Optimal. Leaf size=93 \[ \frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 f \sqrt{a^2-b^2}}+\frac{d x (2 b c-a d)}{b^2}-\frac{d^2 \cos (e+f x)}{b f} \]
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Rubi [A] time = 0.167061, antiderivative size = 93, 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} \[ \frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 f \sqrt{a^2-b^2}}+\frac{d x (2 b c-a d)}{b^2}-\frac{d^2 \cos (e+f x)}{b f} \]
Antiderivative was successfully verified.
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Rule 2746
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{(c+d \sin (e+f x))^2}{a+b \sin (e+f x)} \, dx &=-\frac{d^2 \cos (e+f x)}{b f}+\frac{\int \frac{b c^2+d (2 b c-a d) \sin (e+f x)}{a+b \sin (e+f x)} \, dx}{b}\\ &=\frac{d (2 b c-a d) x}{b^2}-\frac{d^2 \cos (e+f x)}{b f}+\frac{(b c-a d)^2 \int \frac{1}{a+b \sin (e+f x)} \, dx}{b^2}\\ &=\frac{d (2 b c-a d) x}{b^2}-\frac{d^2 \cos (e+f x)}{b f}+\frac{\left (2 (b c-a d)^2\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 )}{b^2 f}\\ &=\frac{d (2 b c-a d) x}{b^2}-\frac{d^2 \cos (e+f x)}{b f}-\frac{\left (4 (b c-a d)^2\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 )}{b^2 f}\\ &=\frac{d (2 b c-a d) x}{b^2}+\frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2} f}-\frac{d^2 \cos (e+f x)}{b f}\\ \end{align*}
Mathematica [A] time = 0.159349, size = 90, normalized size = 0.97 \[ \frac{\frac{2 (b c-a d)^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+d (e+f x) (2 b c-a d)-b d^2 \cos (e+f x)}{b^2 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.065, size = 226, normalized size = 2.4 \begin{align*} -2\,{\frac{{d}^{2}}{bf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{{d}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) a}{{b}^{2}f}}+4\,{\frac{d\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c}{bf}}+2\,{\frac{{a}^{2}{d}^{2}}{{b}^{2}f\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-4\,{\frac{acd}{bf\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{{c}^{2}}{f\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{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 = 1.59945, size = 771, normalized size = 8.29 \begin{align*} \left [-\frac{2 \,{\left (a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right ) - 2 \,{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} f x +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} f}, -\frac{{\left (a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right ) -{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} f x +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (f x + e\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (f x + e\right )}\right )}{{\left (a^{2} b^{2} - b^{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 = 1.35917, size = 181, normalized size = 1.95 \begin{align*} \frac{\frac{{\left (2 \, b c d - a d^{2}\right )}{\left (f x + e\right )}}{b^{2}} - \frac{2 \, d^{2}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} b} + \frac{2 \,{\left (b^{2} c^{2} - 2 \, a b 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 (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 )}}{\sqrt{a^{2} - b^{2}} b^{2}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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