Integrand size = 18, antiderivative size = 256 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=-\frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a-b \arcsin \left (1-d x^2\right )}}-\frac {\left (-\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}+\frac {\left (-\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )} \]
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Time = 0.04 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {4906} \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=-\frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a-b \arcsin \left (1-d x^2\right )}}-\frac {\sqrt {\pi } \left (-\frac {1}{b}\right )^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}+\frac {\sqrt {\pi } \left (-\frac {1}{b}\right )^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )} \]
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Rule 4906
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a-b \arcsin \left (1-d x^2\right )}}-\frac {\left (-\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\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 )}+\frac {\left (-\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\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 )} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=-\frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a-b \arcsin \left (1-d x^2\right )}}-\frac {\left (-\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}+\frac {\left (-\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )} \]
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\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{\frac {3}{2}}}d x\]
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Exception generated. \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^{3/2}} \,d x \]
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