Optimal. Leaf size=161 \[ \text {Int}\left (\frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )},x\right )-\frac {c \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b}-\frac {c \cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b}-\frac {c \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b}-\frac {c \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b}-\frac {15 c \log \left (a+b \sin ^{-1}(c x)\right )}{8 b} \]
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Rubi [A] time = 0.93, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx &=\int \left (-\frac {3 c^2}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}+\frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}+\frac {3 c^4 x^2}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}-\frac {c^6 x^4}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}\right ) \, dx\\ &=-\left (\left (3 c^2\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\right )+\left (3 c^4\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx-c^6 \int \frac {x^4}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac {3 c \log \left (a+b \sin ^{-1}(c x)\right )}{b}-c \operatorname {Subst}\left (\int \frac {\sin ^4(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+(3 c) \operatorname {Subst}\left (\int \frac {\sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac {3 c \log \left (a+b \sin ^{-1}(c x)\right )}{b}-c \operatorname {Subst}\left (\int \left (\frac {3}{8 (a+b x)}-\frac {\cos (2 x)}{2 (a+b x)}+\frac {\cos (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )+(3 c) \operatorname {Subst}\left (\int \left (\frac {1}{2 (a+b x)}-\frac {\cos (2 x)}{2 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac {15 c \log \left (a+b \sin ^{-1}(c x)\right )}{8 b}-\frac {1}{8} c \operatorname {Subst}\left (\int \frac {\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\frac {1}{2} c \operatorname {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )-\frac {1}{2} (3 c) \operatorname {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac {15 c \log \left (a+b \sin ^{-1}(c x)\right )}{8 b}+\frac {1}{2} \left (c \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )-\frac {1}{2} \left (3 c \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )-\frac {1}{8} \left (c \cos \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\frac {1}{2} \left (c \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )-\frac {1}{2} \left (3 c \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )-\frac {1}{8} \left (c \sin \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac {c \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b}-\frac {c \cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b}-\frac {15 c \log \left (a+b \sin ^{-1}(c x)\right )}{8 b}-\frac {c \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b}-\frac {c \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b}+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ \end {align*}
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Mathematica [A] time = 1.43, size = 0, normalized size = 0.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx \]
Verification is Not applicable to the result.
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fricas [A] time = 1.15, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt {-c^{2} x^{2} + 1}}{b x^{2} \arcsin \left (c x\right ) + a x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.61, size = 0, normalized size = 0.00 \[ \int \frac {\left (-c^{2} x^{2}+1\right )^{\frac {5}{2}}}{x^{2} \left (a +b \arcsin \left (c x \right )\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [A] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{x^{2} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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