Optimal. Leaf size=230 \[ \frac{2 i b \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )}{\sqrt{1-a^2}}-\frac{2 i b \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )}{\sqrt{1-a^2}}-\frac{2 b \sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{-\sqrt{1-a^2}+i a}\right )}{\sqrt{1-a^2}}+\frac{2 b \sin ^{-1}(a+b x) \log \left (1-\frac{e^{i \sin ^{-1}(a+b x)}}{\sqrt{1-a^2}+i a}\right )}{\sqrt{1-a^2}}-\frac{\sin ^{-1}(a+b x)^2}{x} \]
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Rubi [A] time = 0.446628, antiderivative size = 208, normalized size of antiderivative = 0.9, number of steps used = 11, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4805, 4743, 4773, 3323, 2264, 2190, 2279, 2391} \[ \frac{2 b \text{PolyLog}\left (2,-\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{a^2-1}}\right )}{\sqrt{a^2-1}}-\frac{2 b \text{PolyLog}\left (2,-\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )}{\sqrt{a^2-1}}+\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{a^2-1}}\right )}{\sqrt{a^2-1}}-\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )}{\sqrt{a^2-1}}-\frac{\sin ^{-1}(a+b x)^2}{x} \]
Warning: Unable to verify antiderivative.
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Rule 4805
Rule 4743
Rule 4773
Rule 3323
Rule 2264
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\sin ^{-1}(a+b x)^2}{x^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)^2}{\left (-\frac{a}{b}+\frac{x}{b}\right )^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sin ^{-1}(a+b x)^2}{x}+2 \operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right ) \sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac{\sin ^{-1}(a+b x)^2}{x}+2 \operatorname{Subst}\left (\int \frac{x}{-\frac{a}{b}+\frac{\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac{\sin ^{-1}(a+b x)^2}{x}+4 \operatorname{Subst}\left (\int \frac{e^{i x} x}{\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)^2}{x}-\frac{(4 i) \operatorname{Subst}\left (\int \frac{e^{i x} x}{-\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{(4 i) \operatorname{Subst}\left (\int \frac{e^{i x} x}{-\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)^2}{x}+\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{-1+a^2}}\right )}{\sqrt{-1+a^2}}-\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt{-1+a^2}}\right )}{\sqrt{-1+a^2}}+\frac{(2 i b) \operatorname{Subst}\left (\int \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{(2 i b) \operatorname{Subst}\left (\int \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)^2}{x}+\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{-1+a^2}}\right )}{\sqrt{-1+a^2}}-\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt{-1+a^2}}\right )}{\sqrt{-1+a^2}}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 i x}{\left (-\frac{2 a}{b}-\frac{2 \sqrt{-1+a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )}{\sqrt{-1+a^2}}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 i x}{\left (-\frac{2 a}{b}+\frac{2 \sqrt{-1+a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \sin ^{-1}(a+b x)}\right )}{\sqrt{-1+a^2}}\\ &=-\frac{\sin ^{-1}(a+b x)^2}{x}+\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{-1+a^2}}\right )}{\sqrt{-1+a^2}}-\frac{2 i b \sin ^{-1}(a+b x) \log \left (1+\frac{i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt{-1+a^2}}\right )}{\sqrt{-1+a^2}}+\frac{2 b \text{Li}_2\left (-\frac{i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt{-1+a^2}}\right )}{\sqrt{-1+a^2}}-\frac{2 b \text{Li}_2\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 = 0.130079, size = 208, normalized size = 0.9 \[ \frac{2 b x \text{PolyLog}\left (2,\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}-a}\right )-2 b x \text{PolyLog}\left (2,-\frac{i e^{i \sin ^{-1}(a+b x)}}{\sqrt{a^2-1}+a}\right )-\sqrt{a^2-1} \sin ^{-1}(a+b x)^2+2 i b x \sin ^{-1}(a+b x) \left (\log \left (\frac{-\sqrt{a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{a-\sqrt{a^2-1}}\right )-\log \left (\frac{\sqrt{a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{\sqrt{a^2-1}+a}\right )\right )}{\sqrt{a^2-1} x} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.282, size = 333, normalized size = 1.5 \begin{align*} -{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}{x}}-2\,{\frac{b\sqrt{-{a}^{2}+1}\arcsin \left ( bx+a \right ) }{{a}^{2}-1}\ln \left ({\frac{ia+\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}}}{ia+\sqrt{-{a}^{2}+1}}} \right ) }+2\,{\frac{b\sqrt{-{a}^{2}+1}\arcsin \left ( bx+a \right ) }{{a}^{2}-1}\ln \left ({\frac{ia-\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}}}{ia-\sqrt{-{a}^{2}+1}}} \right ) }+{\frac{2\,ib}{{a}^{2}-1}\sqrt{-{a}^{2}+1}{\it dilog} \left ({ \left ( ia+\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }-{\frac{2\,ib}{{a}^{2}-1}\sqrt{-{a}^{2}+1}{\it dilog} \left ({ \left ( ia-\sqrt{-{a}^{2}+1}-i \left ( bx+a \right ) -\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) \left ( ia-\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) } \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^{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}^{2}{\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 )^{2}}{x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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