Optimal. Leaf size=107 \[ \frac{16 b^2 c^2 (d x)^{5/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{5}{4},\frac{5}{4}\right \},\left \{\frac{7}{4},\frac{9}{4}\right \},c^2 x^2\right )}{15 d^3}-\frac{8 b c (d x)^{3/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2}+\frac{2 \sqrt{d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d} \]
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Rubi [A] time = 0.130082, antiderivative size = 107, 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 = {4627, 4711} \[ \frac{16 b^2 c^2 (d x)^{5/2} \, _3F_2\left (1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};c^2 x^2\right )}{15 d^3}-\frac{8 b c (d x)^{3/2} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2}+\frac{2 \sqrt{d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d} \]
Antiderivative was successfully verified.
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Rule 4627
Rule 4711
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{d x}} \, dx &=\frac{2 \sqrt{d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac{(4 b c) \int \frac{\sqrt{d x} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{d}\\ &=\frac{2 \sqrt{d x} \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac{8 b c (d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};c^2 x^2\right )}{3 d^2}+\frac{16 b^2 c^2 (d x)^{5/2} \, _3F_2\left (1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};c^2 x^2\right )}{15 d^3}\\ \end{align*}
Mathematica [A] time = 0.0492373, size = 90, normalized size = 0.84 \[ \frac{2 x \left (8 b^2 c^2 x^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{5}{4},\frac{5}{4}\right \},\left \{\frac{7}{4},\frac{9}{4}\right \},c^2 x^2\right )+5 \left (a+b \sin ^{-1}(c x)\right ) \left (3 \left (a+b \sin ^{-1}(c x)\right )-4 b c x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},c^2 x^2\right )\right )\right )}{15 \sqrt{d x}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.358, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}{\frac{1}{\sqrt{dx}}}}\, 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} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt{d x}}{d x}, 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 \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt{d x}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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