Optimal. Leaf size=158 \[ -\frac {6 i b^2 \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac {6 i b^2 \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c}+\frac {6 b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )}{c} \]
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Rubi [A] time = 0.12, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5216, 4409, 4181, 2531, 2282, 6589} \[ -\frac {6 i b^2 \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac {6 i b^2 \text {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac {6 b^3 \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \text {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4181
Rule 4409
Rule 5216
Rule 6589
Rubi steps
\begin {align*} \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int (a+b x)^3 \sec (x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3-\frac {(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}-\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {\left (6 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {\left (6 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )}{c}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 289, normalized size = 1.83 \[ \frac {a^3 c x-3 a^2 b \log \left (c x \left (\sqrt {1-\frac {1}{c^2 x^2}}+1\right )\right )+3 a^2 b c x \sec ^{-1}(c x)-6 i b^2 \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 i b^2 \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+3 a b^2 c x \sec ^{-1}(c x)^2-6 a b^2 \sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+6 a b^2 \sec ^{-1}(c x) \log \left (1+i e^{i \sec ^{-1}(c x)}\right )+6 b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )-6 b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )+b^3 c x \sec ^{-1}(c x)^3-3 b^3 \sec ^{-1}(c x)^2 \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+3 b^3 \sec ^{-1}(c x)^2 \log \left (1+i e^{i \sec ^{-1}(c x)}\right )}{c} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} \operatorname {arcsec}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname {arcsec}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname {arcsec}\left (c x\right ) + a^{3}, x\right ) \]
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 \[ \int {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.62, size = 0, normalized size = 0.00 \[ \int \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )^{3}\, dx \]
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 \[ \text {result too large to display} \]
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 \[ \int {\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3 \,d x \]
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 \[ \int \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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