Integrand size = 10, antiderivative size = 51 \[ \int x (a+b \arccos (c x)) \, dx=-\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} x^2 (a+b \arccos (c x))+\frac {b \arcsin (c x)}{4 c^2} \]
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Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4724, 327, 222} \[ \int x (a+b \arccos (c x)) \, dx=\frac {1}{2} x^2 (a+b \arccos (c x))+\frac {b \arcsin (c x)}{4 c^2}-\frac {b x \sqrt {1-c^2 x^2}}{4 c} \]
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Rule 222
Rule 327
Rule 4724
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 (a+b \arccos (c x))+\frac {1}{2} (b c) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx \\ & = -\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} x^2 (a+b \arccos (c x))+\frac {b \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c} \\ & = -\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} x^2 (a+b \arccos (c x))+\frac {b \arcsin (c x)}{4 c^2} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int x (a+b \arccos (c x)) \, dx=\frac {a x^2}{2}-\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} b x^2 \arccos (c x)+\frac {b \arcsin (c x)}{4 c^2} \]
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Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94
method | result | size |
parts | \(\frac {a \,x^{2}}{2}+\frac {b \left (\frac {c^{2} x^{2} \arccos \left (c x \right )}{2}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}+\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\) | \(48\) |
derivativedivides | \(\frac {\frac {c^{2} x^{2} a}{2}+b \left (\frac {c^{2} x^{2} \arccos \left (c x \right )}{2}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}+\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\) | \(52\) |
default | \(\frac {\frac {c^{2} x^{2} a}{2}+b \left (\frac {c^{2} x^{2} \arccos \left (c x \right )}{2}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}+\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\) | \(52\) |
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Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int x (a+b \arccos (c x)) \, dx=\frac {2 \, a c^{2} x^{2} - \sqrt {-c^{2} x^{2} + 1} b c x + {\left (2 \, b c^{2} x^{2} - b\right )} \arccos \left (c x\right )}{4 \, c^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int x (a+b \arccos (c x)) \, dx=\begin {cases} \frac {a x^{2}}{2} + \frac {b x^{2} \operatorname {acos}{\left (c x \right )}}{2} - \frac {b x \sqrt {- c^{2} x^{2} + 1}}{4 c} - \frac {b \operatorname {acos}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {x^{2} \left (a + \frac {\pi b}{2}\right )}{2} & \text {otherwise} \end {cases} \]
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Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int x (a+b \arccos (c x)) \, dx=\frac {1}{2} \, a x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arccos \left (c x\right ) - c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b \]
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Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int x (a+b \arccos (c x)) \, dx=\frac {1}{2} \, b x^{2} \arccos \left (c x\right ) + \frac {1}{2} \, a x^{2} - \frac {\sqrt {-c^{2} x^{2} + 1} b x}{4 \, c} - \frac {b \arccos \left (c x\right )}{4 \, c^{2}} \]
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Time = 0.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int x (a+b \arccos (c x)) \, dx=\frac {a\,x^2}{2}+\frac {b\,\left (\frac {\mathrm {acos}\left (c\,x\right )\,\left (2\,c^2\,x^2-1\right )}{4}-\frac {c\,x\,\sqrt {1-c^2\,x^2}}{4}\right )}{c^2} \]
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