Optimal. Leaf size=236 \[ -\frac {i b^2 \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac {i b^2 \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac {b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}+\frac {i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c^3}-\frac {b x^2 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac {b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c^3} \]
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Rubi [A] time = 0.19, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5222, 4409, 4186, 3770, 4181, 2531, 2282, 6589} \[ -\frac {i b^2 \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac {i b^2 \text {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac {b^3 \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \text {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac {b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac {b x^2 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c^3}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c^3} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 3770
Rule 4181
Rule 4186
Rule 4409
Rule 5222
Rule 6589
Rubi steps
\begin {align*} \int x^2 \left (a+b \sec ^{-1}(c x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int (a+b x)^3 \sec ^3(x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {b \operatorname {Subst}\left (\int (a+b x)^2 \sec ^3(x) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac {b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {b \operatorname {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sec ^{-1}(c x)\right )}{2 c^3}-\frac {b^3 \operatorname {Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac {b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac {i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c^3}+\frac {b^2 \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}-\frac {b^2 \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac {b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac {i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c^3}-\frac {i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac {i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac {\left (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^3}-\frac {\left (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^3}\\ &=\frac {b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac {i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c^3}-\frac {i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac {i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac {b^3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c^3}\\ &=\frac {b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac {i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c^3}-\frac {i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac {i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac {b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac {b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}\\ \end {align*}
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Mathematica [A] time = 1.44, size = 403, normalized size = 1.71 \[ \frac {2 a^3 c^3 x^3+6 a^2 b c^3 x^3 \sec ^{-1}(c x)-3 a^2 b c^2 x^2 \sqrt {1-\frac {1}{c^2 x^2}}-3 a^2 b \log \left (x \left (\sqrt {1-\frac {1}{c^2 x^2}}+1\right )\right )+6 a b^2 c^3 x^3 \sec ^{-1}(c x)^2-6 a b^2 c^2 x^2 \sqrt {1-\frac {1}{c^2 x^2}} \sec ^{-1}(c x)-6 i b^2 \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 i b^2 \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 a b^2 c x-6 a b^2 \sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+6 a b^2 \sec ^{-1}(c x) \log \left (1+i e^{i \sec ^{-1}(c x)}\right )+2 b^3 c^3 x^3 \sec ^{-1}(c x)^3-3 b^3 c^2 x^2 \sqrt {1-\frac {1}{c^2 x^2}} \sec ^{-1}(c x)^2-6 b^3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )+6 b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )-6 b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )+6 b^3 c x \sec ^{-1}(c x)+6 i b^3 \sec ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{6 c^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} x^{2} \operatorname {arcsec}\left (c x\right )^{3} + 3 \, a b^{2} x^{2} \operatorname {arcsec}\left (c x\right )^{2} + 3 \, a^{2} b x^{2} \operatorname {arcsec}\left (c x\right ) + a^{3} x^{2}, x\right ) \]
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 \[ \int {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.05, size = 687, normalized size = 2.91 \[ \frac {a^{3} x^{3}}{3}+\frac {x^{3} b^{3} \mathrm {arcsec}\left (c x \right )^{3}}{3}-\frac {b^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, \mathrm {arcsec}\left (c x \right )^{2} x^{2}}{2 c}+\frac {b^{3} \mathrm {arcsec}\left (c x \right ) x}{c^{2}}-\frac {b^{3} \mathrm {arcsec}\left (c x \right )^{2} \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{2 c^{3}}-\frac {i a \,b^{2} \dilog \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}-\frac {b^{3} \polylog \left (3, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}+\frac {b^{3} \mathrm {arcsec}\left (c x \right )^{2} \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{2 c^{3}}-\frac {i b^{3} \mathrm {arcsec}\left (c x \right ) \polylog \left (2, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}+\frac {b^{3} \polylog \left (3, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}+\frac {i b^{3} \mathrm {arcsec}\left (c x \right ) \polylog \left (2, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}+b^{2} x^{3} a \mathrm {arcsec}\left (c x \right )^{2}-\frac {a \,b^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, \mathrm {arcsec}\left (c x \right ) x^{2}}{c}-\frac {a \,b^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}+\frac {a \,b^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}+\frac {i a \,b^{2} \dilog \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{c^{3}}+\frac {2 i b^{3} \arctan \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}{c^{3}}+\frac {a \,b^{2} x}{c^{2}}+x^{3} a^{2} b \,\mathrm {arcsec}\left (c x \right )-\frac {a^{2} b \,x^{2}}{2 c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}+\frac {a^{2} b}{2 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {a^{2} b \sqrt {c^{2} x^{2}-1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{2 c^{4} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x} \]
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 = -0.00 \[ \int x^2\,{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3 \,d x \]
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 \[ \int x^{2} \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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