Optimal. Leaf size=124 \[ -\frac{\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{2 a x}{b^3} \]
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Rubi [A] time = 0.218285, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2792, 3023, 2735, 2660, 618, 204} \[ -\frac{\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{2 a x}{b^3} \]
Antiderivative was successfully verified.
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Rule 2792
Rule 3023
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\sin ^3(x)}{(a+b \sin (x))^2} \, dx &=\frac{a^2 \cos (x) \sin (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\int \frac{a^2-a b \sin (x)-\left (2 a^2-b^2\right ) \sin ^2(x)}{a+b \sin (x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\int \frac{a^2 b+2 a \left (a^2-b^2\right ) \sin (x)}{a+b \sin (x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 a x}{b^3}-\frac{\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\left (a^2 \left (2 a^2-3 b^2\right )\right ) \int \frac{1}{a+b \sin (x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 a x}{b^3}-\frac{\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\left (2 a^2 \left (2 a^2-3 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^3 \left (a^2-b^2\right )}\\ &=-\frac{2 a x}{b^3}-\frac{\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\left (4 a^2 \left (2 a^2-3 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^3 \left (a^2-b^2\right )}\\ &=-\frac{2 a x}{b^3}+\frac{2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}-\frac{\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end{align*}
Mathematica [A] time = 0.391117, size = 94, normalized size = 0.76 \[ \frac{\frac{2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+b \cos (x) \left (-\frac{a^3}{(a-b) (a+b) (a+b \sin (x))}-1\right )-2 a x}{b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.05, size = 196, normalized size = 1.6 \begin{align*} -2\,{\frac{1}{{b}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-4\,{\frac{a\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{b}^{3}}}-2\,{\frac{{a}^{2}\tan \left ( x/2 \right ) }{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 ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}+4\,{\frac{{a}^{4}}{{b}^{3} \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 ) }-6\,{\frac{{a}^{2}}{b \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 [A] time = 1.94299, size = 1071, normalized size = 8.64 \begin{align*} \left [-\frac{{\left (2 \, a^{5} - 3 \, a^{3} b^{2} +{\left (2 \, a^{4} b - 3 \, a^{2} 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 ) + 4 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} x + 2 \,{\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (x\right ) + 2 \,{\left (2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} x +{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \,{\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7} +{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sin \left (x\right )\right )}}, -\frac{{\left (2 \, a^{5} - 3 \, a^{3} b^{2} +{\left (2 \, a^{4} b - 3 \, a^{2} 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 ) + 2 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} x +{\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (x\right ) +{\left (2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} x +{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7} +{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\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 = 1.86384, size = 275, normalized size = 2.22 \begin{align*} \frac{2 \,{\left (2 \, a^{4} - 3 \, a^{2} 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^{3} - b^{5}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, a^{3} \tan \left (\frac{1}{2} \, x\right )^{2} - a b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 3 \, a^{2} b \tan \left (\frac{1}{2} \, x\right ) - 2 \, b^{3} \tan \left (\frac{1}{2} \, x\right ) + 2 \, a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, x\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}{\left (a^{2} b^{2} - b^{4}\right )}} - \frac{2 \, a x}{b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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