Optimal. Leaf size=184 \[ -\frac{b c d (3 m+7) \sqrt{1-c^2 x^2} (f x)^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{f^2 (m+1) (m+2) (m+3)^2 \sqrt{c x-1} \sqrt{c x+1}}-\frac{c^2 d (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac{d (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}+\frac{b c d \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2}}{f^2 (m+3)^2} \]
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Rubi [A] time = 0.258224, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {14, 5731, 12, 460, 126, 365, 364} \[ -\frac{c^2 d (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac{d (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}-\frac{b c d (3 m+7) \sqrt{1-c^2 x^2} (f x)^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};c^2 x^2\right )}{f^2 (m+1) (m+2) (m+3)^2 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b c d \sqrt{c x-1} \sqrt{c x+1} (f x)^{m+2}}{f^2 (m+3)^2} \]
Antiderivative was successfully verified.
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Rule 14
Rule 5731
Rule 12
Rule 460
Rule 126
Rule 365
Rule 364
Rubi steps
\begin{align*} \int (f x)^m \left (d-c^2 d x^2\right ) \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=\frac{d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac{c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-(b c) \int \frac{d (f x)^{1+m} \left (3+m-c^2 (1+m) x^2\right )}{f (1+m) (3+m) \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=\frac{d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac{c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac{(b c d) \int \frac{(f x)^{1+m} \left (3+m-c^2 (1+m) x^2\right )}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{f \left (3+4 m+m^2\right )}\\ &=\frac{b c d (f x)^{2+m} \sqrt{-1+c x} \sqrt{1+c x}}{f^2 (3+m)^2}+\frac{d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac{c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac{(b c d (7+3 m)) \int \frac{(f x)^{1+m}}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{f (1+m) (3+m)^2}\\ &=\frac{b c d (f x)^{2+m} \sqrt{-1+c x} \sqrt{1+c x}}{f^2 (3+m)^2}+\frac{d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac{c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac{\left (b c d (7+3 m) \sqrt{-1+c^2 x^2}\right ) \int \frac{(f x)^{1+m}}{\sqrt{-1+c^2 x^2}} \, dx}{f (1+m) (3+m)^2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b c d (f x)^{2+m} \sqrt{-1+c x} \sqrt{1+c x}}{f^2 (3+m)^2}+\frac{d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac{c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac{\left (b c d (7+3 m) \sqrt{1-c^2 x^2}\right ) \int \frac{(f x)^{1+m}}{\sqrt{1-c^2 x^2}} \, dx}{f (1+m) (3+m)^2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b c d (f x)^{2+m} \sqrt{-1+c x} \sqrt{1+c x}}{f^2 (3+m)^2}+\frac{d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac{c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac{b c d (7+3 m) (f x)^{2+m} \sqrt{1-c^2 x^2} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};c^2 x^2\right )}{f^2 (1+m) (2+m) (3+m)^2 \sqrt{-1+c x} \sqrt{1+c x}}\\ \end{align*}
Mathematica [A] time = 0.230071, size = 191, normalized size = 1.04 \[ d x (f x)^m \left (\frac{b c^3 x^3 \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+4}{2},\frac{m+6}{2},c^2 x^2\right )}{\left (m^2+7 m+12\right ) \sqrt{c x-1} \sqrt{c x+1}}-\frac{b c x \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt{c x-1} \sqrt{c x+1}}-\frac{c^2 x^2 \left (a+b \cosh ^{-1}(c x)\right )}{m+3}+\frac{a+b \cosh ^{-1}(c x)}{m+1}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 2.286, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( -{c}^{2}d{x}^{2}+d \right ) \left ( a+b{\rm arccosh} \left (cx\right ) \right ) \, 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 (a c^{2} d x^{2} - a d +{\left (b c^{2} d x^{2} - b d\right )} \operatorname{arcosh}\left (c x\right )\right )} \left (f x\right )^{m}, 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*} - d \left (\int - a \left (f x\right )^{m}\, dx + \int - b \left (f x\right )^{m} \operatorname{acosh}{\left (c x \right )}\, dx + \int a c^{2} x^{2} \left (f x\right )^{m}\, dx + \int b c^{2} x^{2} \left (f x\right )^{m} \operatorname{acosh}{\left (c x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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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