Optimal. Leaf size=92 \[ \frac{2 a^2 (c-d)^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f \sqrt{c^2-d^2}}-\frac{a^2 x (c-2 d)}{d^2}-\frac{a^2 \cos (e+f x)}{d f} \]
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Rubi [A] time = 0.202015, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2746, 2735, 2660, 618, 204} \[ \frac{2 a^2 (c-d)^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f \sqrt{c^2-d^2}}-\frac{a^2 x (c-2 d)}{d^2}-\frac{a^2 \cos (e+f x)}{d f} \]
Antiderivative was successfully verified.
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Rule 2746
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx &=-\frac{a^2 \cos (e+f x)}{d f}+\frac{\int \frac{a^2 d-a^2 (c-2 d) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{d}\\ &=-\frac{a^2 (c-2 d) x}{d^2}-\frac{a^2 \cos (e+f x)}{d f}+\frac{\left (a^2 (c-d)^2\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{d^2}\\ &=-\frac{a^2 (c-2 d) x}{d^2}-\frac{a^2 \cos (e+f x)}{d f}+\frac{\left (2 a^2 (c-d)^2\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^2 f}\\ &=-\frac{a^2 (c-2 d) x}{d^2}-\frac{a^2 \cos (e+f x)}{d f}-\frac{\left (4 a^2 (c-d)^2\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^2 f}\\ &=-\frac{a^2 (c-2 d) x}{d^2}+\frac{2 a^2 (c-d)^2 \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{d^2 \sqrt{c^2-d^2} f}-\frac{a^2 \cos (e+f x)}{d f}\\ \end{align*}
Mathematica [A] time = 0.405786, size = 130, normalized size = 1.41 \[ -\frac{a^2 (\sin (e+f x)+1)^2 \left (\sqrt{c^2-d^2} ((c-2 d) (e+f x)+d \cos (e+f x))-2 (c-d)^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )\right )}{d^2 f \sqrt{c^2-d^2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.095, size = 228, normalized size = 2.5 \begin{align*} 2\,{\frac{{a}^{2}{c}^{2}}{f{d}^{2}\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-4\,{\frac{{a}^{2}c}{df\sqrt{{c}^{2}-{d}^{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{{a}^{2}}{f\sqrt{{c}^{2}-{d}^{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{{a}^{2}}{df \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c}{f{d}^{2}}}+4\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{df}} \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 = 1.72513, size = 668, normalized size = 7.26 \begin{align*} \left [-\frac{2 \, a^{2} d \cos \left (f x + e\right ) + 2 \,{\left (a^{2} c - 2 \, a^{2} d\right )} f x +{\left (a^{2} c - a^{2} d\right )} \sqrt{-\frac{c - d}{c + d}} \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 ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right )}{2 \, d^{2} f}, -\frac{a^{2} d \cos \left (f x + e\right ) +{\left (a^{2} c - 2 \, a^{2} d\right )} f x +{\left (a^{2} c - a^{2} d\right )} \sqrt{\frac{c - d}{c + d}} \arctan \left (-\frac{{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt{\frac{c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right )}{d^{2} 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 = 1.41076, size = 184, normalized size = 2. \begin{align*} -\frac{\frac{{\left (a^{2} c - 2 \, a^{2} d\right )}{\left (f x + e\right )}}{d^{2}} + \frac{2 \, a^{2}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} d} - \frac{2 \,{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\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 )}}{\sqrt{c^{2} - d^{2}} d^{2}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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