Integrand size = 16, antiderivative size = 51 \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \]
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Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3262, 3261} \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )}{f \sqrt {a+b \sin ^2(e+f x)}} \]
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Rule 3261
Rule 3262
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+\frac {b \sin ^2(e+f x)}{a}} \int \frac {1}{\sqrt {1+\frac {b \sin ^2(e+f x)}{a}}} \, dx}{\sqrt {a+b \sin ^2(e+f x)}} \\ & = \frac {\operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )}{f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.55 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02
method | result | size |
default | \(\frac {\sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \operatorname {am}^{-1}\left (f x +e \bigg | \frac {i \sqrt {b}}{\sqrt {a}}\right )}{f \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}\) | \(52\) |
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Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 305, normalized size of antiderivative = 5.98 \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {{\left (2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (-2 i \, a - i \, b\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (-2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (2 i \, a + i \, b\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{b^{2} f} \]
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\[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {1}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]
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\[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
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\[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
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Timed out. \[ \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {1}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]
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