Integrand size = 10, antiderivative size = 99 \[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=-\frac {\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},3,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}+\frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},4,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \]
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Time = 0.09 (sec) , antiderivative size = 99, 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 = {5374, 12, 4559, 2283} \[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},4,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}-\frac {\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},3,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \]
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Rule 12
Rule 2283
Rule 4559
Rule 5374
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {e^x \sec ^3(x) \tan (x)}{a^2} \, dx,x,\sec ^{-1}(a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int e^x \sec ^3(x) \tan (x) \, dx,x,\sec ^{-1}(a x)\right )}{a^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {16 i e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^4}-\frac {8 i e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^3}\right ) \, dx,x,\sec ^{-1}(a x)\right )}{a^3} \\ & = -\frac {(8 i) \text {Subst}\left (\int \frac {e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^3} \, dx,x,\sec ^{-1}(a x)\right )}{a^3}+\frac {(16 i) \text {Subst}\left (\int \frac {e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^4} \, dx,x,\sec ^{-1}(a x)\right )}{a^3} \\ & = -\frac {\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},3,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}+\frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},4,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.96 \[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\frac {e^{\sec ^{-1}(a x)} \left ((-4-4 i) \left (-i+a \sqrt {1-\frac {1}{a^2 x^2}} x\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )+a^4 x^4 \left (5+\cos \left (2 \sec ^{-1}(a x)\right )-\sin \left (2 \sec ^{-1}(a x)\right )\right )\right )}{12 a^4 x} \]
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\[\int {\mathrm e}^{\operatorname {arcsec}\left (a x \right )} x^{2}d x\]
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\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int { x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )} \,d x } \]
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\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int x^{2} e^{\operatorname {asec}{\left (a x \right )}}\, dx \]
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\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int { x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )} \,d x } \]
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\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int { x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )} \,d x } \]
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Timed out. \[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int x^2\,{\mathrm {e}}^{\mathrm {acos}\left (\frac {1}{a\,x}\right )} \,d x \]
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