Optimal. Leaf size=87 \[ -\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{x}{b^2} \]
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Rubi [A] time = 0.125101, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2790, 2735, 2660, 618, 204} \[ -\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{x}{b^2} \]
Antiderivative was successfully verified.
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Rule 2790
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\sin ^2(x)}{(a+b \sin (x))^2} \, dx &=\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\int \frac{a b+\left (a^2-b^2\right ) \sin (x)}{a+b \sin (x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{x}{b^2}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\left (a \left (a^2-2 b^2\right )\right ) \int \frac{1}{a+b \sin (x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac{x}{b^2}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\left (2 a \left (a^2-2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2 \left (a^2-b^2\right )}\\ &=\frac{x}{b^2}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\left (4 a \left (a^2-2 b^2\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{x}{2}\right )\right )}{b^2 \left (a^2-b^2\right )}\\ &=\frac{x}{b^2}-\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end{align*}
Mathematica [A] time = 0.214287, size = 83, normalized size = 0.95 \[ \frac{-\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a^2 b \cos (x)}{(a-b) (a+b) (a+b \sin (x))}+x}{b^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.045, size = 170, normalized size = 2. \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{b}^{2}}}+2\,{\frac{a\tan \left ( x/2 \right ) }{ \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}+2\,{\frac{{a}^{2}}{b \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{{a}^{3}}{{b}^{2} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+4\,{\frac{a}{ \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/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 [B] time = 1.85131, size = 896, normalized size = 10.3 \begin{align*} \left [\frac{2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} x \sin \left (x\right ) -{\left (a^{4} - 2 \, a^{2} b^{2} +{\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (x\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} x + 2 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6} +{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sin \left (x\right )\right )}}, \frac{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} x \sin \left (x\right ) +{\left (a^{4} - 2 \, a^{2} b^{2} +{\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (x\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} x +{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )}{a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6} +{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sin \left (x\right )}\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 = 1.93779, size = 167, normalized size = 1.92 \begin{align*} -\frac{2 \,{\left (a^{3} - 2 \, a b^{2}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (a b \tan \left (\frac{1}{2} \, x\right ) + a^{2}\right )}}{{\left (a^{2} b - b^{3}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}} + \frac{x}{b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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