Integrand size = 8, antiderivative size = 69 \[ \int x^3 \arccos (a x) \, dx=-\frac {3 x \sqrt {1-a^2 x^2}}{32 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{16 a}+\frac {1}{4} x^4 \arccos (a x)+\frac {3 \arcsin (a x)}{32 a^4} \]
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Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4724, 327, 222} \[ \int x^3 \arccos (a x) \, dx=\frac {3 \arcsin (a x)}{32 a^4}-\frac {x^3 \sqrt {1-a^2 x^2}}{16 a}-\frac {3 x \sqrt {1-a^2 x^2}}{32 a^3}+\frac {1}{4} x^4 \arccos (a x) \]
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Rule 222
Rule 327
Rule 4724
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \arccos (a x)+\frac {1}{4} a \int \frac {x^4}{\sqrt {1-a^2 x^2}} \, dx \\ & = -\frac {x^3 \sqrt {1-a^2 x^2}}{16 a}+\frac {1}{4} x^4 \arccos (a x)+\frac {3 \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{16 a} \\ & = -\frac {3 x \sqrt {1-a^2 x^2}}{32 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{16 a}+\frac {1}{4} x^4 \arccos (a x)+\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{32 a^3} \\ & = -\frac {3 x \sqrt {1-a^2 x^2}}{32 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{16 a}+\frac {1}{4} x^4 \arccos (a x)+\frac {3 \arcsin (a x)}{32 a^4} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78 \[ \int x^3 \arccos (a x) \, dx=\frac {-a x \sqrt {1-a^2 x^2} \left (3+2 a^2 x^2\right )+8 a^4 x^4 \arccos (a x)+3 \arcsin (a x)}{32 a^4} \]
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Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(\frac {\frac {a^{4} x^{4} \arccos \left (a x \right )}{4}-\frac {a^{3} x^{3} \sqrt {-a^{2} x^{2}+1}}{16}-\frac {3 a x \sqrt {-a^{2} x^{2}+1}}{32}+\frac {3 \arcsin \left (a x \right )}{32}}{a^{4}}\) | \(60\) |
default | \(\frac {\frac {a^{4} x^{4} \arccos \left (a x \right )}{4}-\frac {a^{3} x^{3} \sqrt {-a^{2} x^{2}+1}}{16}-\frac {3 a x \sqrt {-a^{2} x^{2}+1}}{32}+\frac {3 \arcsin \left (a x \right )}{32}}{a^{4}}\) | \(60\) |
parts | \(\frac {x^{4} \arccos \left (a x \right )}{4}+\frac {a \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )}{4}\) | \(89\) |
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Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int x^3 \arccos (a x) \, dx=\frac {{\left (8 \, a^{4} x^{4} - 3\right )} \arccos \left (a x\right ) - {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \sqrt {-a^{2} x^{2} + 1}}{32 \, a^{4}} \]
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Time = 0.33 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int x^3 \arccos (a x) \, dx=\begin {cases} \frac {x^{4} \operatorname {acos}{\left (a x \right )}}{4} - \frac {x^{3} \sqrt {- a^{2} x^{2} + 1}}{16 a} - \frac {3 x \sqrt {- a^{2} x^{2} + 1}}{32 a^{3}} - \frac {3 \operatorname {acos}{\left (a x \right )}}{32 a^{4}} & \text {for}\: a \neq 0 \\\frac {\pi x^{4}}{8} & \text {otherwise} \end {cases} \]
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Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88 \[ \int x^3 \arccos (a x) \, dx=\frac {1}{4} \, x^{4} \arccos \left (a x\right ) - \frac {1}{32} \, {\left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} x^{3}}{a^{2}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{a^{4}} - \frac {3 \, \arcsin \left (a x\right )}{a^{5}}\right )} a \]
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Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int x^3 \arccos (a x) \, dx=\frac {1}{4} \, x^{4} \arccos \left (a x\right ) - \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{16 \, a} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{32 \, a^{3}} - \frac {3 \, \arccos \left (a x\right )}{32 \, a^{4}} \]
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Timed out. \[ \int x^3 \arccos (a x) \, dx=\int x^3\,\mathrm {acos}\left (a\,x\right ) \,d x \]
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