Integrand size = 10, antiderivative size = 81 \[ \int e^{\arcsin (a x)} x^3 \, dx=-\frac {e^{\arcsin (a x)} \cos (2 \arcsin (a x))}{10 a^4}+\frac {e^{\arcsin (a x)} \cos (4 \arcsin (a x))}{34 a^4}+\frac {e^{\arcsin (a x)} \sin (2 \arcsin (a x))}{20 a^4}-\frac {e^{\arcsin (a x)} \sin (4 \arcsin (a x))}{136 a^4} \]
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Time = 0.04 (sec) , antiderivative size = 81, 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, 4557, 4517} \[ \int e^{\arcsin (a x)} x^3 \, dx=\frac {e^{\arcsin (a x)} \sin (2 \arcsin (a x))}{20 a^4}-\frac {e^{\arcsin (a x)} \sin (4 \arcsin (a x))}{136 a^4}-\frac {e^{\arcsin (a x)} \cos (2 \arcsin (a x))}{10 a^4}+\frac {e^{\arcsin (a x)} \cos (4 \arcsin (a x))}{34 a^4} \]
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Rule 12
Rule 4517
Rule 4557
Rule 4920
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {e^x \cos (x) \sin ^3(x)}{a^3} \, dx,x,\arcsin (a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int e^x \cos (x) \sin ^3(x) \, dx,x,\arcsin (a x)\right )}{a^4} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{4} e^x \sin (2 x)-\frac {1}{8} e^x \sin (4 x)\right ) \, dx,x,\arcsin (a x)\right )}{a^4} \\ & = -\frac {\text {Subst}\left (\int e^x \sin (4 x) \, dx,x,\arcsin (a x)\right )}{8 a^4}+\frac {\text {Subst}\left (\int e^x \sin (2 x) \, dx,x,\arcsin (a x)\right )}{4 a^4} \\ & = -\frac {e^{\arcsin (a x)} \cos (2 \arcsin (a x))}{10 a^4}+\frac {e^{\arcsin (a x)} \cos (4 \arcsin (a x))}{34 a^4}+\frac {e^{\arcsin (a x)} \sin (2 \arcsin (a x))}{20 a^4}-\frac {e^{\arcsin (a x)} \sin (4 \arcsin (a x))}{136 a^4} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int e^{\arcsin (a x)} x^3 \, dx=\frac {e^{\arcsin (a x)} (-68 \cos (2 \arcsin (a x))+20 \cos (4 \arcsin (a x))+34 \sin (2 \arcsin (a x))-5 \sin (4 \arcsin (a x)))}{680 a^4} \]
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\[\int {\mathrm e}^{\arcsin \left (a x \right )} x^{3}d x\]
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none
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.67 \[ \int e^{\arcsin (a x)} x^3 \, dx=\frac {{\left (20 \, a^{4} x^{4} - 3 \, a^{2} x^{2} + {\left (5 \, a^{3} x^{3} + 6 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} - 6\right )} e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{4}} \]
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Time = 0.51 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.23 \[ \int e^{\arcsin (a x)} x^3 \, dx=\begin {cases} \frac {4 x^{4} e^{\operatorname {asin}{\left (a x \right )}}}{17} + \frac {x^{3} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{17 a} - \frac {3 x^{2} e^{\operatorname {asin}{\left (a x \right )}}}{85 a^{2}} + \frac {6 x \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {asin}{\left (a x \right )}}}{85 a^{3}} - \frac {6 e^{\operatorname {asin}{\left (a x \right )}}}{85 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases} \]
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\[ \int e^{\arcsin (a x)} x^3 \, dx=\int { x^{3} e^{\left (\arcsin \left (a x\right )\right )} \,d x } \]
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none
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int e^{\arcsin (a x)} x^3 \, dx=-\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x e^{\left (\arcsin \left (a x\right )\right )}}{17 \, a^{3}} + \frac {11 \, \sqrt {-a^{2} x^{2} + 1} x e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{3}} + \frac {4 \, {\left (a^{2} x^{2} - 1\right )}^{2} e^{\left (\arcsin \left (a x\right )\right )}}{17 \, a^{4}} + \frac {37 \, {\left (a^{2} x^{2} - 1\right )} e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{4}} + \frac {11 \, e^{\left (\arcsin \left (a x\right )\right )}}{85 \, a^{4}} \]
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Timed out. \[ \int e^{\arcsin (a x)} x^3 \, dx=\int x^3\,{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )} \,d x \]
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