Optimal. Leaf size=430 \[ -3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+6 i \text {Li}_4\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 i \text {Li}_4\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {3}{2} \sec ^{-1}(a+b x) \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {3}{4} i \text {Li}_4\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
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Rubi [A] time = 0.53, antiderivative size = 430, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 10, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5258, 4551, 4530, 3719, 2190, 2531, 6609, 2282, 6589, 4520} \[ -3 i \sec ^{-1}(a+b x)^2 \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+6 \sec ^{-1}(a+b x) \text {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 \sec ^{-1}(a+b x) \text {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+6 i \text {PolyLog}\left (4,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 i \text {PolyLog}\left (4,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {3}{2} \sec ^{-1}(a+b x) \text {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {3}{4} i \text {PolyLog}\left (4,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3719
Rule 4520
Rule 4530
Rule 4551
Rule 5258
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {\sec ^{-1}(a+b x)^3}{x} \, dx &=\operatorname {Subst}\left (\int \frac {x^3 \sec (x) \tan (x)}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\operatorname {Subst}\left (\int \frac {x^3 \tan (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=a \operatorname {Subst}\left (\int \frac {x^3 \sin (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )+\operatorname {Subst}\left (\int x^3 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\left (2 i \operatorname {Subst}\left (\int \frac {e^{2 i x} x^3}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\right )-(i a) \operatorname {Subst}\left (\int \frac {e^{i x} x^3}{1-\sqrt {1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )-(i a) \operatorname {Subst}\left (\int \frac {e^{i x} x^3}{1+\sqrt {1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-3 \operatorname {Subst}\left (\int x^2 \log \left (1-\frac {a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )-3 \operatorname {Subst}\left (\int x^2 \log \left (1-\frac {a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+3 \operatorname {Subst}\left (\int x^2 \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-3 i \operatorname {Subst}\left (\int x \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+6 i \operatorname {Subst}\left (\int x \text {Li}_2\left (\frac {a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+6 i \operatorname {Subst}\left (\int x \text {Li}_2\left (\frac {a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{2} \sec ^{-1}(a+b x) \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+\frac {3}{2} \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )-6 \operatorname {Subst}\left (\int \text {Li}_3\left (\frac {a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )-6 \operatorname {Subst}\left (\int \text {Li}_3\left (\frac {a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{2} \sec ^{-1}(a+b x) \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {3}{4} i \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )+6 i \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {a x}{1-\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )+6 i \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {a x}{1+\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )\\ &=\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 \sec ^{-1}(a+b x) \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{2} \sec ^{-1}(a+b x) \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+6 i \text {Li}_4\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 i \text {Li}_4\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{4} i \text {Li}_4\left (-e^{2 i \sec ^{-1}(a+b x)}\right )\\ \end {align*}
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Mathematica [B] time = 4.06, size = 1058, normalized size = 2.46 \[ 2 \log \left (\frac {e^{i \sec ^{-1}(a+b x)} a}{\sqrt {1-a^2}-1}+1\right ) \sec ^{-1}(a+b x)^3+\log \left (\frac {e^{i \sec ^{-1}(a+b x)} \left (\sqrt {1-a^2}-1\right )}{a}+1\right ) \sec ^{-1}(a+b x)^3+2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \sec ^{-1}(a+b x)^3+\log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \sec ^{-1}(a+b x)^3-3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \sec ^{-1}(a+b x)^3+2 \log \left (\frac {2 \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a+b x}\right ) \sec ^{-1}(a+b x)^3-\log \left (\frac {a \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{\sqrt {1-a^2}-1}+1\right ) \sec ^{-1}(a+b x)^3-\log \left (\frac {\left (\sqrt {1-a^2}-1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}+1\right ) \sec ^{-1}(a+b x)^3-\log \left (1-\frac {a \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{\sqrt {1-a^2}+1}\right ) \sec ^{-1}(a+b x)^3-\log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}\right ) \sec ^{-1}(a+b x)^3-6 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (\frac {e^{i \sec ^{-1}(a+b x)} \left (\sqrt {1-a^2}-1\right )}{a}+1\right ) \sec ^{-1}(a+b x)^2+6 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \sec ^{-1}(a+b x)^2+6 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (\frac {\left (\sqrt {1-a^2}-1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}+1\right ) \sec ^{-1}(a+b x)^2-6 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}\right ) \sec ^{-1}(a+b x)^2-3 i \text {Li}_2\left (-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}-1}\right ) \sec ^{-1}(a+b x)^2-3 i \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \sec ^{-1}(a+b x)^2+\frac {3}{2} i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right ) \sec ^{-1}(a+b x)^2+6 \text {Li}_3\left (-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}-1}\right ) \sec ^{-1}(a+b x)+6 \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \sec ^{-1}(a+b x)-\frac {3}{2} \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right ) \sec ^{-1}(a+b x)+6 i \text {Li}_4\left (-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}-1}\right )+6 i \text {Li}_4\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-\frac {3}{4} i \text {Li}_4\left (-e^{2 i \sec ^{-1}(a+b x)}\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x}, x\right ) \]
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 \[ \int \frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.56, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arcsec}\left (b x +a \right )^{3}}{x}\, dx \]
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 \[ \int \frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x}\,{d x} \]
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 \[ \int \frac {{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3}{x} \,d x \]
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 \[ \int \frac {\operatorname {asec}^{3}{\left (a + b x \right )}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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