Integrand size = 10, antiderivative size = 87 \[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=(1+i) a e^{(1+i) \arccos (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},-e^{2 i \arccos (a x)}\right )-(2+2 i) a e^{(1+i) \arccos (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},-e^{2 i \arccos (a x)}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 87, 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 = {4921, 12, 4559, 2283} \[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=(1+i) a e^{(1+i) \arccos (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},-e^{2 i \arccos (a x)}\right )-(2+2 i) a e^{(1+i) \arccos (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},-e^{2 i \arccos (a x)}\right ) \]
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Rule 12
Rule 2283
Rule 4559
Rule 4921
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int a^2 e^x \sec (x) \tan (x) \, dx,x,\arccos (a x)\right )}{a} \\ & = -\left (a \text {Subst}\left (\int e^x \sec (x) \tan (x) \, dx,x,\arccos (a x)\right )\right ) \\ & = -\left (a \text {Subst}\left (\int \left (\frac {4 i e^{(1+i) x}}{\left (1+e^{2 i x}\right )^2}-\frac {2 i e^{(1+i) x}}{1+e^{2 i x}}\right ) \, dx,x,\arccos (a x)\right )\right ) \\ & = (2 i a) \text {Subst}\left (\int \frac {e^{(1+i) x}}{1+e^{2 i x}} \, dx,x,\arccos (a x)\right )-(4 i a) \text {Subst}\left (\int \frac {e^{(1+i) x}}{\left (1+e^{2 i x}\right )^2} \, dx,x,\arccos (a x)\right ) \\ & = (1+i) a e^{(1+i) \arccos (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},-e^{2 i \arccos (a x)}\right )-(2+2 i) a e^{(1+i) \arccos (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},2,\frac {3}{2}-\frac {i}{2},-e^{2 i \arccos (a x)}\right ) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63 \[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=-\frac {e^{\arccos (a x)}}{x}+(1-i) a e^{(1+i) \arccos (a x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},-e^{2 i \arccos (a x)}\right ) \]
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\[\int \frac {{\mathrm e}^{\arccos \left (a x \right )}}{x^{2}}d x\]
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\[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=\int { \frac {e^{\left (\arccos \left (a x\right )\right )}}{x^{2}} \,d x } \]
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\[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=\int \frac {e^{\operatorname {acos}{\left (a x \right )}}}{x^{2}}\, dx \]
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\[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=\int { \frac {e^{\left (\arccos \left (a x\right )\right )}}{x^{2}} \,d x } \]
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\[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=\int { \frac {e^{\left (\arccos \left (a x\right )\right )}}{x^{2}} \,d x } \]
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Timed out. \[ \int \frac {e^{\arccos (a x)}}{x^2} \, dx=\int \frac {{\mathrm {e}}^{\mathrm {acos}\left (a\,x\right )}}{x^2} \,d x \]
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