Integrand size = 10, antiderivative size = 82 \[ \int e^{\arccos (a x)} x^2 \, dx=\frac {e^{\arccos (a x)} x}{8 a^2}-\frac {e^{\arccos (a x)} \sqrt {1-a^2 x^2}}{8 a^3}+\frac {3 e^{\arccos (a x)} \cos (3 \arccos (a x))}{40 a^3}-\frac {e^{\arccos (a x)} \sin (3 \arccos (a x))}{40 a^3} \]
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Time = 0.05 (sec) , antiderivative size = 82, 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, 4557, 4517} \[ \int e^{\arccos (a x)} x^2 \, dx=-\frac {e^{\arccos (a x)} \sin (3 \arccos (a x))}{40 a^3}+\frac {3 e^{\arccos (a x)} \cos (3 \arccos (a x))}{40 a^3}+\frac {x e^{\arccos (a x)}}{8 a^2}-\frac {\sqrt {1-a^2 x^2} e^{\arccos (a x)}}{8 a^3} \]
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Rule 12
Rule 4517
Rule 4557
Rule 4921
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {e^x \cos ^2(x) \sin (x)}{a^2} \, dx,x,\arccos (a x)\right )}{a} \\ & = -\frac {\text {Subst}\left (\int e^x \cos ^2(x) \sin (x) \, dx,x,\arccos (a x)\right )}{a^3} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {1}{4} e^x \sin (x)+\frac {1}{4} e^x \sin (3 x)\right ) \, dx,x,\arccos (a x)\right )}{a^3} \\ & = -\frac {\text {Subst}\left (\int e^x \sin (x) \, dx,x,\arccos (a x)\right )}{4 a^3}-\frac {\text {Subst}\left (\int e^x \sin (3 x) \, dx,x,\arccos (a x)\right )}{4 a^3} \\ & = \frac {e^{\arccos (a x)} x}{8 a^2}-\frac {e^{\arccos (a x)} \sqrt {1-a^2 x^2}}{8 a^3}+\frac {3 e^{\arccos (a x)} \cos (3 \arccos (a x))}{40 a^3}-\frac {e^{\arccos (a x)} \sin (3 \arccos (a x))}{40 a^3} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.61 \[ \int e^{\arccos (a x)} x^2 \, dx=-\frac {e^{\arccos (a x)} \left (-5 a x+5 \sqrt {1-a^2 x^2}-3 \cos (3 \arccos (a x))+\sin (3 \arccos (a x))\right )}{40 a^3} \]
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\[\int {\mathrm e}^{\arccos \left (a x \right )} x^{2}d x\]
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none
Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.56 \[ \int e^{\arccos (a x)} x^2 \, dx=\frac {{\left (3 \, a^{3} x^{3} - a x - {\left (a^{2} x^{2} + 1\right )} \sqrt {-a^{2} x^{2} + 1}\right )} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{3}} \]
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Time = 0.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int e^{\arccos (a x)} x^2 \, dx=\begin {cases} \frac {3 x^{3} e^{\operatorname {acos}{\left (a x \right )}}}{10} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{10 a} - \frac {x e^{\operatorname {acos}{\left (a x \right )}}}{10 a^{2}} - \frac {\sqrt {- a^{2} x^{2} + 1} e^{\operatorname {acos}{\left (a x \right )}}}{10 a^{3}} & \text {for}\: a \neq 0 \\\frac {x^{3} e^{\frac {\pi }{2}}}{3} & \text {otherwise} \end {cases} \]
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\[ \int e^{\arccos (a x)} x^2 \, dx=\int { x^{2} e^{\left (\arccos \left (a x\right )\right )} \,d x } \]
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none
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int e^{\arccos (a x)} x^2 \, dx=\frac {3}{10} \, x^{3} e^{\left (\arccos \left (a x\right )\right )} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{2} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a} - \frac {x e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} e^{\left (\arccos \left (a x\right )\right )}}{10 \, a^{3}} \]
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Timed out. \[ \int e^{\arccos (a x)} x^2 \, dx=\int x^2\,{\mathrm {e}}^{\mathrm {acos}\left (a\,x\right )} \,d x \]
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