Optimal. Leaf size=300 \[ -\frac{b c m \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )}{d f^2 (m+1) (m+2) \sqrt{d-c^2 d x^2}}-\frac{m \sqrt{1-c^2 x^2} (f x)^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d f (m+1) \sqrt{d-c^2 d x^2}}+\frac{b c \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{d f^2 (m+2) \sqrt{d-c^2 d x^2}}+\frac{(f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{d f \sqrt{d-c^2 d x^2}} \]
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Rubi [A] time = 0.660031, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {5798, 5756, 5763, 364} \[ -\frac{b c m \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \, _3F_2\left (1,\frac{m}{2}+1,\frac{m}{2}+1;\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2;c^2 x^2\right )}{d f^2 (m+1) (m+2) \sqrt{d-c^2 d x^2}}-\frac{m \sqrt{1-c^2 x^2} (f x)^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d f (m+1) \sqrt{d-c^2 d x^2}}+\frac{(f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{d f \sqrt{d-c^2 d x^2}}+\frac{b c \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};c^2 x^2\right )}{d f^2 (m+2) \sqrt{d-c^2 d x^2}} \]
Antiderivative was successfully verified.
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Rule 5798
Rule 5756
Rule 5763
Rule 364
Rubi steps
\begin{align*} \int \frac{(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=-\frac{\left (\sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{d \sqrt{d-c^2 d x^2}}\\ &=\frac{(f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{d f \sqrt{d-c^2 d x^2}}-\frac{\left (b c \sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{(f x)^{1+m}}{-1+c^2 x^2} \, dx}{d f \sqrt{d-c^2 d x^2}}-\frac{\left (m \sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{d \sqrt{d-c^2 d x^2}}\\ &=\frac{(f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{d f \sqrt{d-c^2 d x^2}}-\frac{m (f x)^{1+m} \sqrt{1-c^2 x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};c^2 x^2\right )}{d f (1+m) \sqrt{d-c^2 d x^2}}+\frac{b c (f x)^{2+m} \sqrt{-1+c x} \sqrt{1+c x} \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};c^2 x^2\right )}{d f^2 (2+m) \sqrt{d-c^2 d x^2}}-\frac{b c m (f x)^{2+m} \sqrt{-1+c x} \sqrt{1+c x} \, _3F_2\left (1,1+\frac{m}{2},1+\frac{m}{2};\frac{3}{2}+\frac{m}{2},2+\frac{m}{2};c^2 x^2\right )}{d f^2 (1+m) (2+m) \sqrt{d-c^2 d x^2}}\\ \end{align*}
Mathematica [A] time = 0.241915, size = 216, normalized size = 0.72 \[ \frac{x (f x)^m \left (-b c m x \sqrt{c x-1} \sqrt{c x+1} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+1,\frac{m}{2}+1\right \},\left \{\frac{m}{2}+\frac{3}{2},\frac{m}{2}+2\right \},c^2 x^2\right )-m (m+2) \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )+(m+1) \left (b c x \sqrt{c x-1} \sqrt{c x+1} \text{Hypergeometric2F1}\left (1,\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )+(m+2) \left (a+b \cosh ^{-1}(c x)\right )\right )\right )}{d (m+1) (m+2) \sqrt{d-c^2 d x^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.598, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx \right ) ^{m} \left ( a+b{\rm arccosh} \left (cx\right ) \right ) \left ( -{c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \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{\sqrt{-c^{2} d x^{2} + d}{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, 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 \frac{\left (f x\right )^{m} \left (a + b \operatorname{acosh}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{3}{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 \frac{{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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