Optimal. Leaf size=108 \[ \frac{a \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac{\text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}-\frac{a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}-\frac{1-2 (a+b x)^2}{2 b^2 \sin ^{-1}(a+b x)}-\frac{x \sqrt{1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]
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Rubi [A] time = 0.269426, antiderivative size = 151, normalized size of antiderivative = 1.4, number of steps used = 14, number of rules used = 12, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.2, Rules used = {4805, 4745, 4621, 4719, 4623, 3302, 4633, 4635, 4406, 12, 3299, 4641} \[ \frac{a \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac{\text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}+\frac{(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}-\frac{a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}-\frac{\sqrt{1-(a+b x)^2} (a+b x)}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sin ^{-1}(a+b x)}+\frac{a \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 4745
Rule 4621
Rule 4719
Rule 4623
Rule 3302
Rule 4633
Rule 4635
Rule 4406
Rule 12
Rule 3299
Rule 4641
Rubi steps
\begin{align*} \int \frac{x}{\sin ^{-1}(a+b x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-\frac{a}{b}+\frac{x}{b}}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b \sin ^{-1}(x)^3}+\frac{x}{b \sin ^{-1}(x)^3}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^2}\\ &=\frac{a \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^2}-\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^2}+\frac{a \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{a \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sin ^{-1}(a+b x)}-\frac{a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac{(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}-\frac{2 \operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^2}+\frac{a \operatorname{Subst}\left (\int \frac{1}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{a \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sin ^{-1}(a+b x)}-\frac{a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac{(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}-\frac{2 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}+\frac{a \operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac{a \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sin ^{-1}(a+b x)}-\frac{a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac{(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}+\frac{a \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac{2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{a \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sin ^{-1}(a+b x)}-\frac{a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac{(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}+\frac{a \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac{\operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{a \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sin ^{-1}(a+b x)}-\frac{a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}+\frac{(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}+\frac{a \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac{\text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}\\ \end{align*}
Mathematica [A] time = 0.10872, size = 121, normalized size = 1.12 \[ -\frac{x \sqrt{-a^2-2 a b x-b^2 x^2+1}}{2 b \sin ^{-1}(a+b x)^2}+\frac{a^2+3 a b x+2 b^2 x^2-1}{2 b^2 \sin ^{-1}(a+b x)}-2 \left (\frac{\text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{b^2}\right )-\frac{3 a \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.04, size = 109, normalized size = 1. \begin{align*}{\frac{1}{{b}^{2}} \left ( -{\frac{\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) }{4\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}}-{\frac{\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) }{2\,\arcsin \left ( bx+a \right ) }}-{\it Si} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) +{\frac{a}{2\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}} \left ({\it Ci} \left ( \arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}- \left ( bx+a \right ) \arcsin \left ( bx+a \right ) +\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{x}{\arcsin \left (b x + a\right )^{3}}, 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{x}{\operatorname{asin}^{3}{\left (a + b x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.21216, size = 188, normalized size = 1.74 \begin{align*} \frac{a \operatorname{Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b^{2}} - \frac{{\left (b x + a\right )} a}{2 \, b^{2} \arcsin \left (b x + a\right )} - \frac{\operatorname{Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{2}} + \frac{{\left (b x + a\right )}^{2} - 1}{b^{2} \arcsin \left (b x + a\right )} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )}}{2 \, b^{2} \arcsin \left (b x + a\right )^{2}} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a}{2 \, b^{2} \arcsin \left (b x + a\right )^{2}} + \frac{1}{2 \, b^{2} \arcsin \left (b x + a\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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