Optimal. Leaf size=128 \[ \frac{16 b^2 (e (c+d x))^{5/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{5}{4},\frac{5}{4}\right \},\left \{\frac{7}{4},\frac{9}{4}\right \},(c+d x)^2\right )}{15 d e^3}-\frac{8 b (e (c+d x))^{3/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^2}+\frac{2 \sqrt{e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e} \]
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Rubi [A] time = 0.193197, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4805, 4627, 4711} \[ \frac{16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};(c+d x)^2\right )}{15 d e^3}-\frac{8 b (e (c+d x))^{3/2} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^2}+\frac{2 \sqrt{e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 4627
Rule 4711
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{\sqrt{c e+d e x}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^2}{\sqrt{e x}} \, dx,x,c+d x\right )}{d}\\ &=\frac{2 \sqrt{e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{\sqrt{e x} \left (a+b \sin ^{-1}(x)\right )}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \sqrt{e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}-\frac{8 b (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};(c+d x)^2\right )}{3 d e^2}+\frac{16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};(c+d x)^2\right )}{15 d e^3}\\ \end{align*}
Mathematica [A] time = 0.080617, size = 107, normalized size = 0.84 \[ \frac{2 \sqrt{e (c+d x)} \left (8 b^2 (c+d x)^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{5}{4},\frac{5}{4}\right \},\left \{\frac{7}{4},\frac{9}{4}\right \},(c+d x)^2\right )+5 \left (a+b \sin ^{-1}(c+d x)\right ) \left (3 \left (a+b \sin ^{-1}(c+d x)\right )-4 b (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},(c+d x)^2\right )\right )\right )}{15 d e} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.42, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{2}{\frac{1}{\sqrt{dex+ce}}}}\, 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{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{\sqrt{d e x + c e}}, 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 (d x + c\right ) + a\right )}^{2}}{\sqrt{d e x + c e}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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