Optimal. Leaf size=142 \[ -\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}+\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4} \]
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Rubi [A] time = 0.285445, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {2967, 2857, 2750, 2650, 2648} \[ -\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}+\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4} \]
Antiderivative was successfully verified.
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Rule 2967
Rule 2857
Rule 2750
Rule 2650
Rule 2648
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}+\frac{a \int \frac{-A c-8 B c-7 B c \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx}{7 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}-\frac{(a (2 A-5 B)) \int \frac{1}{(c-c \sin (e+f x))^2} \, dx}{35 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{(a (2 A-5 B)) \int \frac{1}{c-c \sin (e+f x)} \, dx}{105 c^3}\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^4-c^4 \sin (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.825032, size = 174, normalized size = 1.23 \[ \frac{a \left (35 (4 A-B) \cos \left (e+\frac{f x}{2}\right )+14 A \sin \left (2 e+\frac{5 f x}{2}\right )-42 A \cos \left (e+\frac{3 f x}{2}\right )+2 A \cos \left (3 e+\frac{7 f x}{2}\right )+70 A \sin \left (\frac{f x}{2}\right )+105 B \sin \left (2 e+\frac{3 f x}{2}\right )-35 B \sin \left (2 e+\frac{5 f x}{2}\right )-5 B \cos \left (3 e+\frac{7 f x}{2}\right )+140 B \sin \left (\frac{f x}{2}\right )\right )}{420 c^4 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.119, size = 159, normalized size = 1.1 \begin{align*} 2\,{\frac{a}{f{c}^{4}} \left ( -1/6\,{\frac{48\,A+48\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/4\,{\frac{56\,A+40\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-1/5\,{\frac{68\,A+60\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-1/2\,{\frac{8\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-1/3\,{\frac{28\,A+14\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-1/7\,{\frac{16\,A+16\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-{\frac{A}{\tan \left ( 1/2\,fx+e/2 \right ) -1}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.09437, size = 1458, normalized size = 10.27 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.3599, size = 637, normalized size = 4.49 \begin{align*} \frac{{\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{4} + 4 \,{\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{3} - 3 \,{\left (3 \, A + 10 \, B\right )} a \cos \left (f x + e\right )^{2} + 15 \,{\left (A + B\right )} a \cos \left (f x + e\right ) + 30 \,{\left (A + B\right )} a -{\left ({\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{3} - 3 \,{\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{2} - 15 \,{\left (A + B\right )} a \cos \left (f x + e\right ) - 30 \,{\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{105 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19253, size = 252, normalized size = 1.77 \begin{align*} -\frac{2 \,{\left (105 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 210 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 105 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 455 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 35 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 350 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 140 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 273 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 56 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 35 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 23 \, A a - 5 \, B a\right )}}{105 \, c^{4} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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