Integrand size = 25, antiderivative size = 512 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}+\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \]
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Time = 1.24 (sec) , antiderivative size = 512, 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 = {2772, 2942, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {5 e^{7/2} \left (3 a^2-2 b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac {5 e^{7/2} \left (3 a^2-2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac {5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{8 b^4 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{8 b^4 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}-\frac {15 a e^4 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \]
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Rule 211
Rule 214
Rule 218
Rule 335
Rule 2720
Rule 2721
Rule 2772
Rule 2781
Rule 2884
Rule 2886
Rule 2942
Rule 2946
Rubi steps \begin{align*} \text {integral}& = \frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (5 e^2\right ) \int \frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx}{4 b} \\ & = \frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}+\frac {\left (5 e^4\right ) \int \frac {-b-\frac {3}{2} a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{4 b^3} \\ & = \frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (15 a e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{8 b^4}+\frac {\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{8 b^4} \\ & = \frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 b^4 \sqrt {-a^2+b^2}}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 b^4 \sqrt {-a^2+b^2}}-\frac {\left (5 \left (3 a^2-2 b^2\right ) e^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{8 b^3 d}-\frac {\left (15 a e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{8 b^4 \sqrt {e \sin (c+d x)}} \\ & = -\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (5 \left (3 a^2-2 b^2\right ) e^5\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{4 b^3 d}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4 \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}{16 b^4 \sqrt {-a^2+b^2} \sqrt {e \sin (c+d x)}}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4 \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}{16 b^4 \sqrt {-a^2+b^2} \sqrt {e \sin (c+d x)}} \\ & = -\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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)}}{8 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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)}}{8 b^4 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}+\frac {\left (5 \left (3 a^2-2 b^2\right ) e^4\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 )}{8 b^3 \sqrt {-a^2+b^2} d}+\frac {\left (5 \left (3 a^2-2 b^2\right ) e^4\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 )}{8 b^3 \sqrt {-a^2+b^2} d} \\ & = \frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}+\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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)}}{8 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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)}}{8 b^4 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.94 (sec) , antiderivative size = 946, normalized size of antiderivative = 1.85 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {\csc ^3(c+d x) (e \sin (c+d x))^{7/2} \left (7 a^2+2 b^2+9 a b \cos (c+d x)+\frac {(a+b \cos (c+d x)) (-6 b-7 a \cos (c+d x)+4 b \cos (2 (c+d x))) \left (8 (a+b)-5 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}+(3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{\frac {1}{9} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sin (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-45 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+18 (3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {9 (3 a+2 b) \left (2 (a-b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}+9 (3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-\frac {5 (3 a-2 b) \left (2 (a-b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {1}{2},2,\frac {13}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {3}{2},1,\frac {13}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+\cos (c+d x) \left (8 (a+b)-5 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}+(3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right )}{4 b^3 d (a+b \cos (c+d x))^2} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(2588\) vs. \(2(536)=1072\).
Time = 98.09 (sec) , antiderivative size = 2589, normalized size of antiderivative = 5.06
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Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
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\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]
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