Integrand size = 18, antiderivative size = 109 \[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\frac {2 (d x)^{7/2} (a+b \arccos (c x))^2}{7 d}+\frac {8 b c (d x)^{9/2} (a+b \arccos (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{4},\frac {13}{4},c^2 x^2\right )}{63 d^2}+\frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3} \]
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Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4724, 4806} \[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3}+\frac {8 b c (d x)^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{4},\frac {13}{4},c^2 x^2\right ) (a+b \arccos (c x))}{63 d^2}+\frac {2 (d x)^{7/2} (a+b \arccos (c x))^2}{7 d} \]
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Rule 4724
Rule 4806
Rubi steps \begin{align*} \text {integral}& = \frac {2 (d x)^{7/2} (a+b \arccos (c x))^2}{7 d}+\frac {(4 b c) \int \frac {(d x)^{7/2} (a+b \arccos (c x))}{\sqrt {1-c^2 x^2}} \, dx}{7 d} \\ & = \frac {2 (d x)^{7/2} (a+b \arccos (c x))^2}{7 d}+\frac {8 b c (d x)^{9/2} (a+b \arccos (c x)) \operatorname {Hypergeometric2F1}\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*}
Leaf count is larger than twice the leaf count of optimal. \(234\) vs. \(2(109)=218\).
Time = 10.98 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.15 \[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\frac {(d x)^{5/2} \left (882 a^2 x^3+\frac {84 a b \left (-2 \sqrt {1-c^2 x^2} \left (5+3 c^2 x^2\right )+21 c^3 x^3 \arccos (c x)+10 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},c^2 x^2\right )\right )}{c^3}+\frac {b^2 \left (-16 c x \left (35+9 c^2 x^2\right )-168 \sqrt {1-c^2 x^2} \left (5+3 c^2 x^2\right ) \arccos (c x)+882 c^3 x^3 \arccos (c x)^2+840 \sqrt {1-c^2 x^2} \arccos (c x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {5}{4},c^2 x^2\right )+\frac {105 \sqrt {2} c \pi x \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};c^2 x^2\right )}{\operatorname {Gamma}\left (\frac {5}{4}\right ) \operatorname {Gamma}\left (\frac {7}{4}\right )}\right )}{c^3}\right )}{3087 x^2} \]
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\[\int \left (d x \right )^{\frac {5}{2}} \left (a +b \arccos \left (c x \right )\right )^{2}d x\]
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\[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\int { \left (d x\right )^{\frac {5}{2}} {\left (b \arccos \left (c x\right ) + a\right )}^{2} \,d x } \]
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Timed out. \[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\text {Timed out} \]
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\[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\int { \left (d x\right )^{\frac {5}{2}} {\left (b \arccos \left (c x\right ) + a\right )}^{2} \,d x } \]
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Exception generated. \[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\text {Exception raised: RuntimeError} \]
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Timed out. \[ \int (d x)^{5/2} (a+b \arccos (c x))^2 \, dx=\int {\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2\,{\left (d\,x\right )}^{5/2} \,d x \]
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