Integrand size = 37, antiderivative size = 320 \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} d^{3/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} d^{3/2} f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{a d f g \sqrt {d \sin (e+f x)}}-\frac {b \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a^2 d f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \]
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Time = 0.55 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {2989, 2643, 2653, 2720, 2987, 2986, 1232} \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{a^2 d^{3/2} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{a^2 d^{3/2} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{a d f g \sqrt {d \sin (e+f x)}} \]
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Rule 1232
Rule 2643
Rule 2653
Rule 2720
Rule 2986
Rule 2987
Rule 2989
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}} \, dx}{a}-\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx}{a d} \\ & = -\frac {2 \sqrt {g \cos (e+f x)}}{a d f g \sqrt {d \sin (e+f x)}}+\frac {b^2 \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^2 d^2}-\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{a^2 d} \\ & = -\frac {2 \sqrt {g \cos (e+f x)}}{a d f g \sqrt {d \sin (e+f x)}}+\frac {\left (b^2 \sqrt {\cos (e+f x)}\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {\left (b \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{a^2 d \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \\ & = -\frac {2 \sqrt {g \cos (e+f x)}}{a d f g \sqrt {d \sin (e+f x)}}-\frac {b \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a^2 d f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}+\frac {\left (2 \sqrt {2} b^2 \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) \sqrt {\cos (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\left (b-\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a^2 d f \sqrt {g \cos (e+f x)}}+\frac {\left (2 \sqrt {2} b^2 \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) \sqrt {\cos (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\left (b+\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a^2 d f \sqrt {g \cos (e+f x)}} \\ & = -\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} d^{3/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} d^{3/2} f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{a d f g \sqrt {d \sin (e+f x)}}-\frac {b \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a^2 d f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 14.25 (sec) , antiderivative size = 715, normalized size of antiderivative = 2.23 \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \cos (e+f x) \sin (e+f x)}{a f \sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}}+\frac {2 b \sqrt {\cos (e+f x)} \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\left (1-\cos ^2(e+f x)\right )^{3/4} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-4 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+3 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) b \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )+\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}-\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )-\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}+\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )\right )}{\sqrt {a} \left (-a^2+b^2\right )^{3/4}}\right ) \sin ^2(e+f x)}{a f \sqrt {g \cos (e+f x)} \sqrt [4]{1-\cos ^2(e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(1134\) vs. \(2(298)=596\).
Time = 1.72 (sec) , antiderivative size = 1135, normalized size of antiderivative = 3.55
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Timed out. \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\left (d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}} \sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
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\[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\sqrt {g\,\cos \left (e+f\,x\right )}\,{\left (d\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]
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