Optimal. Leaf size=60 \[ \frac{a^2 \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac{\text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{\text{CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3} \]
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Rubi [A] time = 0.611634, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4805, 4747, 6741, 12, 6742, 3302, 4406, 3299} \[ \frac{a^2 \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac{\text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{\text{CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 4747
Rule 6741
Rule 12
Rule 6742
Rule 3302
Rule 4406
Rule 3299
Rubi steps
\begin{align*} \int \frac{x^2}{\sin ^{-1}(a+b x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^2}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cos (x) \left (-\frac{a}{b}+\frac{\sin (x)}{b}\right )^2}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cos (x) (a-\sin (x))^2}{b^2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cos (x) (a-\sin (x))^2}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2 \cos (x)}{x}-\frac{2 a \cos (x) \sin (x)}{x}+\frac{\cos (x) \sin ^2(x)}{x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cos (x) \sin ^2(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac{(2 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 )}{b^3}\\ &=\frac{a^2 \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac{\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 )}{b^3}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{a^2 \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac{\operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{\operatorname{Subst}\left (\int \frac{\cos (3 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{a \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{\text{Ci}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}+\frac{a^2 \text{Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac{\text{Ci}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}\\ \end{align*}
Mathematica [A] time = 0.155845, size = 45, normalized size = 0.75 \[ -\frac{-\left (4 a^2+1\right ) \text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )+\text{CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )+4 a \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{4 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.041, size = 49, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{3}} \left ( -a{\it Si} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) +{\frac{{\it Ci} \left ( \arcsin \left ( bx+a \right ) \right ) }{4}}-{\frac{{\it Ci} \left ( 3\,\arcsin \left ( bx+a \right ) \right ) }{4}}+{a}^{2}{\it Ci} \left ( \arcsin \left ( bx+a \right ) \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\arcsin \left (b x + a\right )}\,{d x} \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 )}, 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}{\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 = 1.26167, size = 76, normalized size = 1.27 \begin{align*} \frac{a^{2} \operatorname{Ci}\left (\arcsin \left (b x + a\right )\right )}{b^{3}} - \frac{a \operatorname{Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} - \frac{\operatorname{Ci}\left (3 \, \arcsin \left (b x + a\right )\right )}{4 \, b^{3}} + \frac{\operatorname{Ci}\left (\arcsin \left (b x + a\right )\right )}{4 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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