Optimal. Leaf size=316 \[ \frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}+\frac {6 i b \text {Li}_3\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {6 i b \text {Li}_3\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}+\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\sqrt {a^2-1}}-\frac {\sin ^{-1}(a+b x)^3}{x} \]
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Rubi [A] time = 0.63, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {4805, 4743, 4773, 3323, 2264, 2190, 2531, 2282, 6589} \[ \frac {6 b \sin ^{-1}(a+b x) \text {PolyLog}\left (2,-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {6 b \sin ^{-1}(a+b x) \text {PolyLog}\left (2,-\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\sqrt {a^2-1}}+\frac {6 i b \text {PolyLog}\left (3,-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {6 i b \text {PolyLog}\left (3,-\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\sqrt {a^2-1}}+\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\sqrt {a^2-1}}-\frac {\sin ^{-1}(a+b x)^3}{x} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3323
Rule 4743
Rule 4773
Rule 4805
Rule 6589
Rubi steps
\begin {align*} \int \frac {\sin ^{-1}(a+b x)^3}{x^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)^3}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}+3 \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)^2}{\left (-\frac {a}{b}+\frac {x}{b}\right ) \sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}+3 \operatorname {Subst}\left (\int \frac {x^2}{-\frac {a}{b}+\frac {\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}+6 \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{\frac {i}{b}-\frac {2 a e^{i x}}{b}-\frac {i e^{2 i x}}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{-\frac {2 a}{b}-\frac {2 \sqrt {-1+a^2}}{b}-\frac {2 i e^{i x}}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{-\frac {2 a}{b}+\frac {2 \sqrt {-1+a^2}}{b}-\frac {2 i e^{i x}}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}+\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {(6 i b) \operatorname {Subst}\left (\int x \log \left (1-\frac {2 i e^{i x}}{\left (-\frac {2 a}{b}-\frac {2 \sqrt {-1+a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}-\frac {(6 i b) \operatorname {Subst}\left (\int x \log \left (1-\frac {2 i e^{i x}}{\left (-\frac {2 a}{b}+\frac {2 \sqrt {-1+a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}+\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {(6 b) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i e^{i x}}{\left (-\frac {2 a}{b}-\frac {2 \sqrt {-1+a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}-\frac {(6 b) \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {2 i e^{i x}}{\left (-\frac {2 a}{b}+\frac {2 \sqrt {-1+a^2}}{b}\right ) b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}+\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {(6 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i x}{-a+\sqrt {-1+a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )}{\sqrt {-1+a^2}}-\frac {(6 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {i x}{a+\sqrt {-1+a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )}{\sqrt {-1+a^2}}\\ &=-\frac {\sin ^{-1}(a+b x)^3}{x}+\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {6 b \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {6 i b \text {Li}_3\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {6 i b \text {Li}_3\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}\\ \end {align*}
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Mathematica [A] time = 0.14, size = 309, normalized size = 0.98 \[ -\frac {-6 b x \sin ^{-1}(a+b x) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}-a}\right )+6 b x \sin ^{-1}(a+b x) \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )-6 i b x \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}-a}\right )+6 i b x \text {Li}_3\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )+\sqrt {a^2-1} \sin ^{-1}(a+b x)^3-3 i b x \sin ^{-1}(a+b x)^2 \log \left (\frac {-\sqrt {a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{a-\sqrt {a^2-1}}\right )+3 i b x \sin ^{-1}(a+b x)^2 \log \left (\frac {\sqrt {a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{\sqrt {a^2-1}+a}\right )}{\sqrt {a^2-1} x} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (b x + a\right )^{3}}{x^{2}}, x\right ) \]
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 \[ \int \frac {\arcsin \left (b x + a\right )^{3}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (b x +a \right )^{3}}{x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 \[ \int \frac {{\mathrm {asin}\left (a+b\,x\right )}^3}{x^2} \,d x \]
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 \[ \int \frac {\operatorname {asin}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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