Optimal. Leaf size=68 \[ \frac{a (b x)^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{b^2 (m+1) (m+2)}+\frac{\cos ^{-1}(a x) (b x)^{m+1}}{b (m+1)} \]
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Rubi [A] time = 0.0256728, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4628, 364} \[ \frac{a (b x)^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{b^2 (m+1) (m+2)}+\frac{\cos ^{-1}(a x) (b x)^{m+1}}{b (m+1)} \]
Antiderivative was successfully verified.
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Rule 4628
Rule 364
Rubi steps
\begin{align*} \int (b x)^m \cos ^{-1}(a x) \, dx &=\frac{(b x)^{1+m} \cos ^{-1}(a x)}{b (1+m)}+\frac{a \int \frac{(b x)^{1+m}}{\sqrt{1-a^2 x^2}} \, dx}{b (1+m)}\\ &=\frac{(b x)^{1+m} \cos ^{-1}(a x)}{b (1+m)}+\frac{a (b x)^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{b^2 (1+m) (2+m)}\\ \end{align*}
Mathematica [A] time = 0.0352791, size = 54, normalized size = 0.79 \[ \frac{x (b x)^m \left (a x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2}+1,\frac{m}{2}+2,a^2 x^2\right )+(m+2) \cos ^{-1}(a x)\right )}{(m+1) (m+2)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.866, size = 0, normalized size = 0. \begin{align*} \int \left ( bx \right ) ^{m}\arccos \left ( ax \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b x\right )^{m} \arccos \left (a x\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{m} \operatorname{acos}{\left (a x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{m} \arccos \left (a x\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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