Optimal. Leaf size=134 \[ -\frac{16 b^2 \text{HypergeometricPFQ}\left (\left \{-\frac{1}{4},-\frac{1}{4},1\right \},\left \{\frac{1}{4},\frac{3}{4}\right \},-(c+d x)^2\right )}{15 d e^3 \sqrt{e (c+d x)}}-\frac{8 b \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{1}{2},\frac{1}{4},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}} \]
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Rubi [A] time = 0.229618, 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} \[ -\frac{16 b^2 \, _3F_2\left (-\frac{1}{4},-\frac{1}{4},1;\frac{1}{4},\frac{3}{4};-(c+d x)^2\right )}{15 d e^3 \sqrt{e (c+d x)}}-\frac{8 b \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 5661
Rule 5762
Rubi steps
\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{(c e+d e x)^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{(e x)^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{a+b \sinh ^{-1}(x)}{(e x)^{5/2} \sqrt{1+x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}}-\frac{8 b \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};-(c+d x)^2\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac{16 b^2 \, _3F_2\left (-\frac{1}{4},-\frac{1}{4},1;\frac{1}{4},\frac{3}{4};-(c+d x)^2\right )}{15 d e^3 \sqrt{e (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.0863127, size = 110, normalized size = 0.82 \[ -\frac{2 \left (8 b^2 (c+d x)^2 \text{HypergeometricPFQ}\left (\left \{-\frac{1}{4},-\frac{1}{4},1\right \},\left \{\frac{1}{4},\frac{3}{4}\right \},-(c+d x)^2\right )+\left (a+b \sinh ^{-1}(c+d x)\right ) \left (4 b (c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{1}{2},\frac{1}{4},-(c+d x)^2\right )+3 \left (a+b \sinh ^{-1}(c+d x)\right )\right )\right )}{15 d e (e (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.25, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dex+ce \right ) ^{-{\frac{7}{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} \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}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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