Optimal. Leaf size=154 \[ -\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac {6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b} \]
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Rubi [A] time = 0.10, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5252, 5216, 3757, 4181, 2531, 2282, 6589} \[ -\frac {6 i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 \text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac {6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 3757
Rule 4181
Rule 5216
Rule 5252
Rule 6589
Rubi steps
\begin {align*} \int \sec ^{-1}(a+b x)^3 \, dx &=\frac {\operatorname {Subst}\left (\int \sec ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int x^3 \sec (x) \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}-\frac {3 \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac {6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 \operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}-\frac {6 \operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac {6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}-\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac {6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac {6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 160, normalized size = 1.04 \[ \frac {-6 i \sec ^{-1}(a+b x) \left (\text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )-\text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )\right )+6 \left (\text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )-\text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )\right )+(a+b x) \sec ^{-1}(a+b x)^3-3 \sec ^{-1}(a+b x)^2 \left (\log \left (1-i e^{i \sec ^{-1}(a+b x)}\right )-\log \left (1+i e^{i \sec ^{-1}(a+b x)}\right )\right )}{b} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arcsec}\left (b x + a\right )^{3}, x\right ) \]
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 \[ \int \operatorname {arcsec}\left (b x + a\right )^{3}\,{d x} \]
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 \[ \int \mathrm {arcsec}\left (b x +a \right )^{3}\, dx \]
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 \[ x \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{3} - \frac {3}{4} \, x \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {3 \, {\left ({\left (4 \, b x \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - b x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 4 \, {\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + {\left (a^{2} - 1\right )} b x + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )}}{4 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]
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 \[ \int {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]
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 \[ \int \operatorname {asec}^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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