Integrand size = 33, antiderivative size = 380 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \]
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Time = 0.67 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {2988, 2645, 30, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^2 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^2 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g} \]
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Rule 30
Rule 211
Rule 214
Rule 218
Rule 335
Rule 2645
Rule 2720
Rule 2721
Rule 2781
Rule 2884
Rule 2886
Rule 2946
Rule 2988
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{b}-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b} \\ & = -\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b^2}+\frac {a^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^2}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,g \cos (e+f x)\right )}{b f g} \\ & = -\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {a^3 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {-a^2+b^2}}-\frac {a^3 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {-a^2+b^2}}+\frac {\left (a^2 g\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) g^2+b^2 x^2\right )} \, dx,x,g \cos (e+f x)\right )}{b f}-\frac {\left (a \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b^2 \sqrt {g \cos (e+f x)}} \\ & = -\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {\left (2 a^2 g\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) g^2+b^2 x^4} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b f}-\frac {\left (a^3 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}-\frac {\left (a^3 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}} \\ & = -\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \sqrt {-a^2+b^2} f}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \sqrt {-a^2+b^2} f} \\ & = -\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 14.43 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {\cos ^2(e+f x) \left (a^2-b^2+b^2 \cos ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4} \left (\frac {a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{a^2-b^2}+\frac {\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt {b}}\right )}{\left (-a^2+b^2\right )^{3/4}}-\frac {a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt {b}}\right )}{\left (-a^2+b^2\right )^{3/4}}-\frac {2 b}{\sqrt [4]{\sec ^2(e+f x)}}+\frac {a^3 \operatorname {EllipticPi}\left (-\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \tan (e+f x)}{\left (a^2-b^2\right ) \sqrt {-\tan ^2(e+f x)}}+\frac {a^3 \operatorname {EllipticPi}\left (\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \tan (e+f x)}{\left (a^2-b^2\right ) \sqrt {-\tan ^2(e+f x)}}}{b^2}\right )}{f \sqrt {g \cos (e+f x)} (a+b \sin (e+f x)) \left (a-b \cos (e+f x) \sqrt {\sec ^2(e+f x)} \sin (e+f x)\right )} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.91 (sec) , antiderivative size = 995, normalized size of antiderivative = 2.62
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Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
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\[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
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\[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
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Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]
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