Integrand size = 12, antiderivative size = 101 \[ \int e^{\arcsin (a x)^2} x^3 \, dx=\frac {e \sqrt {\pi } \text {erf}(1-i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2-i \arcsin (a x))}{32 a^4}+\frac {e \sqrt {\pi } \text {erf}(1+i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2+i \arcsin (a x))}{32 a^4} \]
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Time = 0.09 (sec) , antiderivative size = 101, 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} \[ \int e^{\arcsin (a x)^2} x^3 \, dx=\frac {e \sqrt {\pi } \text {erf}(1-i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2-i \arcsin (a x))}{32 a^4}+\frac {e \sqrt {\pi } \text {erf}(1+i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2+i \arcsin (a x))}{32 a^4} \]
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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 ^3(x)}{a^3} \, dx,x,\arcsin (a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int e^{x^2} \cos (x) \sin ^3(x) \, dx,x,\arcsin (a x)\right )}{a^4} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{8} i e^{-2 i x+x^2}-\frac {1}{8} i e^{2 i x+x^2}-\frac {1}{16} i e^{-4 i x+x^2}+\frac {1}{16} i e^{4 i x+x^2}\right ) \, dx,x,\arcsin (a x)\right )}{a^4} \\ & = -\frac {i \text {Subst}\left (\int e^{-4 i x+x^2} \, dx,x,\arcsin (a x)\right )}{16 a^4}+\frac {i \text {Subst}\left (\int e^{4 i x+x^2} \, dx,x,\arcsin (a x)\right )}{16 a^4}+\frac {i \text {Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^4}-\frac {i \text {Subst}\left (\int e^{2 i x+x^2} \, dx,x,\arcsin (a x)\right )}{8 a^4} \\ & = \frac {(i e) \text {Subst}\left (\int e^{\frac {1}{4} (-2 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^4}-\frac {(i e) \text {Subst}\left (\int e^{\frac {1}{4} (2 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{8 a^4}-\frac {\left (i e^4\right ) \text {Subst}\left (\int e^{\frac {1}{4} (-4 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{16 a^4}+\frac {\left (i e^4\right ) \text {Subst}\left (\int e^{\frac {1}{4} (4 i+2 x)^2} \, dx,x,\arcsin (a x)\right )}{16 a^4} \\ & = \frac {e \sqrt {\pi } \text {erf}(1-i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2-i \arcsin (a x))}{32 a^4}+\frac {e \sqrt {\pi } \text {erf}(1+i \arcsin (a x))}{16 a^4}-\frac {e^4 \sqrt {\pi } \text {erf}(2+i \arcsin (a x))}{32 a^4} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int e^{\arcsin (a x)^2} x^3 \, dx=\frac {e \sqrt {\pi } \left (2 (\text {erf}(1-i \arcsin (a x))+\text {erf}(1+i \arcsin (a x)))-e^3 (\text {erf}(2-i \arcsin (a x))+\text {erf}(2+i \arcsin (a x)))\right )}{32 a^4} \]
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\[\int {\mathrm e}^{\arcsin \left (a x \right )^{2}} x^{3}d x\]
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\[ \int e^{\arcsin (a x)^2} x^3 \, dx=\int { x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )} \,d x } \]
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\[ \int e^{\arcsin (a x)^2} x^3 \, dx=\int x^{3} e^{\operatorname {asin}^{2}{\left (a x \right )}}\, dx \]
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\[ \int e^{\arcsin (a x)^2} x^3 \, dx=\int { x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )} \,d x } \]
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\[ \int e^{\arcsin (a x)^2} x^3 \, dx=\int { x^{3} e^{\left (\arcsin \left (a x\right )^{2}\right )} \,d x } \]
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Timed out. \[ \int e^{\arcsin (a x)^2} x^3 \, dx=\int x^3\,{\mathrm {e}}^{{\mathrm {asin}\left (a\,x\right )}^2} \,d x \]
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