Optimal. Leaf size=109 \[ \frac{16 b^2 c^2 (d x)^{11/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{11}{4},\frac{11}{4}\right \},\left \{\frac{13}{4},\frac{15}{4}\right \},c^2 x^2\right )}{693 d^3}+\frac{8 b c (d x)^{9/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{9}{4},\frac{13}{4},c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{63 d^2}+\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d} \]
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Rubi [A] time = 0.13904, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4628, 4712} \[ \frac{16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac{11}{4},\frac{11}{4};\frac{13}{4},\frac{15}{4};c^2 x^2\right )}{693 d^3}+\frac{8 b c (d x)^{9/2} \, _2F_1\left (\frac{1}{2},\frac{9}{4};\frac{13}{4};c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{63 d^2}+\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d} \]
Antiderivative was successfully verified.
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Rule 4628
Rule 4712
Rubi steps
\begin{align*} \int (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )^2 \, dx &=\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d}+\frac{(4 b c) \int \frac{(d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{7 d}\\ &=\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d}+\frac{8 b c (d x)^{9/2} \left (a+b \cos ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{9}{4};\frac{13}{4};c^2 x^2\right )}{63 d^2}+\frac{16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac{11}{4},\frac{11}{4};\frac{13}{4},\frac{15}{4};c^2 x^2\right )}{693 d^3}\\ \end{align*}
Mathematica [B] time = 1.26082, size = 234, normalized size = 2.15 \[ \frac{(d x)^{5/2} \left (\frac{b^2 \left (\frac{105 \sqrt{2} \pi c x \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},\frac{3}{4},1\right \},\left \{\frac{5}{4},\frac{7}{4}\right \},c^2 x^2\right )}{\text{Gamma}\left (\frac{5}{4}\right ) \text{Gamma}\left (\frac{7}{4}\right )}+840 \sqrt{1-c^2 x^2} \cos ^{-1}(c x) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{5}{4},c^2 x^2\right )-16 c x \left (9 c^2 x^2+35\right )+882 c^3 x^3 \cos ^{-1}(c x)^2-168 \sqrt{1-c^2 x^2} \left (3 c^2 x^2+5\right ) \cos ^{-1}(c x)\right )}{c^3}+\frac{84 a b \left (10 \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},c^2 x^2\right )-2 \sqrt{1-c^2 x^2} \left (3 c^2 x^2+5\right )+21 c^3 x^3 \cos ^{-1}(c x)\right )}{c^3}+882 a^2 x^3\right )}{3087 x^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.358, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{{\frac{5}{2}}} \left ( a+b\arccos \left ( cx \right ) \right ) ^{2}\, 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^{2} d^{2} x^{2} \arccos \left (c x\right )^{2} + 2 \, a b d^{2} x^{2} \arccos \left (c x\right ) + a^{2} d^{2} x^{2}\right )} \sqrt{d x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 (d x\right )^{\frac{5}{2}}{\left (b \arccos \left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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