Optimal. Leaf size=316 \[ \frac{6 b \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{a^2-1}}\right )}{\sqrt{a^2-1}}-\frac{6 b \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )}{\sqrt{a^2-1}}+\frac{6 i b \text{PolyLog}\left (3,-\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{a^2-1}}\right )}{\sqrt{a^2-1}}-\frac{6 i b \text{PolyLog}\left (3,-\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )}{\sqrt{a^2-1}}+\frac{3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{a^2-1}}\right )}{\sqrt{a^2-1}}-\frac{3 i b \sin ^{-1}(a+b x)^2 \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )}{\sqrt{a^2-1}}-\frac{\sin ^{-1}(a+b x)^3}{x} \]
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Rubi [A] time = 0.633776, antiderivative size = 316, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, 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 4805
Rule 4743
Rule 4773
Rule 3323
Rule 2264
Rule 2190
Rule 2531
Rule 2282
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*}
Mathematica [A] time = 0.112617, size = 309, normalized size = 0.98 \[ -\frac{-6 b x \sin ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}-a}\right )+6 b x \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )-6 i b x \text{PolyLog}\left (3,\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}-a}\right )+6 i b x \text{PolyLog}\left (3,-\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )+\sqrt{a^2-1} \sin ^{-1}(a+b x)^3-3 i b x \sin ^{-1}(a+b x)^2 \log \left (\frac{-\sqrt{a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{a-\sqrt{a^2-1}}\right )+3 i b x \sin ^{-1}(a+b x)^2 \log \left (\frac{\sqrt{a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{\sqrt{a^2-1}+a}\right )}{\sqrt{a^2-1} x} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.545, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}}{{x}^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arcsin \left (b x + a\right )^{3}}{x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asin}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (b x + a\right )^{3}}{x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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