Integrand size = 38, antiderivative size = 188 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f} \]
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Time = 0.47 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2819, 2817} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt {a \sin (e+f x)+a}}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}-\frac {3 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f} \]
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Rule 2817
Rule 2819
Rule 2920
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx}{a c} \\ & = -\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac {3 \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2} \, dx}{4 c} \\ & = -\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac {(3 a) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2} \, dx}{7 c} \\ & = -\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac {a^2 \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2} \, dx}{7 c} \\ & = -\frac {a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f} \\ \end{align*}
Time = 9.28 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.94 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3 (-1+\sin (e+f x))^3 (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)} (1960 \cos (2 (e+f x))+980 \cos (4 (e+f x))+280 \cos (6 (e+f x))+35 \cos (8 (e+f x))+19600 \sin (e+f x)+3920 \sin (3 (e+f x))+784 \sin (5 (e+f x))+80 \sin (7 (e+f x)))}{35840 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
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Time = 214.70 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.59
method | result | size |
default | \(\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a^{2} c^{3} \left (35 \left (\cos ^{7}\left (f x +e \right )\right )+40 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )+48 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+64 \cos \left (f x +e \right ) \sin \left (f x +e \right )+128 \tan \left (f x +e \right )-35 \sec \left (f x +e \right )\right )}{280 f}\) | \(110\) |
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Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.68 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\left (35 \, a^{2} c^{3} \cos \left (f x + e\right )^{8} - 35 \, a^{2} c^{3} + 8 \, {\left (5 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 6 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 8 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{280 \, f \cos \left (f x + e\right )} \]
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Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \]
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\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \cos \left (f x + e\right )^{2} \,d x } \]
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Time = 0.34 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {32 \, {\left (35 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{16} - 120 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} + 140 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 56 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10}\right )} \sqrt {a} \sqrt {c}}{35 \, f} \]
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Time = 13.29 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.00 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {35\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{128\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{512\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]
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