Optimal. Leaf size=158 \[ x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 b^3 \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \text {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c} \]
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Rubi [A]
time = 0.09, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5324, 4494,
4266, 2611, 2320, 6724} \begin {gather*} \frac {6 i b \text {ArcTan}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c}-\frac {6 i b^2 \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac {6 i b^2 \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )}{c} \end {gather*}
Antiderivative was successfully verified.
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Rule 2320
Rule 2611
Rule 4266
Rule 4494
Rule 5324
Rule 6724
Rubi steps
\begin {align*} \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx &=\frac {\text {Subst}\left (\int (a+b x)^3 \sec (x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3-\frac {(3 b) \text {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {\left (6 b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}-\frac {\left (6 b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {\left (6 i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}-\frac {\left (6 i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {\left (6 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {\left (6 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c}\\ &=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 b^3 \text {Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \text {Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )}{c}\\ \end {align*}
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Mathematica [A]
time = 0.19, size = 289, normalized size = 1.83 \begin {gather*} \frac {a^3 c x+3 a^2 b c x \sec ^{-1}(c x)+3 a b^2 c x \sec ^{-1}(c x)^2+b^3 c x \sec ^{-1}(c x)^3-6 a b^2 \sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )-3 b^3 \sec ^{-1}(c x)^2 \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 )+3 b^3 \sec ^{-1}(c x)^2 \log \left (1+i e^{i \sec ^{-1}(c x)}\right )-3 a^2 b \log \left (c \left (1+\sqrt {1-\frac {1}{c^2 x^2}}\right ) x\right )-6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )+6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )+6 b^3 \text {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )-6 b^3 \text {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )^{3}\, dx\]
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 \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 \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}\, 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 = -0.01 \begin {gather*} \int {\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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