Optimal. Leaf size=206 \[ -\frac{b \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}+\frac{(2 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]
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Rubi [A] time = 0.166938, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3192, 412, 525, 418, 411} \[ \frac{\tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}-\frac{b \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(2 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]
Antiderivative was successfully verified.
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Rule 3192
Rule 412
Rule 525
Rule 418
Rule 411
Rubi steps
\begin{align*} \int \text{sech}^4(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \, dx &=\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\left (1+x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}-\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{-2 a-b x^2}{\left (1+x^2\right )^{3/2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 f}\\ &=\frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}+\frac{\left ((2 a-b) \sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 (a-b) f}-\frac{\left (a b \sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 (a-b) f}\\ &=\frac{(2 a-b) E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 (a-b) f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{b F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 (a-b) f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}\\ \end{align*}
Mathematica [C] time = 2.83669, size = 204, normalized size = 0.99 \[ \frac{-16 i a (a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\sqrt{2} \tanh (e+f x) \text{sech}^2(e+f x) \left (\left (8 a^2-4 b^2\right ) \cosh (2 (e+f x))+(2 a-b) (8 a+b \cosh (4 (e+f x))-5 b)\right )+8 i a (2 a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{24 f (a-b) \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.132, size = 318, normalized size = 1.5 \begin{align*}{\frac{1}{3\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3} \left ( a-b \right ) f} \left ( \left ( 2\,\sqrt{-{\frac{b}{a}}}ab-\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \sinh \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( 2\,\sqrt{-{\frac{b}{a}}}{a}^{2}-2\,\sqrt{-{\frac{b}{a}}}ab \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) + \left ( \sqrt{-{\frac{b}{a}}}{a}^{2}-2\,\sqrt{-{\frac{b}{a}}}ab+\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \sinh \left ( fx+e \right ) +\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}b \left ( a{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -b{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -2\,{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) a+b{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{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 \sqrt{b \sinh \left (f x + e\right )^{2} + a} \operatorname{sech}\left (f x + e\right )^{4}\,{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 (\sqrt{b \sinh \left (f x + e\right )^{2} + a} \operatorname{sech}\left (f x + e\right )^{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 \sqrt{b \sinh \left (f x + e\right )^{2} + a} \operatorname{sech}\left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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