Integrand size = 28, antiderivative size = 28 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=-\frac {c \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{b}-\frac {c \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 b}-\frac {15 c \log (a+b \arcsin (c x))}{8 b}-\frac {c \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{b}-\frac {c \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 b}+\text {Int}\left (\frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))},x\right ) \]
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Not integrable
Time = 0.54 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=\int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {3 c^2}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}+\frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}+\frac {3 c^4 x^2}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}-\frac {c^6 x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}\right ) \, dx \\ & = -\left (\left (3 c^2\right ) \int \frac {1}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx\right )+\left (3 c^4\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx-c^6 \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = -\frac {3 c \log (a+b \arcsin (c x))}{b}-\frac {c \text {Subst}\left (\int \frac {\sin ^4\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{b}+\frac {(3 c) \text {Subst}\left (\int \frac {\sin ^2\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{b}+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = -\frac {3 c \log (a+b \arcsin (c x))}{b}-\frac {c \text {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cos \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{8 x}-\frac {\cos \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{2 x}\right ) \, dx,x,a+b \arcsin (c x)\right )}{b}+\frac {(3 c) \text {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cos \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{2 x}\right ) \, dx,x,a+b \arcsin (c x)\right )}{b}+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = -\frac {15 c \log (a+b \arcsin (c x))}{8 b}-\frac {c \text {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{8 b}+\frac {c \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{2 b}-\frac {(3 c) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{2 b}+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = -\frac {15 c \log (a+b \arcsin (c x))}{8 b}+\frac {\left (c \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{2 b}-\frac {\left (3 c \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{2 b}-\frac {\left (c \cos \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{8 b}+\frac {\left (c \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{2 b}-\frac {\left (3 c \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{2 b}-\frac {\left (c \sin \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c x)\right )}{8 b}+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ & = -\frac {c \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{b}-\frac {c \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 b}-\frac {15 c \log (a+b \arcsin (c x))}{8 b}-\frac {c \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{b}-\frac {c \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 b}+\int \frac {1}{x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx \\ \end{align*}
Not integrable
Time = 2.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=\int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx \]
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Not integrable
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
\[\int \frac {\left (-c^{2} x^{2}+1\right )^{\frac {5}{2}}}{x^{2} \left (a +b \arcsin \left (c x \right )\right )}d x\]
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Not integrable
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}} \,d x } \]
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Not integrable
Time = 3.87 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=\int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{x^{2} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}\, dx \]
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Not integrable
Time = 0.50 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}} \,d x } \]
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Not integrable
Time = 0.56 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}} \,d x } \]
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Not integrable
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{x^2 (a+b \arcsin (c x))} \, dx=\int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )} \,d x \]
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