Optimal. Leaf size=116 \[ \frac{2 (a \sinh (x)+b \cosh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}-\frac{2 i \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} \text{EllipticF}\left (\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right ),2\right )}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}} \]
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Rubi [A] time = 0.0481128, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3076, 3078, 2641} \[ \frac{2 (a \sinh (x)+b \cosh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}-\frac{2 i \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} F\left (\left .\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}} \]
Antiderivative was successfully verified.
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Rule 3076
Rule 3078
Rule 2641
Rubi steps
\begin{align*} \int \frac{1}{(a \cosh (x)+b \sinh (x))^{5/2}} \, dx &=\frac{2 (b \cosh (x)+a \sinh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}+\frac{\int \frac{1}{\sqrt{a \cosh (x)+b \sinh (x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{2 (b \cosh (x)+a \sinh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}+\frac{\sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} \int \frac{1}{\sqrt{\cosh \left (x+i \tan ^{-1}(a,-i b)\right )}} \, dx}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}}\\ &=\frac{2 (b \cosh (x)+a \sinh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}-\frac{2 i F\left (\left .\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}}}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}}\\ \end{align*}
Mathematica [C] time = 0.55822, size = 133, normalized size = 1.15 \[ -\frac{2 \left ((a \cosh (x)+b \sinh (x))^2 \sqrt{\cosh ^2\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right )} \text{sech}\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},-\sinh ^2\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right )\right )+b \sqrt{1-\frac{a^2}{b^2}} (a \sinh (x)+b \cosh (x))\right )}{3 b \sqrt{1-\frac{a^2}{b^2}} (b-a) (a+b) (a \cosh (x)+b \sinh (x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.144, size = 37, normalized size = 0.3 \begin{align*} -{\frac{\cosh \left ( x \right ) }{ \left ({a}^{2}-{b}^{2} \right ) \sinh \left ( x \right ) }{\frac{1}{\sqrt{-\sinh \left ( x \right ) \sqrt{{a}^{2}-{b}^{2}}}}}} \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{1}{{\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}^{\frac{5}{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{a \cosh \left (x\right ) + b \sinh \left (x\right )}}{a^{3} \cosh \left (x\right )^{3} + 3 \, a^{2} b \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, a b^{2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + b^{3} \sinh \left (x\right )^{3}}, 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{1}{{\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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