Optimal. Leaf size=176 \[ -\frac{\left (4 a^2+1\right ) \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}+\frac{9 \text{CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{2 a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac{2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}-\frac{3 \sin \left (3 \sin ^{-1}(a+b x)\right )}{8 b^3 \sin ^{-1}(a+b x)}+\frac{9 a+b x}{8 b^3 \sin ^{-1}(a+b x)}-\frac{x^2 \sqrt{1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]
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Rubi [A] time = 0.506512, antiderivative size = 263, normalized size of antiderivative = 1.49, number of steps used = 24, number of rules used = 12, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {4805, 4745, 4621, 4719, 4623, 3302, 4633, 4635, 4406, 12, 3299, 4641} \[ -\frac{a^2 \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac{a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac{a^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}-\frac{\text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{9 \text{CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{2 a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}+\frac{3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}-\frac{2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}-\frac{\sqrt{1-(a+b x)^2} (a+b x)^2}{2 b^3 \sin ^{-1}(a+b x)^2}-\frac{a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac{a \sqrt{1-(a+b x)^2} (a+b x)}{b^3 \sin ^{-1}(a+b x)^2}+\frac{a}{b^3 \sin ^{-1}(a+b x)} \]
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^2}{\sin ^{-1}(a+b x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^2}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2}{b^2 \sin ^{-1}(x)^3}-\frac{2 a x}{b^2 \sin ^{-1}(x)^3}+\frac{x^2}{b^2 \sin ^{-1}(x)^3}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^3}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^3}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^3}\\ &=-\frac{a^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a (a+b x) \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac{(a+b x)^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}-\frac{3 \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^3}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^3}-\frac{a^2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^3}\\ &=-\frac{a^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a (a+b x) \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac{(a+b x)^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a}{b^3 \sin ^{-1}(a+b x)}-\frac{a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac{a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac{2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac{3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^3}-\frac{9 \operatorname{Subst}\left (\int \frac{x^2}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{2 b^3}+\frac{(4 a) \operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b^3}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{2 b^3}\\ &=-\frac{a^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a (a+b x) \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac{(a+b x)^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a}{b^3 \sin ^{-1}(a+b x)}-\frac{a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac{a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac{2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac{3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac{\operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac{9 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^2(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac{(4 a) \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac{a^2 \operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}\\ &=-\frac{a^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a (a+b x) \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac{(a+b x)^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a}{b^3 \sin ^{-1}(a+b x)}-\frac{a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac{a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac{2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac{3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac{\text{Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac{a^2 \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac{9 \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{4 x}-\frac{\cos (3 x)}{4 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac{(4 a) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac{a^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a (a+b x) \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac{(a+b x)^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a}{b^3 \sin ^{-1}(a+b x)}-\frac{a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac{a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac{2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac{3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}+\frac{\text{Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac{a^2 \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac{9 \operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac{a^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a (a+b x) \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)^2}-\frac{(a+b x)^2 \sqrt{1-(a+b x)^2}}{2 b^3 \sin ^{-1}(a+b x)^2}+\frac{a}{b^3 \sin ^{-1}(a+b x)}-\frac{a+b x}{b^3 \sin ^{-1}(a+b x)}+\frac{a^2 (a+b x)}{2 b^3 \sin ^{-1}(a+b x)}-\frac{2 a (a+b x)^2}{b^3 \sin ^{-1}(a+b x)}+\frac{3 (a+b x)^3}{2 b^3 \sin ^{-1}(a+b x)}-\frac{\text{Ci}\left (\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac{a^2 \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac{9 \text{Ci}\left (3 \sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{2 a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}\\ \end{align*}
Mathematica [A] time = 0.511304, size = 115, normalized size = 0.65 \[ \frac{\frac{4 b x \left (\left (2 a^2+5 a b x+3 b^2 x^2-2\right ) \sin ^{-1}(a+b x)-b x \sqrt{-a^2-2 a b x-b^2 x^2+1}\right )}{\sin ^{-1}(a+b x)^2}-\left (4 a^2+1\right ) \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )+9 \text{CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )+16 a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{8 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.058, size = 215, normalized size = 1.2 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{a \left ( 4\,{\it Si} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}+2\,\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) +\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \right ) }{2\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}}-{\frac{1}{8\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}\sqrt{1- \left ( bx+a \right ) ^{2}}}+{\frac{bx+a}{8\,\arcsin \left ( bx+a \right ) }}-{\frac{{\it Ci} \left ( \arcsin \left ( bx+a \right ) \right ) }{8}}+{\frac{\cos \left ( 3\,\arcsin \left ( bx+a \right ) \right ) }{8\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}}-{\frac{3\,\sin \left ( 3\,\arcsin \left ( bx+a \right ) \right ) }{8\,\arcsin \left ( bx+a \right ) }}+{\frac{9\,{\it Ci} \left ( 3\,\arcsin \left ( bx+a \right ) \right ) }{8}}-{\frac{{a}^{2}}{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^{2}}{\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^{2}}{\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 = 1.26139, size = 367, normalized size = 2.09 \begin{align*} -\frac{a^{2} \operatorname{Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b^{3}} + \frac{{\left (b x + a\right )} a^{2}}{2 \, b^{3} \arcsin \left (b x + a\right )} + \frac{2 \, a \operatorname{Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} + \frac{3 \,{\left ({\left (b x + a\right )}^{2} - 1\right )}{\left (b x + a\right )}}{2 \, b^{3} \arcsin \left (b x + a\right )} - \frac{2 \,{\left ({\left (b x + a\right )}^{2} - 1\right )} a}{b^{3} \arcsin \left (b x + a\right )} + \frac{9 \, \operatorname{Ci}\left (3 \, \arcsin \left (b x + a\right )\right )}{8 \, b^{3}} - \frac{\operatorname{Ci}\left (\arcsin \left (b x + a\right )\right )}{8 \, b^{3}} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} a}{b^{3} \arcsin \left (b x + a\right )^{2}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a^{2}}{2 \, b^{3} \arcsin \left (b x + a\right )^{2}} + \frac{b x + a}{2 \, b^{3} \arcsin \left (b x + a\right )} - \frac{a}{b^{3} \arcsin \left (b x + a\right )} + \frac{{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}}{2 \, b^{3} \arcsin \left (b x + a\right )^{2}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{2 \, b^{3} \arcsin \left (b x + a\right )^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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