Optimal. Leaf size=328 \[ \frac{\sqrt{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}+1\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}} \]
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Rubi [A] time = 0.192076, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2585, 2575, 297, 1162, 617, 204, 1165, 628} \[ \frac{\sqrt{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}+1\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2585
Rule 2575
Rule 297
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{b \sec (e+f x)} \sqrt{\sin (e+f x)}} \, dx &=\frac{\int \frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}} \, dx}{\sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{(2 b) \operatorname{Subst}\left (\int \frac{x^2}{b^2+x^4} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=\frac{b \operatorname{Subst}\left (\int \frac{b-x^2}{b^2+x^4} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \operatorname{Subst}\left (\int \frac{b+x^2}{b^2+x^4} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{b}+2 x}{-b-\sqrt{2} \sqrt{b} x-x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{b}-2 x}{-b+\sqrt{2} \sqrt{b} x-x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \operatorname{Subst}\left (\int \frac{1}{b-\sqrt{2} \sqrt{b} x+x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \operatorname{Subst}\left (\int \frac{1}{b+\sqrt{2} \sqrt{b} x+x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=\frac{\sqrt{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{\sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.938711, size = 166, normalized size = 0.51 \[ \frac{\sqrt{\sin (e+f x)} \sqrt{b \sec (e+f x)} \left (-2 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}\right )+2 \tan ^{-1}\left (\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}+1\right )-\log \left (\sqrt{\tan ^2(e+f x)}-\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}+1\right )+\log \left (\sqrt{\tan ^2(e+f x)}+\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}+1\right )\right )}{2 \sqrt{2} b f \sqrt [4]{\tan ^2(e+f x)}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.11, size = 304, normalized size = 0.9 \begin{align*} -{\frac{\sqrt{2}}{2\,f\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}} \left ( i{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\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{1-\cos \left ( fx+e \right ) +\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{1-\cos \left ( fx+e \right ) +\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{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -2\,{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},1/2\,\sqrt{2} \right ) \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{{\frac{b}{\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{1}{\sqrt{b \sec \left (f x + e\right )} \sqrt{\sin \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec{\left (e + f x \right )}} \sqrt{\sin{\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 \frac{1}{\sqrt{b \sec \left (f x + e\right )} \sqrt{\sin \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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