Optimal. Leaf size=103 \[ -\frac{a b^2 \tanh ^{-1}\left (\frac{1-a (a+b x)}{\sqrt{1-a^2} \sqrt{1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )^{3/2}}-\frac{b \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac{\sin ^{-1}(a+b x)}{2 x^2} \]
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Rubi [A] time = 0.116231, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4805, 4743, 731, 725, 206} \[ -\frac{a b^2 \tanh ^{-1}\left (\frac{1-a (a+b x)}{\sqrt{1-a^2} \sqrt{1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )^{3/2}}-\frac{b \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac{\sin ^{-1}(a+b x)}{2 x^2} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 4743
Rule 731
Rule 725
Rule 206
Rubi steps
\begin{align*} \int \frac{\sin ^{-1}(a+b x)}{x^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right )^3} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sin ^{-1}(a+b x)}{2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac{b \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac{\sin ^{-1}(a+b x)}{2 x^2}+\frac{(a b) \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right ) \sqrt{1-x^2}} \, dx,x,a+b x\right )}{2 \left (1-a^2\right )}\\ &=-\frac{b \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac{\sin ^{-1}(a+b x)}{2 x^2}-\frac{(a b) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{b^2}-\frac{a^2}{b^2}-x^2} \, dx,x,\frac{\frac{1}{b}-\frac{a (a+b x)}{b}}{\sqrt{1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )}\\ &=-\frac{b \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac{\sin ^{-1}(a+b x)}{2 x^2}-\frac{a b^2 \tanh ^{-1}\left (\frac{1-a (a+b x)}{\sqrt{1-a^2} \sqrt{1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.199077, size = 125, normalized size = 1.21 \[ -\frac{\frac{b x \left (\sqrt{1-a^2} \sqrt{-a^2-2 a b x-b^2 x^2+1}+a b x \log \left (\sqrt{1-a^2} \sqrt{-a^2-2 a b x-b^2 x^2+1}-a^2-a b x+1\right )-a b x \log (x)\right )}{\left (1-a^2\right )^{3/2}}+\sin ^{-1}(a+b x)}{2 x^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.007, size = 118, normalized size = 1.2 \begin{align*} -{\frac{\arcsin \left ( bx+a \right ) }{2\,{x}^{2}}}-{\frac{b}{ \left ( -2\,{a}^{2}+2 \right ) x}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{{b}^{2}a}{2}\ln \left ({\frac{1}{bx} \left ( -2\,{a}^{2}+2-2\,xab+2\,\sqrt{-{a}^{2}+1}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \right ) \left ( -{a}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \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 [A] time = 3.43971, size = 765, normalized size = 7.43 \begin{align*} \left [-\frac{\sqrt{-a^{2} + 1} a b^{2} x^{2} \log \left (\frac{{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \,{\left (a^{3} - a\right )} b x - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a^{2} - 1\right )} b x + 2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} \arcsin \left (b x + a\right )}{4 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}, -\frac{\sqrt{a^{2} - 1} a b^{2} x^{2} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \,{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a^{2} - 1\right )} b x +{\left (a^{4} - 2 \, a^{2} + 1\right )} \arcsin \left (b x + a\right )}{2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}\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}{\left (a + b x \right )}}{x^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.22368, size = 328, normalized size = 3.18 \begin{align*} -{\left (\frac{a b^{2} \arctan \left (\frac{\frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt{a^{2} - 1}}\right )}{{\left (a^{2}{\left | b \right |} -{\left | b \right |}\right )} \sqrt{a^{2} - 1}} - \frac{a b^{2} - \frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )} b^{2}}{b^{2} x + a b}}{{\left (a^{3}{\left | b \right |} - a{\left | b \right |}\right )}{\left (\frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + a - \frac{2 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}}{b^{2} x + a b}\right )}}\right )} b - \frac{\arcsin \left (b x + a\right )}{2 \, x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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