Optimal. Leaf size=168 \[ \frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{f}-\frac{2 \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]
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Rubi [A] time = 0.177173, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3196, 467, 531, 418, 492, 411} \[ \frac{\tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{f}+\frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]
Antiderivative was successfully verified.
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Rule 3196
Rule 467
Rule 531
Rule 418
Rule 492
Rule 411
Rubi steps
\begin{align*} \int \sqrt{a+b \sinh ^2(e+f x)} \tanh ^2(e+f x) \, dx &=\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2 \sqrt{a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{\sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{a+2 b x^2}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{\sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}+\frac{\left (a \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 )}{f}+\frac{\left (2 b \sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{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)}}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}-\frac{\left (2 \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 )}{f}\\ &=-\frac{2 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)}}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{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)}}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}\\ \end{align*}
Mathematica [C] time = 0.493876, size = 150, normalized size = 0.89 \[ \frac{i \sqrt{2} a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\tanh (e+f x) (-2 a-b \cosh (2 (e+f x))+b)-2 i \sqrt{2} a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{f \sqrt{4 a+2 b \cosh (2 (e+f x))-2 b}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.135, size = 233, normalized size = 1.4 \begin{align*} -{\frac{1}{f\cosh \left ( fx+e \right ) } \left ( \sqrt{-{\frac{b}{a}}}b \left ( \sinh \left ( fx+e \right ) \right ) ^{3}-a\sqrt{{\frac{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) +2\,b\sqrt{{\frac{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -2\,b\sqrt{{\frac{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) +\sqrt{-{\frac{b}{a}}}a\sinh \left ( fx+e \right ) \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} \tanh \left (f x + e\right )^{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 (\sqrt{b \sinh \left (f x + e\right )^{2} + a} \tanh \left (f x + e\right )^{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 \sqrt{a + b \sinh ^{2}{\left (e + f x \right )}} \tanh ^{2}{\left (e + f x \right )}\, 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 \sqrt{b \sinh \left (f x + e\right )^{2} + a} \tanh \left (f x + e\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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