Optimal. Leaf size=174 \[ \frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{\sqrt{2} b}-\frac{\log \left (\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b}+\frac{\log \left (\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b} \]
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Rubi [A] time = 0.0889395, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2575, 297, 1162, 617, 204, 1165, 628} \[ \frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{\sqrt{2} b}-\frac{\log \left (\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b}+\frac{\log \left (\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b} \]
Antiderivative was successfully verified.
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Rule 2575
Rule 297
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}} \, dx &=-\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{b}-\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}\\ &=-\frac{\log \left (1+\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}+\frac{\log \left (1+\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}\\ &=\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\log \left (1+\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}+\frac{\log \left (1+\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}\\ \end{align*}
Mathematica [C] time = 0.025568, size = 55, normalized size = 0.32 \[ \frac{2 \sqrt{\sin (a+b x)} \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{4};\frac{5}{4};\sin ^2(a+b x)\right )}{b \sqrt{\cos (a+b x)}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.086, size = 298, normalized size = 1.7 \begin{align*} -{\frac{\sqrt{2}}{2\,b \left ( -1+\cos \left ( bx+a \right ) \right ) }\sqrt{-{\frac{-1+\cos \left ( bx+a \right ) -\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}} \left ( i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( bx+a \right ) -\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( bx+a \right ) -\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( bx+a \right ) -\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( bx+a \right ) -\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -2\,{\it EllipticF} \left ( \sqrt{-{\frac{-1+\cos \left ( bx+a \right ) -\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{\cos \left ( bx+a \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{\sqrt{\cos \left (b x + a\right )}}{\sqrt{\sin \left (b x + a\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{\sqrt{\cos{\left (a + b x \right )}}}{\sqrt{\sin{\left (a + b 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{\sqrt{\cos \left (b x + a\right )}}{\sqrt{\sin \left (b x + a\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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