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 = 0.19, antiderivative size = 328, normalized size of antiderivative = 1.00, 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 204
Rule 297
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2575
Rule 2585
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*}
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Mathematica [A] time = 0.86, 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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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sqrt {\sin \left (f x + e\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.14, size = 304, normalized size = 0.93 \[ -\frac {\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \left (i \EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-2 \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )\right ) \left (\sin ^{\frac {3}{2}}\left (f x +e \right )\right ) \sqrt {2}}{2 f \sqrt {\frac {b}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sqrt {\sin \left (f x + e\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\sqrt {\sin \left (e+f\,x\right )}\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec {\left (e + f x \right )}} \sqrt {\sin {\left (e + f x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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