Optimal. Leaf size=128 \[ -\frac{3}{2} b^2 \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{3}{2} i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-\frac{3}{4} i b^3 \text{PolyLog}\left (4,-e^{2 i \sec ^{-1}(c x)}\right )+\frac{i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^3 \]
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Rubi [A] time = 0.140944, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5222, 3719, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3}{2} b^2 \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{3}{2} i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-\frac{3}{4} i b^3 \text{PolyLog}\left (4,-e^{2 i \sec ^{-1}(c x)}\right )+\frac{i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^3 \]
Antiderivative was successfully verified.
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Rule 5222
Rule 3719
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sec ^{-1}(c x)\right )^3}{x} \, dx &=\operatorname{Subst}\left (\int (a+b x)^3 \tan (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^3}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \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 )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac{3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\left (3 i b^2\right ) \operatorname{Subst}\left (\int (a+b x) \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 )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac{3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac{3}{2} b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )+\frac{1}{2} \left (3 b^3\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac{3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac{3}{2} b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac{1}{4} \left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )\\ &=\frac{i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac{3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac{3}{2} b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac{3}{4} i b^3 \text{Li}_4\left (-e^{2 i \sec ^{-1}(c x)}\right )\\ \end{align*}
Mathematica [A] time = 0.17005, size = 204, normalized size = 1.59 \[ \frac{1}{4} \left (-6 b^2 \text{PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-3 i b^3 \text{PolyLog}\left (4,-e^{2 i \sec ^{-1}(c x)}\right )+6 i a^2 b \sec ^{-1}(c x)^2-12 a^2 b \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+4 a^3 \log (c x)+4 i a b^2 \sec ^{-1}(c x)^3-12 a b^2 \sec ^{-1}(c x)^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b^3 \sec ^{-1}(c x)^4-4 b^3 \sec ^{-1}(c x)^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right ) \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.431, size = 390, normalized size = 3.1 \begin{align*}{a}^{3}\ln \left ( cx \right ) +{\frac{i}{4}}{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{4}-{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{3}\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) +{\frac{3\,i}{2}}{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) -{\frac{3\,{b}^{3}{\rm arcsec} \left (cx\right )}{2}{\it polylog} \left ( 3,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) }-{\frac{3\,i}{4}}{b}^{3}{\it polylog} \left ( 4,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) +ia{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{3}-3\,a{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) +3\,ia{b}^{2}{\rm arcsec} \left (cx\right ){\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) -{\frac{3\,a{b}^{2}}{2}{\it polylog} \left ( 3,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) }+{\frac{3\,i}{2}}{a}^{2}b \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}-3\,{a}^{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{3\,i}{2}}{a}^{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*} \text{result too large to display} \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^{3} \operatorname{arcsec}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{arcsec}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{arcsec}\left (c x\right ) + a^{3}}{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{\left (a + b \operatorname{asec}{\left (c x \right )}\right )^{3}}{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{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{3}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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