Optimal. Leaf size=176 \[ -\frac{\text{Unintegrable}\left (\frac{1-c^2 x^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )},x\right )}{b c}-\frac{9 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{4 b^2}-\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2}-\frac{9 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{4 b^2}-\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2}-\frac{\left (1-c^2 x^2\right )^2}{b c x \left (a+b \sin ^{-1}(c x)\right )} \]
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Rubi [A] time = 0.403718, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\left (1-c^2 x^2\right )^{3/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin{align*} \int \frac{\left (1-c^2 x^2\right )^{3/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac{\left (1-c^2 x^2\right )^2}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{\int \frac{1-c^2 x^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}-\frac{(3 c) \int \frac{1-c^2 x^2}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac{\left (1-c^2 x^2\right )^2}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{3 \operatorname{Subst}\left (\int \frac{\cos ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b}-\frac{\int \frac{1-c^2 x^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1-c^2 x^2\right )^2}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{3 \operatorname{Subst}\left (\int \left (\frac{3 \cos (x)}{4 (a+b x)}+\frac{\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b}-\frac{\int \frac{1-c^2 x^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1-c^2 x^2\right )^2}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{3 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b}-\frac{9 \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b}-\frac{\int \frac{1-c^2 x^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1-c^2 x^2\right )^2}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{\int \frac{1-c^2 x^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}-\frac{\left (9 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b}-\frac{\left (3 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b}-\frac{\left (9 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b}-\frac{\left (3 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b}\\ &=-\frac{\left (1-c^2 x^2\right )^2}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{9 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b^2}-\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2}-\frac{9 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b^2}-\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2}-\frac{\int \frac{1-c^2 x^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ \end{align*}
Mathematica [A] time = 10.3348, size = 0, normalized size = 0. \[ \int \frac{\left (1-c^2 x^2\right )^{3/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.343, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}} \left ( -{c}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{c^{4} x^{4} - 2 \, c^{2} x^{2} - \frac{{\left (b^{2} c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a b c x\right )}{\left (3 \, c^{4} \int \frac{x^{4}}{b x^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a x^{2}}\,{d x} - 2 \, c^{2} \int \frac{x^{2}}{b x^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a x^{2}}\,{d x} - \int \frac{1}{b x^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a x^{2}}\,{d x}\right )}}{b c} + 1}{b^{2} c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a b c x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{b^{2} x \arcsin \left (c x\right )^{2} + 2 \, a b x \arcsin \left (c x\right ) + a^{2} x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{3}{2}}}{x \left (a + b \operatorname{asin}{\left (c x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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