Integrand size = 16, antiderivative size = 240 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {x \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \]
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Time = 0.03 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4912, 4900} \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=-\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {x}{8 b^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^2} \]
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Rule 4900
Rule 4912
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {\int \frac {1}{a-b \arcsin \left (1-d x^2\right )} \, dx}{8 b^2} \\ & = -\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {x \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \\ \end{align*}
Time = 0.48 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=-\frac {\frac {4 b^2 \sqrt {-d x^2 \left (-2+d x^2\right )}}{d \left (a-b \arcsin \left (1-d x^2\right )\right )^2}-\frac {2 b x^2}{a-b \arcsin \left (1-d x^2\right )}+\frac {\left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right ) \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (-\frac {a}{b}+\arcsin \left (1-d x^2\right )\right )\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )+\left (-\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )\right )}{d}}{16 b^3 x} \]
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\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{3}}d x\]
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\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}} \,d x } \]
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\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{3}}\, dx \]
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\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}} \,d x } \]
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\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^3} \,d x \]
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