Integrand size = 10, antiderivative size = 83 \[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=(1-i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right )-(2-2 i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4920, 12, 4559, 2283} \[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=(1-i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right )-(2-2 i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right ) \]
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Rule 12
Rule 2283
Rule 4559
Rule 4920
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int a^2 e^x \cot (x) \csc (x) \, dx,x,\arcsin (a x)\right )}{a} \\ & = a \text {Subst}\left (\int e^x \cot (x) \csc (x) \, dx,x,\arcsin (a x)\right ) \\ & = a \text {Subst}\left (\int \left (\frac {2 e^{(1+i) x}}{1-e^{2 i x}}-\frac {4 e^{(1+i) x}}{\left (-1+e^{2 i x}\right )^2}\right ) \, dx,x,\arcsin (a x)\right ) \\ & = (2 a) \text {Subst}\left (\int \frac {e^{(1+i) x}}{1-e^{2 i x}} \, dx,x,\arcsin (a x)\right )-(4 a) \text {Subst}\left (\int \frac {e^{(1+i) x}}{\left (-1+e^{2 i x}\right )^2} \, dx,x,\arcsin (a x)\right ) \\ & = (1-i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right )-(2-2 i) a e^{(1+i) \arcsin (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right ) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=-\frac {e^{\arcsin (a x)}+(1+i) a e^{(1+i) \arcsin (a x)} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},e^{2 i \arcsin (a x)}\right )}{x} \]
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\[\int \frac {{\mathrm e}^{\arcsin \left (a x \right )}}{x^{2}}d x\]
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\[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{x^{2}} \,d x } \]
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\[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=\int \frac {e^{\operatorname {asin}{\left (a x \right )}}}{x^{2}}\, dx \]
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\[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{x^{2}} \,d x } \]
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\[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=\int { \frac {e^{\left (\arcsin \left (a x\right )\right )}}{x^{2}} \,d x } \]
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Timed out. \[ \int \frac {e^{\arcsin (a x)}}{x^2} \, dx=\int \frac {{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )}}{x^2} \,d x \]
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