Optimal. Leaf size=198 \[ -\frac{2 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^3 f (c+d) \sqrt{c^2-d^2}}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 f (c+d)}-\frac{a^2 x (-A d+2 B c-2 B d)}{d^3}+\frac{(B c-A d) \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{d f (c+d) (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.580826, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2975, 2968, 3023, 2735, 2660, 618, 204} \[ -\frac{2 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^3 f (c+d) \sqrt{c^2-d^2}}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 f (c+d)}-\frac{a^2 x (-A d+2 B c-2 B d)}{d^3}+\frac{(B c-A d) \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{d f (c+d) (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 2975
Rule 2968
Rule 3023
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx &=\frac{(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))}+\frac{\int \frac{(a+a \sin (e+f x)) (-a (B (c-d)-2 A d)-a (A d-B (2 c+d)) \sin (e+f x))}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=\frac{(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))}+\frac{\int \frac{-a^2 (B (c-d)-2 A d)+\left (-a^2 (B (c-d)-2 A d)-a^2 (A d-B (2 c+d))\right ) \sin (e+f x)-a^2 (A d-B (2 c+d)) \sin ^2(e+f x)}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 (c+d) f}+\frac{(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))}+\frac{\int \frac{-a^2 d (B (c-d)-2 A d)-a^2 (c+d) (2 B (c-d)-A d) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{d^2 (c+d)}\\ &=-\frac{a^2 (2 B c-A d-2 B d) x}{d^3}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 (c+d) f}+\frac{(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))}-\frac{\left (a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{d^3 (c+d)}\\ &=-\frac{a^2 (2 B c-A d-2 B d) x}{d^3}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 (c+d) f}+\frac{(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))}-\frac{\left (2 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{d^3 (c+d) f}\\ &=-\frac{a^2 (2 B c-A d-2 B d) x}{d^3}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 (c+d) f}+\frac{(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))}+\frac{\left (4 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac{1}{2} (e+f x)\right )\right )}{d^3 (c+d) f}\\ &=-\frac{a^2 (2 B c-A d-2 B d) x}{d^3}-\frac{2 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{d^3 (c+d) \sqrt{c^2-d^2} f}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 (c+d) f}+\frac{(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.00535, size = 192, normalized size = 0.97 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\frac{2 (c-d) \left (B \left (2 c^2+2 c d-d^2\right )-A d (c+2 d)\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}+(e+f x) (A d-2 B c+2 B d)-\frac{d (d-c) (A d-B c) \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}-B d \cos (e+f x)\right )}{d^3 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.161, size = 848, normalized size = 4.3 \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 [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.56487, size = 1589, normalized size = 8.03 \begin{align*} \text{result too large to display} \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 [B] time = 1.35175, size = 672, normalized size = 3.39 \begin{align*} \frac{\frac{2 \,{\left (2 \, B a^{2} c^{3} - A a^{2} c^{2} d - A a^{2} c d^{2} - 3 \, B a^{2} c d^{2} + 2 \, A a^{2} d^{3} + B a^{2} d^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )}}{{\left (c d^{3} + d^{4}\right )} \sqrt{c^{2} - d^{2}}} - \frac{2 \,{\left (B a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - A a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - B a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + A a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, B a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - A a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + A a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - A a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + B a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + A a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, B a^{2} c^{3} - A a^{2} c^{2} d + A a^{2} c d^{2}\right )}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c\right )}{\left (c^{2} d^{2} + c d^{3}\right )}} - \frac{{\left (2 \, B a^{2} c - A a^{2} d - 2 \, B a^{2} d\right )}{\left (f x + e\right )}}{d^{3}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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