Optimal. Leaf size=365 \[ -\frac {1}{4} i \text {ArcSin}(a+b x)^4+\text {ArcSin}(a+b x)^3 \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\text {ArcSin}(a+b x)^3 \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-3 i \text {ArcSin}(a+b x)^2 \text {PolyLog}\left (2,\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-3 i \text {ArcSin}(a+b x)^2 \text {PolyLog}\left (2,\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 \text {ArcSin}(a+b x) \text {PolyLog}\left (3,\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 \text {ArcSin}(a+b x) \text {PolyLog}\left (3,\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 i \text {PolyLog}\left (4,\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 i \text {PolyLog}\left (4,\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right ) \]
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Rubi [A]
time = 0.31, antiderivative size = 365, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4889, 4825,
4617, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -3 i \text {ArcSin}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-3 i \text {ArcSin}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 \text {ArcSin}(a+b x) \text {Li}_3\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 \text {ArcSin}(a+b x) \text {Li}_3\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 i \text {Li}_4\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 i \text {Li}_4\left (\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+\text {ArcSin}(a+b x)^3 \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )+\text {ArcSin}(a+b x)^3 \log \left (1-\frac {e^{i \text {ArcSin}(a+b x)}}{\sqrt {1-a^2}+i a}\right )-\frac {1}{4} i \text {ArcSin}(a+b x)^4 \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 4617
Rule 4825
Rule 4889
Rule 6724
Rule 6744
Rubi steps
\begin {align*} \int \frac {\sin ^{-1}(a+b x)^3}{x} \, dx &=\frac {\text {Subst}\left (\int \frac {\sin ^{-1}(x)^3}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {x^3 \cos (x)}{-\frac {a}{b}+\frac {\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{4} i \sin ^{-1}(a+b x)^4+\frac {i \text {Subst}\left (\int \frac {e^{i x} x^3}{-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}+\frac {e^{i x}}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}+\frac {i \text {Subst}\left (\int \frac {e^{i x} x^3}{-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}+\frac {e^{i x}}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{4} i \sin ^{-1}(a+b x)^4+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-3 \text {Subst}\left (\int x^2 \log \left (1+\frac {e^{i x}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )-3 \text {Subst}\left (\int x^2 \log \left (1+\frac {e^{i x}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac {1}{4} i \sin ^{-1}(a+b x)^4+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 i \text {Subst}\left (\int x \text {Li}_2\left (-\frac {e^{i x}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )+6 i \text {Subst}\left (\int x \text {Li}_2\left (-\frac {e^{i x}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac {1}{4} i \sin ^{-1}(a+b x)^4+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 \sin ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 \sin ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-6 \text {Subst}\left (\int \text {Li}_3\left (-\frac {e^{i x}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )-6 \text {Subst}\left (\int \text {Li}_3\left (-\frac {e^{i x}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac {1}{4} i \sin ^{-1}(a+b x)^4+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 \sin ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 \sin ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 i \text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{i a-\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )+6 i \text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{i a+\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )\\ &=-\frac {1}{4} i \sin ^{-1}(a+b x)^4+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+\sin ^{-1}(a+b x)^3 \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )-3 i \sin ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 \sin ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 \sin ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )+6 i \text {Li}_4\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 i \text {Li}_4\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 424, normalized size = 1.16 \begin {gather*} -\frac {1}{4} i \text {ArcSin}(a+b x)^4+\text {ArcSin}(a+b x)^3 \log \left (1+\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )+\text {ArcSin}(a+b x)^3 \log \left (1+\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )-3 i \text {ArcSin}(a+b x)^2 \text {PolyLog}\left (2,-\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )-3 i \text {ArcSin}(a+b x)^2 \text {PolyLog}\left (2,-\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )+6 \text {ArcSin}(a+b x) \text {PolyLog}\left (3,-\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )+6 \text {ArcSin}(a+b x) \text {PolyLog}\left (3,-\frac {e^{i \text {ArcSin}(a+b x)}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )+6 i \text {PolyLog}\left (4,\frac {e^{i \text {ArcSin}(a+b x)}}{i a-\sqrt {1-a^2}}\right )+6 i \text {PolyLog}\left (4,\frac {e^{i \text {ArcSin}(a+b x)}}{i a+\sqrt {1-a^2}}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 1.70, size = 0, normalized size = 0.00 \[\int \frac {\arcsin \left (b x +a \right )^{3}}{x}\, dx\]
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 \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 \frac {\operatorname {asin}^{3}{\left (a + b x \right )}}{x}\, 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 \frac {{\mathrm {asin}\left (a+b\,x\right )}^3}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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