Integrand size = 25, antiderivative size = 461 \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=-\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d} \]
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Time = 1.53 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2774, 2944, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=-\frac {e^{9/2} \left (b^2-a^2\right )^{7/4} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac {e^{9/2} \left (b^2-a^2\right )^{7/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac {a e^5 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b^5 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a e^5 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b^5 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}-\frac {2 a e^4 \left (5 a^2-8 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 (e \sin (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right )}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d} \]
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Rule 211
Rule 214
Rule 304
Rule 335
Rule 2719
Rule 2721
Rule 2774
Rule 2780
Rule 2884
Rule 2886
Rule 2944
Rule 2946
Rubi steps \begin{align*} \text {integral}& = -\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {e^2 \int \frac {(-b-a \cos (c+d x)) (e \sin (c+d x))^{5/2}}{a+b \cos (c+d x)} \, dx}{b} \\ & = \frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (2 e^4\right ) \int \frac {\left (\frac {1}{2} b \left (2 a^2-5 b^2\right )+\frac {1}{2} a \left (5 a^2-8 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{5 b^3} \\ & = \frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (a \left (5 a^2-8 b^2\right ) e^4\right ) \int \sqrt {e \sin (c+d x)} \, dx}{5 b^4}+\frac {\left (\left (a^2-b^2\right )^2 e^4\right ) \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{b^4} \\ & = \frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (a \left (a^2-b^2\right )^2 e^5\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b^5}+\frac {\left (a \left (a^2-b^2\right )^2 e^5\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b^5}-\frac {\left (\left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \sin (c+d x)\right )}{b^3 d}-\frac {\left (a \left (5 a^2-8 b^2\right ) e^4 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 b^4 \sqrt {\sin (c+d x)}} \\ & = -\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (2 \left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^3 d}-\frac {\left (a \left (a^2-b^2\right )^2 e^5 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b^5 \sqrt {e \sin (c+d x)}}+\frac {\left (a \left (a^2-b^2\right )^2 e^5 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b^5 \sqrt {e \sin (c+d x)}} \\ & = \frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}+\frac {\left (\left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^4 d}-\frac {\left (\left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^4 d} \\ & = -\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 36.31 (sec) , antiderivative size = 834, normalized size of antiderivative = 1.81 \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=-\frac {(e \sin (c+d x))^{9/2} \left (\frac {\left (5 a^3-8 a b^2\right ) \cos ^2(c+d x) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d 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 {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )+8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{12 b^{3/2} \left (-a^2+b^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {2 \left (2 a^2 b-5 b^3\right ) \cos (c+d x) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d 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 {\sin (c+d x)}+i b \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}+\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}\right ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{5 b^3 d \sin ^{\frac {9}{2}}(c+d x)}+\frac {\csc ^4(c+d x) (e \sin (c+d x))^{9/2} \left (-\frac {\left (-28 a^2+37 b^2\right ) \sin (c+d x)}{42 b^3}-\frac {a \sin (2 (c+d x))}{5 b^2}+\frac {\sin (3 (c+d x))}{14 b}\right )}{d} \]
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Time = 5.37 (sec) , antiderivative size = 851, normalized size of antiderivative = 1.85
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Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{9/2}}{a+b\,\cos \left (c+d\,x\right )} \,d x \]
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