Integrand size = 12, antiderivative size = 129 \[ \int e^{\arcsin (a x)^2} x^2 \, dx=\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-i+2 \arcsin (a x))\right )}{16 a^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (i+2 \arcsin (a x))\right )}{16 a^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-3 i+2 \arcsin (a x))\right )}{16 a^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (3 i+2 \arcsin (a x))\right )}{16 a^3} \]
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Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4920, 12, 4562, 2266, 2235} \[ \int e^{\arcsin (a x)^2} x^2 \, dx=\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 \arcsin (a x)-i)\right )}{16 a^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 \arcsin (a x)+i)\right )}{16 a^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 \arcsin (a x)-3 i)\right )}{16 a^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 \arcsin (a x)+3 i)\right )}{16 a^3} \]
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Rule 12
Rule 2235
Rule 2266
Rule 4562
Rule 4920
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {e^{x^2} \cos (x) \sin ^2(x)}{a^2} \, dx,x,\arcsin (a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int e^{x^2} \cos (x) \sin ^2(x) \, dx,x,\arcsin (a x)\right )}{a^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{8} e^{-i x+x^2}+\frac {1}{8} e^{i x+x^2}-\frac {1}{8} e^{-3 i x+x^2}-\frac {1}{8} e^{3 i x+x^2}\right ) \, dx,x,\arcsin (a x)\right )}{a^3} \\ & = \frac {\text {Subst}\left (\int e^{-i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^3}+\frac {\text {Subst}\left (\int e^{i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^3}-\frac {\text {Subst}\left (\int e^{-3 i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^3}-\frac {\text {Subst}\left (\int e^{3 i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^3} \\ & = \frac {\sqrt [4]{e} \text {Subst}\left (\int e^{\frac {1}{4} (-i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^3}+\frac {\sqrt [4]{e} \text {Subst}\left (\int e^{\frac {1}{4} (i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^3}-\frac {e^{9/4} \text {Subst}\left (\int e^{\frac {1}{4} (-3 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^3}-\frac {e^{9/4} \text {Subst}\left (\int e^{\frac {1}{4} (3 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^3} \\ & = \frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-i+2 \arcsin (a x))\right )}{16 a^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (i+2 \arcsin (a x))\right )}{16 a^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-3 i+2 \arcsin (a x))\right )}{16 a^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (3 i+2 \arcsin (a x))\right )}{16 a^3} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.65 \[ \int e^{\arcsin (a x)^2} x^2 \, dx=\frac {\sqrt [4]{e} \sqrt {\pi } \left (\text {erfi}\left (\frac {1}{2} (-i+2 \arcsin (a x))\right )+\text {erfi}\left (\frac {1}{2} (i+2 \arcsin (a x))\right )-e^2 \left (\text {erfi}\left (\frac {1}{2} (-3 i+2 \arcsin (a x))\right )+\text {erfi}\left (\frac {1}{2} (3 i+2 \arcsin (a x))\right )\right )\right )}{16 a^3} \]
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\[\int {\mathrm e}^{\arcsin \left (a x \right )^{2}} x^{2}d x\]
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\[ \int e^{\arcsin (a x)^2} x^2 \, dx=\int { x^{2} e^{\left (\arcsin \left (a x\right )^{2}\right )} \,d x } \]
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\[ \int e^{\arcsin (a x)^2} x^2 \, dx=\int x^{2} e^{\operatorname {asin}^{2}{\left (a x \right )}}\, dx \]
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\[ \int e^{\arcsin (a x)^2} x^2 \, dx=\int { x^{2} e^{\left (\arcsin \left (a x\right )^{2}\right )} \,d x } \]
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\[ \int e^{\arcsin (a x)^2} x^2 \, dx=\int { x^{2} e^{\left (\arcsin \left (a x\right )^{2}\right )} \,d x } \]
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Timed out. \[ \int e^{\arcsin (a x)^2} x^2 \, dx=\int x^2\,{\mathrm {e}}^{{\mathrm {asin}\left (a\,x\right )}^2} \,d x \]
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