Integrand size = 33, antiderivative size = 453 \[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {2 b}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}-\frac {2 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}} \]
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Time = 0.59 (sec) , antiderivative size = 453, 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 = {2981, 2716, 2721, 2719, 2645, 30, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f g^{3/2} \left (b^2-a^2\right )^{5/4}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f g^{3/2} \left (b^2-a^2\right )^{5/4}}-\frac {2 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)}}-\frac {2 b}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f g \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f g \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}} \]
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Rule 30
Rule 211
Rule 214
Rule 304
Rule 335
Rule 2645
Rule 2716
Rule 2719
Rule 2721
Rule 2780
Rule 2884
Rule 2886
Rule 2981
Rubi steps \begin{align*} \text {integral}& = \frac {a \int \frac {1}{(g \cos (e+f x))^{3/2}} \, dx}{a^2-b^2}-\frac {b \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{3/2}} \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{\left (a^2-b^2\right ) g^2} \\ & = \frac {2 a \sin (e+f x)}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}-\frac {a \int \sqrt {g \cos (e+f x)} \, dx}{\left (a^2-b^2\right ) g^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 \left (a^2-b^2\right ) 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 \left (a^2-b^2\right ) g}+\frac {b \text {Subst}\left (\int \frac {1}{x^{3/2}} \, dx,x,g \cos (e+f x)\right )}{\left (a^2-b^2\right ) f g}-\frac {\left (a^2 b\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) g^2+b^2 x^2} \, dx,x,g \cos (e+f x)\right )}{\left (a^2-b^2\right ) f g} \\ & = -\frac {2 b}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}-\frac {\left (2 a^2 b\right ) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) g^2+b^2 x^4} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{\left (a^2-b^2\right ) f g}+\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 \left (a^2-b^2\right ) g \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 \left (a^2-b^2\right ) g \sqrt {g \cos (e+f x)}}-\frac {\left (a \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{\left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)}} \\ & = -\frac {2 b}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}-\frac {2 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\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 \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \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 \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{\left (a^2-b^2\right ) f g \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 )}{\left (a^2-b^2\right ) f g}-\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 )}{\left (a^2-b^2\right ) f g} \\ & = \frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\sqrt {b} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {2 b}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}-\frac {2 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\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 \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \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 \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 14.28 (sec) , antiderivative size = 785, normalized size of antiderivative = 1.73 \[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {2 \cos (e+f x) (-b+a \sin (e+f x))}{\left (a^2-b^2\right ) f (g \cos (e+f x))^{3/2}}-\frac {a \cos ^{\frac {3}{2}}(e+f x) \left (-\frac {4 a \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \sin ^2(e+f x)}{12 \sqrt {b} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{(a-b) (a+b) f (g \cos (e+f x))^{3/2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 4.32 (sec) , antiderivative size = 1209, normalized size of antiderivative = 2.67
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\[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
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Timed out. \[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
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\[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
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\[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
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Timed out. \[ \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]
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