Optimal. Leaf size=132 \[ \frac{\sqrt{b} d \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{\sqrt{b} d \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}} \]
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Rubi [A] time = 0.0979791, antiderivative size = 132, 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 = {2616, 2564, 329, 298, 203, 206} \[ \frac{\sqrt{b} d \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{\sqrt{b} d \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2616
Rule 2564
Rule 329
Rule 298
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)} \, dx &=\frac{\left (d \sqrt{b \tan (e+f x)}\right ) \int \sec (e+f x) \sqrt{b \sin (e+f x)} \, dx}{\sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{\left (d \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-\frac{x^2}{b^2}} \, dx,x,b \sin (e+f x)\right )}{b f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{\left (2 d \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{b f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{\left (b d \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}-\frac{\left (b d \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{\sqrt{b} d \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{b \tan (e+f x)}}{f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}+\frac{\sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{b \tan (e+f x)}}{f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.788373, size = 136, normalized size = 1.03 \[ \frac{b \sqrt [4]{\tan ^2(e+f x)} \sqrt{d \sec (e+f x)} \left (2 \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\log \left (1-\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+\log \left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}+1\right )\right )}{2 f \sqrt{\sec (e+f x)} \sqrt{b \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.232, size = 306, normalized size = 2.3 \begin{align*}{\frac{\sin \left ( fx+e \right ) \sqrt{2}\cos \left ( fx+e \right ) }{2\,f \left ( \cos \left ( fx+e \right ) -1 \right ) }\sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{-{\frac{i\cos \left ( fx+e \right ) -i-\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }}} \left ( i{\it EllipticPi} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -{\it EllipticPi} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -{\it EllipticPi} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \right ) } \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{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.80112, size = 1701, normalized size = 12.89 \begin{align*} \text{result too large to display} \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{b \tan{\left (e + f x \right )}} \sqrt{d \sec{\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{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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