Optimal. Leaf size=134 \[ \frac{16 b^2 (e (c+d x))^{7/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},-(c+d x)^2\right )}{105 d e^3}-\frac{8 b (e (c+d x))^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{15 d e^2}+\frac{2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e} \]
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Rubi [A] time = 0.209408, antiderivative size = 134, 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 = {5865, 5661, 5762} \[ \frac{16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac{7}{4},\frac{7}{4};\frac{9}{4},\frac{11}{4};-(c+d x)^2\right )}{105 d e^3}-\frac{8 b (e (c+d x))^{5/2} \, _2F_1\left (\frac{1}{2},\frac{5}{4};\frac{9}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{15 d e^2}+\frac{2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 5661
Rule 5762
Rubi steps
\begin{align*} \int \sqrt{c e+d e x} \left (a+b \sinh ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{e x} \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{(e x)^{3/2} \left (a+b \sinh ^{-1}(x)\right )}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d e}\\ &=\frac{2 (e (c+d x))^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e}-\frac{8 b (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac{1}{2},\frac{5}{4};\frac{9}{4};-(c+d x)^2\right )}{15 d e^2}+\frac{16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac{7}{4},\frac{7}{4};\frac{9}{4},\frac{11}{4};-(c+d x)^2\right )}{105 d e^3}\\ \end{align*}
Mathematica [A] time = 0.0848566, size = 110, normalized size = 0.82 \[ \frac{2 (e (c+d x))^{3/2} \left (8 b^2 (c+d x)^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{7}{4},\frac{7}{4}\right \},\left \{\frac{9}{4},\frac{11}{4}\right \},-(c+d x)^2\right )-28 b (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{4},\frac{9}{4},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )+35 \left (a+b \sinh ^{-1}(c+d x)\right )^2\right )}{105 d e} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.275, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}\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 ({\left (b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname{arsinh}\left (d x + c\right ) + a^{2}\right )} \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \left (c + d x\right )} \left (a + b \operatorname{asinh}{\left (c + d x \right )}\right )^{2}\, dx \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 \sqrt{d e x + c e}{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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