Optimal. Leaf size=98 \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac{(b c-a d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.0964343, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac{(b c-a d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 2754
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{a+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx &=-\frac{(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{\int \frac{-a c+b d}{c+d \sin (e+f x)} \, dx}{-c^2+d^2}\\ &=-\frac{(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{(a c-b d) \int \frac{1}{c+d \sin (e+f x)} \, dx}{c^2-d^2}\\ &=-\frac{(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{(2 (a c-b d)) \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 )}{\left (c^2-d^2\right ) f}\\ &=-\frac{(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}-\frac{(4 (a c-b d)) \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 )}{\left (c^2-d^2\right ) f}\\ &=\frac{2 (a c-b d) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2} f}-\frac{(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.288538, size = 96, normalized size = 0.98 \[ \frac{\frac{2 (a c-b d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac{(a d-b c) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.073, size = 309, normalized size = 3.2 \begin{align*} 2\,{\frac{{d}^{2}\tan \left ( 1/2\,fx+e/2 \right ) a}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) c}}-2\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) db}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) }}+2\,{\frac{da}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) }}-2\,{\frac{cb}{f \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ({c}^{2}-{d}^{2} \right ) }}+2\,{\frac{ca}{f \left ({c}^{2}-{d}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{bd}{f \left ({c}^{2}-{d}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) } \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 = 1.74316, size = 871, normalized size = 8.89 \begin{align*} \left [-\frac{{\left (a c^{2} - b c d +{\left (a c d - b d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c^{2} + d^{2}} \log \left (\frac{{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \,{\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt{-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \,{\left (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) +{\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f\right )}}, -\frac{{\left (a c^{2} - b c d +{\left (a c d - b d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{c^{2} - d^{2}} \arctan \left (-\frac{c \sin \left (f x + e\right ) + d}{\sqrt{c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) +{\left (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{{\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) +{\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f}\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 = 1.22791, size = 213, normalized size = 2.17 \begin{align*} \frac{2 \,{\left (\frac{{\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 (a c - b d\right )}}{{\left (c^{2} - d^{2}\right )}^{\frac{3}{2}}} - \frac{b c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b c^{2} - a c d}{{\left (c^{3} - c d^{2}\right )}{\left (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 )}}\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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