Optimal. Leaf size=494 \[ \frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}-\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 a^2 \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 a^2 \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {3 i a \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {\text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^3}+\frac {6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3 \]
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Rubi [A] time = 0.40, antiderivative size = 494, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 14, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.167, Rules used = {5258, 4426, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391, 4186, 3770} \[ -\frac {6 i a^2 \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 a^2 \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 a^2 \text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {3 i a \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {\text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^3}+\frac {6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3719
Rule 3770
Rule 4181
Rule 4184
Rule 4186
Rule 4190
Rule 4426
Rule 5258
Rule 6589
Rubi steps
\begin {align*} \int x^2 \sec ^{-1}(a+b x)^3 \, dx &=\frac {\operatorname {Subst}\left (\int x^3 \sec (x) (-a+\sec (x))^2 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3-\frac {\operatorname {Subst}\left (\int x^2 (-a+\sec (x))^3 \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3-\frac {\operatorname {Subst}\left (\int \left (-a^3 x^2+3 a^2 x^2 \sec (x)-3 a x^2 \sec ^2(x)+x^2 \sec ^3(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3-\frac {\operatorname {Subst}\left (\int x^2 \sec ^3(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac {(3 a) \operatorname {Subst}\left (\int x^2 \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^3}-\frac {\operatorname {Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {(6 a) \operatorname {Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac {3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac {i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^3}-\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {\operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {\operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac {(12 i a) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac {\left (6 i a^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {\left (6 i a^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac {3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac {i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^3}+\frac {6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {i \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {(6 a) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac {3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac {i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^3}+\frac {6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 a^2 \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 a^2 \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {(3 i a) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac {3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac {i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^3}+\frac {6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {3 i a \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {\text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 a^2 \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 a^2 \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}\\ \end {align*}
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Mathematica [A] time = 0.53, size = 442, normalized size = 0.89 \[ \frac {\frac {1}{3} a^3 \sec ^{-1}(a+b x)^3+6 a^2 \left (\text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )-i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )\right )+6 i a^2 \left (\sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )+i \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )\right )+6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )+\frac {1}{3} b^3 x^3 \sec ^{-1}(a+b x)^3-i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )+i \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )+\text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )-\text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )-3 i a \left (\text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \left (\sec ^{-1}(a+b x)+2 i \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )\right )-\frac {1}{2} (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2+3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2+(a+b x) \sec ^{-1}(a+b x)-\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )+i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 3.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {arcsec}\left (b x + a\right )^{3}, x\right ) \]
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 \[ \int x^{2} \operatorname {arcsec}\left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.79, size = 770, normalized size = 1.56 \[ \frac {x^{3} \mathrm {arcsec}\left (b x +a \right )^{3}}{3}-\frac {\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}}{2 b}+\frac {2 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right )^{2} x a}{b^{2}}+\frac {a^{3} \mathrm {arcsec}\left (b x +a \right )^{3}}{3 b^{3}}+\frac {5 \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}}{2 b^{3}}-\frac {3 i a \mathrm {arcsec}\left (b x +a \right )^{2}}{b^{3}}+\frac {\mathrm {arcsec}\left (b x +a \right ) x}{b^{2}}+\frac {\mathrm {arcsec}\left (b x +a \right ) a}{b^{3}}+\frac {2 i \arctan \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{3}}-\frac {3 a^{2} \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^{3}}-\frac {i \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^{3}}-\frac {6 a^{2} \polylog \left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{3}}+\frac {i \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^{3}}+\frac {3 a^{2} \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^{3}}+\frac {6 a^{2} \polylog \left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{3}}+\frac {6 a \,\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^{3}}-\frac {3 i a \polylog \left (2, -\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{b^{3}}-\frac {6 i a^{2} \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^{3}}-\frac {\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 )}{2 b^{3}}+\frac {6 i a^{2} \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^{3}}-\frac {\polylog \left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{3}}+\frac {\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 )}{2 b^{3}}+\frac {\polylog \left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{3}} \]
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 \[ \frac {1}{3} \, x^{3} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{3} - \frac {1}{4} \, x^{3} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {{\left (4 \, b x^{3} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - b x^{3} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 4 \, {\left (3 \, {\left (b^{3} x^{5} + 3 \, a b^{2} x^{4} + {\left (3 \, a^{2} - 1\right )} b x^{3} + {\left (a^{3} - a\right )} x^{2}\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{5} + 2 \, a b^{2} x^{4} + {\left (a^{2} - 1\right )} b x^{3} + 3 \, {\left (b^{3} x^{5} + 3 \, a b^{2} x^{4} + {\left (3 \, a^{2} - 1\right )} b x^{3} + {\left (a^{3} - a\right )} x^{2}\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )}{4 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]
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 \[ \int x^{2} \operatorname {asec}^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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