Optimal. Leaf size=376 \[ -\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}\right )}{\sqrt{2} \sqrt{b} f}+\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}+1\right )}{\sqrt{2} \sqrt{b} f}+\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \log \left (-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)+\sqrt{a}\right )}{2 \sqrt{2} \sqrt{b} f}-\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \log \left (\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)+\sqrt{a}\right )}{2 \sqrt{2} \sqrt{b} f} \]
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Rubi [A] time = 0.264938, antiderivative size = 376, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2585, 2574, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}\right )}{\sqrt{2} \sqrt{b} f}+\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}+1\right )}{\sqrt{2} \sqrt{b} f}+\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \log \left (-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)+\sqrt{a}\right )}{2 \sqrt{2} \sqrt{b} f}-\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)} \log \left (\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)+\sqrt{a}\right )}{2 \sqrt{2} \sqrt{b} f} \]
Antiderivative was successfully verified.
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Rule 2585
Rule 2574
Rule 297
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \sqrt{b \sec (e+f x)} \sqrt{a \sin (e+f x)} \, dx &=\left (\sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}} \, dx\\ &=\frac{\left (2 a b \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{a^2+b^2 x^4} \, dx,x,\frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}\right )}{f}\\ &=-\frac{\left (a \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{a-b x^2}{a^2+b^2 x^4} \, dx,x,\frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}\right )}{f}+\frac{\left (a \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{a+b x^2}{a^2+b^2 x^4} \, dx,x,\frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}\right )}{f}\\ &=\frac{\left (a \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{a}{b}-\frac{\sqrt{2} \sqrt{a} x}{\sqrt{b}}+x^2} \, dx,x,\frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}\right )}{2 b f}+\frac{\left (a \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{a}{b}+\frac{\sqrt{2} \sqrt{a} x}{\sqrt{b}}+x^2} \, dx,x,\frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}\right )}{2 b f}+\frac{\left (\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{a}}{\sqrt{b}}+2 x}{-\frac{a}{b}-\frac{\sqrt{2} \sqrt{a} x}{\sqrt{b}}-x^2} \, dx,x,\frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}\right )}{2 \sqrt{2} \sqrt{b} f}+\frac{\left (\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{a}}{\sqrt{b}}-2 x}{-\frac{a}{b}+\frac{\sqrt{2} \sqrt{a} x}{\sqrt{b}}-x^2} \, dx,x,\frac{\sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}\right )}{2 \sqrt{2} \sqrt{b} f}\\ &=\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \log \left (\sqrt{a}-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)\right ) \sqrt{b \sec (e+f x)}}{2 \sqrt{2} \sqrt{b} f}-\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \log \left (\sqrt{a}+\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)\right ) \sqrt{b \sec (e+f x)}}{2 \sqrt{2} \sqrt{b} f}+\frac{\left (\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}\right )}{\sqrt{2} \sqrt{b} f}-\frac{\left (\sqrt{a} \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}\right )}{\sqrt{2} \sqrt{b} f}\\ &=-\frac{\sqrt{a} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}\right ) \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}{\sqrt{2} \sqrt{b} f}+\frac{\sqrt{a} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{a} \sqrt{b \cos (e+f x)}}\right ) \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}{\sqrt{2} \sqrt{b} f}+\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \log \left (\sqrt{a}-\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)\right ) \sqrt{b \sec (e+f x)}}{2 \sqrt{2} \sqrt{b} f}-\frac{\sqrt{a} \sqrt{b \cos (e+f x)} \log \left (\sqrt{a}+\frac{\sqrt{2} \sqrt{b} \sqrt{a \sin (e+f x)}}{\sqrt{b \cos (e+f x)}}+\sqrt{a} \tan (e+f x)\right ) \sqrt{b \sec (e+f x)}}{2 \sqrt{2} \sqrt{b} f}\\ \end{align*}
Mathematica [C] time = 0.128543, size = 55, normalized size = 0.15 \[ \frac{2 \tan (e+f x) \sqrt{a \sin (e+f x)} \sqrt{b \sec (e+f x)} \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(e+f x)\right )}{3 f} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.105, size = 273, normalized size = 0.7 \begin{align*} -{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{2\,f \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{a\sin \left ( fx+e \right ) }\sqrt{{\frac{b}{\cos \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 ) }{\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 ) \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{b \sec \left (f x + e\right )} \sqrt{a \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 \sqrt{a \sin{\left (e + f x \right )}} \sqrt{b \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{b \sec \left (f x + e\right )} \sqrt{a \sin \left (f x + e\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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