Optimal. Leaf size=162 \[ \frac{\left (2 a^2 c-3 a b d+b^2 c\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac{\left (a^2 (-d)+3 a b c-2 b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac{(b c-a d) \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \]
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Rubi [A] time = 0.174177, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2754, 12, 2660, 618, 204} \[ \frac{\left (2 a^2 c-3 a b d+b^2 c\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac{\left (a^2 (-d)+3 a b c-2 b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac{(b c-a d) \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 2754
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{c+d \sin (e+f x)}{(a+b \sin (e+f x))^3} \, dx &=\frac{(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}-\frac{\int \frac{-2 (a c-b d)+(b c-a d) \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac{(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\int \frac{2 a^2 c+b^2 c-3 a b d}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\left (2 a^2 c+b^2 c-3 a b d\right ) \int \frac{1}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\left (2 a^2 c+b^2 c-3 a b d\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right )^2 f}\\ &=\frac{(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac{\left (2 \left (2 a^2 c+b^2 c-3 a b d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right )^2 f}\\ &=\frac{\left (2 a^2 c+b^2 c-3 a b d\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} f}+\frac{(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.619793, size = 157, normalized size = 0.97 \[ \frac{\frac{2 \left (2 a^2 c-3 a b d+b^2 c\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{\left (a^2 d-3 a b c+2 b^2 d\right ) \cos (e+f x)}{(a-b)^2 (a+b)^2 (a+b \sin (e+f x))}+\frac{(b c-a d) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))^2}}{2 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.087, size = 1291, normalized size = 8. \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 [B] time = 1.84008, size = 1696, normalized size = 10.47 \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 = 1.37491, size = 579, normalized size = 3.57 \begin{align*} \frac{\frac{{\left (2 \, a^{2} c + b^{2} c - 3 \, a b d\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{5 \, a^{3} b^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a b^{4} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, a^{4} b d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 4 \, a^{4} b c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 7 \, a^{2} b^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, b^{5} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a^{5} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, a^{3} b^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a b^{4} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 11 \, a^{3} b^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a b^{4} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 5 \, a^{4} b d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 4 \, a^{2} b^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, a^{4} b c - a^{2} b^{3} c - 2 \, a^{5} d - a^{3} b^{2} d}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a\right )}^{2}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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