Optimal. Leaf size=271 \[ -2 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )-2 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+2 \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+2 \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )-\frac{1}{3} i \sin ^{-1}(a+b x)^3 \]
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Rubi [A] time = 0.410025, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {4805, 4741, 4521, 2190, 2531, 2282, 6589} \[ -2 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )-2 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+2 \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+2 \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )-\frac{1}{3} i \sin ^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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Rule 4805
Rule 4741
Rule 4521
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\sin ^{-1}(a+b x)^2}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)^2}{-\frac{a}{b}+\frac{x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \cos (x)}{-\frac{a}{b}+\frac{\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{1}{3} i \sin ^{-1}(a+b x)^3+\frac{i \operatorname{Subst}\left (\int \frac{e^{i x} x^2}{-\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 \operatorname{Subst}\left (\int \frac{e^{i x} x^2}{-\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}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )-2 \operatorname{Subst}\left (\int x \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 )-2 \operatorname{Subst}\left (\int x \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}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )+2 i \operatorname{Subst}\left (\int \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 )+2 i \operatorname{Subst}\left (\int \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}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )+2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{x}{i a-\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )+2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{x}{i a+\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )\\ &=-\frac{1}{3} i \sin ^{-1}(a+b x)^3+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+\sin ^{-1}(a+b x)^2 \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )-2 i \sin ^{-1}(a+b x) \text{Li}_2\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )+2 \text{Li}_3\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt{1-a^2}}\right )+2 \text{Li}_3\left (\frac{e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt{1-a^2}}\right )\\ \end{align*}
Mathematica [A] time = 0.0354203, size = 309, normalized size = 1.14 \[ -2 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )-2 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )+2 \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )+2 \text{PolyLog}\left (3,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )+\sin ^{-1}(a+b x)^2 \log \left (1+\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )+\sin ^{-1}(a+b x)^2 \log \left (1+\frac{e^{i \sin ^{-1}(a+b x)}}{b \left (\frac{\sqrt{1-a^2}}{b}-\frac{i a}{b}\right )}\right )-\frac{1}{3} i \sin ^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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Maple [F] time = 0.84, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}{x}}\, 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 )^{2}}{x}, 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}^{2}{\left (a + b x \right )}}{x}\, 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 )^{2}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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