Optimal. Leaf size=278 \[ -\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3 \]
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Rubi [A] time = 0.26, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {5258, 4426, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac {6 i a \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3719
Rule 4181
Rule 4184
Rule 4190
Rule 4426
Rule 5258
Rule 6589
Rubi steps
\begin {align*} \int x \sec ^{-1}(a+b x)^3 \, dx &=\frac {\operatorname {Subst}\left (\int x^3 \sec (x) (-a+\sec (x)) \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int x^2 (-a+\sec (x))^2 \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int \left (a^2 x^2-2 a x^2 \sec (x)+x^2 \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int x^2 \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}+\frac {(3 a) \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {3 \operatorname {Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}-\frac {(6 a) \operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}+\frac {(6 a) \operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}-\frac {(6 i a) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}+\frac {(6 i a) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {3 \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}-\frac {(6 a) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {(6 a) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^2}\\ &=\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 a \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}\\ &=\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}\\ \end {align*}
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Mathematica [A] time = 0.44, size = 248, normalized size = 0.89 \[ \frac {1}{2} \left (x^2 \sec ^{-1}(a+b x)^3-\frac {3 \left (\frac {1}{3} a^2 \sec ^{-1}(a+b x)^3+4 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )-i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+4 a \left (\text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )-i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )\right )-4 a \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )+(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2-i \sec ^{-1}(a+b x)^2+2 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )+4 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )\right )}{b^2}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 1.19, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arcsec}\left (b x + a\right )^{3}, x\right ) \]
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 \[ \int x \operatorname {arcsec}\left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.53, size = 429, normalized size = 1.54 \[ \frac {x^{2} \mathrm {arcsec}\left (b x +a \right )^{3}}{2}-\frac {3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right )^{2} x}{2 b}-\frac {a^{2} \mathrm {arcsec}\left (b x +a \right )^{3}}{2 b^{2}}-\frac {3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right )^{2} a}{2 b^{2}}+\frac {3 i \mathrm {arcsec}\left (b x +a \right )^{2}}{2 b^{2}}+\frac {3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}+\frac {6 a \polylog \left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}+\frac {6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {6 a \polylog \left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {3 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{b^{2}}+\frac {3 i \polylog \left (2, -\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{2 b^{2}} \]
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 \[ \frac {1}{2} \, x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{3} - \frac {3}{8} \, x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {3 \, {\left ({\left (4 \, b x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - b x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 4 \, {\left (2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + {\left (a^{2} - 1\right )} b x^{2} + 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )}}{8 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]
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 \[ \int x \operatorname {asec}^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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