Optimal. Leaf size=105 \[ -\frac{16 b^2 c^2 (d x)^{3/2} \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},\frac{3}{4},1\right \},\left \{\frac{5}{4},\frac{7}{4}\right \},c^2 x^2\right )}{3 d^3}-\frac{8 b c \sqrt{d x} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{d^2}-\frac{2 \left (a+b \cos ^{-1}(c x)\right )^2}{d \sqrt{d x}} \]
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Rubi [A] time = 0.130147, antiderivative size = 105, 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} \[ -\frac{16 b^2 c^2 (d x)^{3/2} \, _3F_2\left (\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{7}{4};c^2 x^2\right )}{3 d^3}-\frac{8 b c \sqrt{d x} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{d^2}-\frac{2 \left (a+b \cos ^{-1}(c x)\right )^2}{d \sqrt{d x}} \]
Antiderivative was successfully verified.
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Rule 4628
Rule 4712
Rubi steps
\begin{align*} \int \frac{\left (a+b \cos ^{-1}(c x)\right )^2}{(d x)^{3/2}} \, dx &=-\frac{2 \left (a+b \cos ^{-1}(c x)\right )^2}{d \sqrt{d x}}-\frac{(4 b c) \int \frac{a+b \cos ^{-1}(c x)}{\sqrt{d x} \sqrt{1-c^2 x^2}} \, dx}{d}\\ &=-\frac{2 \left (a+b \cos ^{-1}(c x)\right )^2}{d \sqrt{d x}}-\frac{8 b c \sqrt{d x} \left (a+b \cos ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};c^2 x^2\right )}{d^2}-\frac{16 b^2 c^2 (d x)^{3/2} \, _3F_2\left (\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{7}{4};c^2 x^2\right )}{3 d^3}\\ \end{align*}
Mathematica [A] time = 0.432541, size = 129, normalized size = 1.23 \[ \frac{x \left (-\frac{\sqrt{2} \pi b^2 c^2 x^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},\frac{3}{4},1\right \},\left \{\frac{5}{4},\frac{7}{4}\right \},c^2 x^2\right )}{\text{Gamma}\left (\frac{5}{4}\right ) \text{Gamma}\left (\frac{7}{4}\right )}-2 \left (4 a b c x \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},c^2 x^2\right )+2 b^2 \cos ^{-1}(c x) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{5}{4},c^2 x^2\right ) \sin \left (2 \cos ^{-1}(c x)\right )+\left (a+b \cos ^{-1}(c x)\right )^2\right )\right )}{(d x)^{3/2}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.339, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\arccos \left ( cx \right ) \right ) ^{2} \left ( dx \right ) ^{-{\frac{3}{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 (\frac{{\left (b^{2} \arccos \left (c x\right )^{2} + 2 \, a b \arccos \left (c x\right ) + a^{2}\right )} \sqrt{d x}}{d^{2} x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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 \frac{{\left (b \arccos \left (c x\right ) + a\right )}^{2}}{\left (d x\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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