Optimal. Leaf size=93 \[ \frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} b^2 \text {PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \]
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Rubi [A]
time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5330, 3800,
2221, 2611, 2320, 6724} \begin {gather*} i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-\frac {1}{2} b^2 \text {Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 3800
Rule 5330
Rule 6724
Rubi steps
\begin {align*} \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx &=\text {Subst}\left (\int (a+b x)^2 \tan (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-2 i \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+(2 b) \text {Subst}\left (\int (a+b x) \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 )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\left (i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} b^2 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} b^2 \text {Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 129, normalized size = 1.39 \begin {gather*} i a b \sec ^{-1}(c x)^2+\frac {1}{3} i b^2 \sec ^{-1}(c x)^3-2 a b \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )-b^2 \sec ^{-1}(c x)^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+a^2 \log (c x)+i b \left (a+b \sec ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} b^2 \text {PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.22, size = 215, normalized size = 2.31
method | result | size |
derivativedivides | \(a^{2} \ln \left (c x \right )+\frac {i b^{2} \mathrm {arcsec}\left (c x \right )^{3}}{3}-b^{2} \mathrm {arcsec}\left (c x \right )^{2} \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+i b^{2} \mathrm {arcsec}\left (c x \right ) \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )-\frac {b^{2} \polylog \left (3, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}+i a b \mathrm {arcsec}\left (c x \right )^{2}-2 a b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+i a b \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )\) | \(215\) |
default | \(a^{2} \ln \left (c x \right )+\frac {i b^{2} \mathrm {arcsec}\left (c x \right )^{3}}{3}-b^{2} \mathrm {arcsec}\left (c x \right )^{2} \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+i b^{2} \mathrm {arcsec}\left (c x \right ) \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )-\frac {b^{2} \polylog \left (3, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}+i a b \mathrm {arcsec}\left (c x \right )^{2}-2 a b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+i a b \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )\) | \(215\) |
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 {\left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}{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 = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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