Optimal. Leaf size=153 \[ \frac{2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d \sqrt{a^2-b^2}}-\frac{x \left (6 a^2-b^2\right )}{2 b^4}-\frac{3 a \cos (c+d x)}{b^3 d}-\frac{\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac{3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]
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Rubi [A] time = 0.491838, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2889, 3048, 3050, 3023, 2735, 2660, 618, 204} \[ \frac{2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d \sqrt{a^2-b^2}}-\frac{x \left (6 a^2-b^2\right )}{2 b^4}-\frac{3 a \cos (c+d x)}{b^3 d}-\frac{\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac{3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]
Antiderivative was successfully verified.
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Rule 2889
Rule 3048
Rule 3050
Rule 3023
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \frac{\sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx\\ &=-\frac{\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\int \frac{\sin (c+d x) \left (-2 \left (a^2-b^2\right )+3 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\int \frac{3 a \left (a^2-b^2\right )-b \left (a^2-b^2\right ) \sin (c+d x)-6 a \left (a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{3 a \cos (c+d x)}{b^3 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\int \frac{3 a b \left (a^2-b^2\right )+\left (a^2-b^2\right ) \left (6 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (6 a^2-b^2\right ) x}{2 b^4}-\frac{3 a \cos (c+d x)}{b^3 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))}+\frac{\left (a \left (3 a^2-2 b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^4}\\ &=-\frac{\left (6 a^2-b^2\right ) x}{2 b^4}-\frac{3 a \cos (c+d x)}{b^3 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))}+\frac{\left (2 a \left (3 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{1}{2} (c+d x)\right )\right )}{b^4 d}\\ &=-\frac{\left (6 a^2-b^2\right ) x}{2 b^4}-\frac{3 a \cos (c+d x)}{b^3 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (4 a \left (3 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{1}{2} (c+d x)\right )\right )}{b^4 d}\\ &=-\frac{\left (6 a^2-b^2\right ) x}{2 b^4}+\frac{2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^4 \sqrt{a^2-b^2} d}-\frac{3 a \cos (c+d x)}{b^3 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{b d (a+b \sin (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.376863, size = 129, normalized size = 0.84 \[ \frac{2 \left (b^2-6 a^2\right ) (c+d x)+\frac{8 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{4 a^2 b \cos (c+d x)}{a+b \sin (c+d x)}-8 a b \cos (c+d x)+b^2 \sin (2 (c+d x))}{4 b^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.117, size = 353, normalized size = 2.3 \begin{align*} -{\frac{1}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{d{b}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-4\,{\frac{a}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{d{b}^{4}}}+{\frac{1}{d{b}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-2\,{\frac{{a}^{2}}{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+6\,{\frac{{a}^{3}}{d{b}^{4}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-4\,{\frac{a}{d{b}^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/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 = 1.66269, size = 1237, normalized size = 8.08 \begin{align*} \left [-\frac{{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x -{\left (3 \, a^{4} - 2 \, a^{2} b^{2} +{\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) +{\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) +{\left ({\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x + 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{2} b^{5} - b^{7}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}, -\frac{{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x + 2 \,{\left (3 \, a^{4} - 2 \, a^{2} b^{2} +{\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) +{\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) +{\left ({\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x + 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{2} b^{5} - b^{7}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{4} - a b^{6}\right )} d\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 = 1.30521, size = 285, normalized size = 1.86 \begin{align*} -\frac{\frac{{\left (6 \, a^{2} - b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{4 \,{\left (3 \, a^{3} - 2 \, a b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} b^{4}} + \frac{4 \,{\left (a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )} b^{3}} + \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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