \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.19, size = 86, normalized size = 1.34
method | result | size |
derivativedivides | \(a \ln \left (c x \right )+\frac {i b \mathrm {arcsec}\left (c x \right )^{2}}{2}-b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+\frac {i b \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}\) | \(86\) |
default | \(a \ln \left (c x \right )+\frac {i b \mathrm {arcsec}\left (c x \right )^{2}}{2}-b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+\frac {i b \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}\) | \(86\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {a + b \operatorname {asec}{\left (c x \right )}}{x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
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 \begin {gather*} \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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