Optimal. Leaf size=494 \[ \frac {(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac {3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac {3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac {i \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^3}+\frac {6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 i a^2 \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {3 i a \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {\text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {6 a^2 \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {6 a^2 \text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3} \]
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Rubi [A]
time = 0.31, 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 =
{5366, 4511, 4275, 4266, 2611, 2320, 6724, 4269, 3800, 2221, 2317, 2438, 4271, 3855}
\begin {gather*} \frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (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 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 {i \sec ^{-1}(a+b x)^2 \text {ArcTan}\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 {1}{3} x^3 \sec ^{-1}(a+b x)^3 \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 3800
Rule 3855
Rule 4266
Rule 4269
Rule 4271
Rule 4275
Rule 4511
Rule 5366
Rule 6724
Rubi steps
\begin {align*} \int x^2 \sec ^{-1}(a+b x)^3 \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {Subst}\left (\int x^2 \sec ^3(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac {(3 a) \text {Subst}\left (\int x^2 \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {\left (3 a^2\right ) \text {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 {\text {Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^3}-\frac {\text {Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {(6 a) \text {Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac {\left (6 a^2\right ) \text {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 ) \text {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 {\text {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {\text {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) \text {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 ) \text {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 ) \text {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 \text {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac {i \text {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) \text {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 ) \text {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 ) \text {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 {\text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {\text {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) \text {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 = 0.36, size = 442, normalized size = 0.89 \begin {gather*} \frac {(a+b x) \sec ^{-1}(a+b x)+3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2-\frac {1}{2} (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2+\frac {1}{3} a^3 \sec ^{-1}(a+b x)^3+\frac {1}{3} b^3 x^3 \sec ^{-1}(a+b x)^3+i \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )+6 i a^2 \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )-\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )-i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )+i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-3 i a \left (\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 )+\text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )+6 a^2 \left (-i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )+\text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )\right )+6 i a^2 \left (\sec ^{-1}(a+b x) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )+i \text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )\right )-\text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.10, size = 716, normalized size = 1.45 Too large to display
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^{2} \operatorname {asec}^{3}{\left (a + b x \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^2\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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