Optimal. Leaf size=336 \[ \frac{a^2 \left (5 A d \left (-8 c^2 d+c^3-20 c d^2-8 d^3\right )-2 B \left (16 c^2 d^2-5 c^3 d+c^4+40 c d^3+18 d^4\right )\right ) \cos (e+f x)}{30 d^2 f}+\frac{a^2 \left (5 A d (c-8 d)-2 B \left (c^2-5 c d+18 d^2\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 d^2 f}+\frac{a^2 \left (5 A d \left (2 c^2-16 c d-21 d^2\right )-B \left (-20 c^2 d+4 c^3+66 c d^2+90 d^3\right )\right ) \sin (e+f x) \cos (e+f x)}{120 d f}+\frac{1}{8} a^2 x \left (12 A c^2+16 A c d+7 A d^2+8 B c^2+14 B c d+6 B d^2\right )+\frac{a^2 (2 B (c-3 d)-5 A d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 d^2 f}-\frac{B \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right ) (c+d \sin (e+f x))^3}{5 d f} \]
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Rubi [A] time = 0.703095, antiderivative size = 336, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2976, 2968, 3023, 2753, 2734} \[ \frac{a^2 \left (5 A d \left (-8 c^2 d+c^3-20 c d^2-8 d^3\right )-2 B \left (16 c^2 d^2-5 c^3 d+c^4+40 c d^3+18 d^4\right )\right ) \cos (e+f x)}{30 d^2 f}+\frac{a^2 \left (5 A d (c-8 d)-2 B \left (c^2-5 c d+18 d^2\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 d^2 f}+\frac{a^2 \left (5 A d \left (2 c^2-16 c d-21 d^2\right )-B \left (-20 c^2 d+4 c^3+66 c d^2+90 d^3\right )\right ) \sin (e+f x) \cos (e+f x)}{120 d f}+\frac{1}{8} a^2 x \left (12 A c^2+16 A c d+7 A d^2+8 B c^2+14 B c d+6 B d^2\right )+\frac{a^2 (2 B (c-3 d)-5 A d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 d^2 f}-\frac{B \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right ) (c+d \sin (e+f x))^3}{5 d f} \]
Antiderivative was successfully verified.
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Rule 2976
Rule 2968
Rule 3023
Rule 2753
Rule 2734
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx &=-\frac{B \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right ) (c+d \sin (e+f x))^3}{5 d f}+\frac{\int (a+a \sin (e+f x)) (c+d \sin (e+f x))^2 (a (5 A d+B (c+3 d))-a (2 B c-5 A d-6 B d) \sin (e+f x)) \, dx}{5 d}\\ &=-\frac{B \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right ) (c+d \sin (e+f x))^3}{5 d f}+\frac{\int (c+d \sin (e+f x))^2 \left (a^2 (5 A d+B (c+3 d))+\left (-a^2 (2 B c-5 A d-6 B d)+a^2 (5 A d+B (c+3 d))\right ) \sin (e+f x)-a^2 (2 B c-5 A d-6 B d) \sin ^2(e+f x)\right ) \, dx}{5 d}\\ &=\frac{a^2 (2 B (c-3 d)-5 A d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 d^2 f}-\frac{B \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right ) (c+d \sin (e+f x))^3}{5 d f}+\frac{\int (c+d \sin (e+f x))^2 \left (-a^2 d (2 B c-35 A d-30 B d)-a^2 \left (5 A (c-8 d) d-2 B \left (c^2-5 c d+18 d^2\right )\right ) \sin (e+f x)\right ) \, dx}{20 d^2}\\ &=\frac{a^2 \left (5 A (c-8 d) d-2 B \left (c^2-5 c d+18 d^2\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 d^2 f}+\frac{a^2 (2 B (c-3 d)-5 A d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 d^2 f}-\frac{B \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right ) (c+d \sin (e+f x))^3}{5 d f}+\frac{\int (c+d \sin (e+f x)) \left (a^2 d \left (5 A d (19 c+16 d)-B \left (2 c^2-70 c d-72 d^2\right )\right )-a^2 \left (5 A d \left (2 c^2-16 c d-21 d^2\right )-2 B \left (2 c^3-10 c^2 d+33 c d^2+45 d^3\right )\right ) \sin (e+f x)\right ) \, dx}{60 d^2}\\ &=\frac{1}{8} a^2 \left (12 A c^2+8 B c^2+16 A c d+14 B c d+7 A d^2+6 B d^2\right ) x+\frac{a^2 \left (5 A d \left (c^3-8 c^2 d-20 c d^2-8 d^3\right )-2 B \left (c^4-5 c^3 d+16 c^2 d^2+40 c d^3+18 d^4\right )\right ) \cos (e+f x)}{30 d^2 f}+\frac{a^2 \left (5 A d \left (2 c^2-16 c d-21 d^2\right )-B \left (4 c^3-20 c^2 d+66 c d^2+90 d^3\right )\right ) \cos (e+f x) \sin (e+f x)}{120 d f}+\frac{a^2 \left (5 A (c-8 d) d-2 B \left (c^2-5 c d+18 d^2\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 d^2 f}+\frac{a^2 (2 B (c-3 d)-5 A d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 d^2 f}-\frac{B \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right ) (c+d \sin (e+f x))^3}{5 d f}\\ \end{align*}
Mathematica [A] time = 1.52556, size = 296, normalized size = 0.88 \[ -\frac{a^2 \cos (e+f x) \left (60 \left (A \left (12 c^2+16 c d+7 d^2\right )+2 B \left (4 c^2+7 c d+3 d^2\right )\right ) \sin ^{-1}\left (\frac{\sqrt{1-\sin (e+f x)}}{\sqrt{2}}\right )+\sqrt{\cos ^2(e+f x)} \left (-8 \left (10 A d (c+d)+B \left (5 c^2+20 c d+12 d^2\right )\right ) \cos (2 (e+f x))+120 A c^2 \sin (e+f x)+480 A c^2+480 A c d \sin (e+f x)+880 A c d+255 A d^2 \sin (e+f x)-15 A d^2 \sin (3 (e+f x))+400 A d^2+240 B c^2 \sin (e+f x)+440 B c^2+510 B c d \sin (e+f x)-30 B c d \sin (3 (e+f x))+800 B c d+270 B d^2 \sin (e+f x)-30 B d^2 \sin (3 (e+f x))+6 B d^2 \cos (4 (e+f x))+378 B d^2\right )\right )}{240 f \sqrt{\cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.066, size = 496, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.992176, size = 645, normalized size = 1.92 \begin{align*} \frac{120 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c^{2} + 480 \,{\left (f x + e\right )} A a^{2} c^{2} + 160 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c^{2} + 240 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c^{2} + 320 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} c d + 480 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c d + 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c d + 30 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c d + 240 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c d + 320 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} d^{2} + 15 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} d^{2} + 120 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} d^{2} - 32 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a^{2} d^{2} + 160 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} d^{2} + 30 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} d^{2} - 960 \, A a^{2} c^{2} \cos \left (f x + e\right ) - 480 \, B a^{2} c^{2} \cos \left (f x + e\right ) - 960 \, A a^{2} c d \cos \left (f x + e\right )}{480 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.26476, size = 574, normalized size = 1.71 \begin{align*} -\frac{24 \, B a^{2} d^{2} \cos \left (f x + e\right )^{5} - 40 \,{\left (B a^{2} c^{2} + 2 \,{\left (A + 2 \, B\right )} a^{2} c d +{\left (2 \, A + 3 \, B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (4 \,{\left (3 \, A + 2 \, B\right )} a^{2} c^{2} + 2 \,{\left (8 \, A + 7 \, B\right )} a^{2} c d +{\left (7 \, A + 6 \, B\right )} a^{2} d^{2}\right )} f x + 240 \,{\left ({\left (A + B\right )} a^{2} c^{2} + 2 \,{\left (A + B\right )} a^{2} c d +{\left (A + B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right ) - 15 \,{\left (2 \,{\left (2 \, B a^{2} c d +{\left (A + 2 \, B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (4 \,{\left (A + 2 \, B\right )} a^{2} c^{2} + 2 \,{\left (8 \, A + 9 \, B\right )} a^{2} c d +{\left (9 \, A + 10 \, B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 5.84911, size = 1129, normalized size = 3.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.30365, size = 420, normalized size = 1.25 \begin{align*} -\frac{B a^{2} d^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{1}{8} \,{\left (12 \, A a^{2} c^{2} + 8 \, B a^{2} c^{2} + 16 \, A a^{2} c d + 14 \, B a^{2} c d + 7 \, A a^{2} d^{2} + 6 \, B a^{2} d^{2}\right )} x + \frac{{\left (4 \, B a^{2} c^{2} + 8 \, A a^{2} c d + 16 \, B a^{2} c d + 8 \, A a^{2} d^{2} + 9 \, B a^{2} d^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (16 \, A a^{2} c^{2} + 14 \, B a^{2} c^{2} + 28 \, A a^{2} c d + 24 \, B a^{2} c d + 12 \, A a^{2} d^{2} + 11 \, B a^{2} d^{2}\right )} \cos \left (f x + e\right )}{8 \, f} + \frac{{\left (2 \, B a^{2} c d + A a^{2} d^{2} + 2 \, B a^{2} d^{2}\right )} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} - \frac{{\left (A a^{2} c^{2} + 2 \, B a^{2} c^{2} + 4 \, A a^{2} c d + 4 \, B a^{2} c d + 2 \, A a^{2} d^{2} + 2 \, B a^{2} d^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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