Optimal. Leaf size=131 \[ \frac{d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{\sqrt{b} f \sqrt{b \tan (e+f x)}}+\frac{d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{\sqrt{b} f \sqrt{b \tan (e+f x)}} \]
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Rubi [A] time = 0.106436, antiderivative size = 131, 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, 212, 206, 203} \[ \frac{d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{\sqrt{b} f \sqrt{b \tan (e+f x)}}+\frac{d \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{\sqrt{b} f \sqrt{b \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2616
Rule 2564
Rule 329
Rule 212
Rule 206
Rule 203
Rubi steps
\begin{align*} \int \frac{(d \sec (e+f x))^{3/2}}{\sqrt{b \tan (e+f x)}} \, dx &=\frac{\left (d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \int \frac{\sec (e+f x)}{\sqrt{b \sin (e+f x)}} \, dx}{\sqrt{b \tan (e+f x)}}\\ &=\frac{\left (d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{b^2}\right )} \, dx,x,b \sin (e+f x)\right )}{b f \sqrt{b \tan (e+f x)}}\\ &=\frac{\left (2 d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{b f \sqrt{b \tan (e+f x)}}\\ &=\frac{\left (d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{f \sqrt{b \tan (e+f x)}}+\frac{\left (d \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{f \sqrt{b \tan (e+f x)}}\\ &=\frac{d \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}{\sqrt{b} f \sqrt{b \tan (e+f x)}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}{\sqrt{b} f \sqrt{b \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 4.25268, size = 105, normalized size = 0.8 \[ -\frac{\sqrt{b \tan (e+f x)} (d \sec (e+f x))^{3/2} \left (\tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\tanh ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right )}{b f \sqrt [4]{\tan ^2(e+f x)} \sec ^{\frac{3}{2}}(e+f x)} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.214, size = 344, normalized size = 2.6 \begin{align*}{\frac{\cos \left ( fx+e \right ) \sqrt{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{2\,f \left ( \cos \left ( fx+e \right ) -1 \right ) } \left ( 2\,i{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\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 ) -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 ) \sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \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\cos \left ( fx+e \right ) -i-\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \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 \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}{\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 = 3.01246, size = 1662, normalized size = 12.69 \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(-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{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}{\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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