Optimal. Leaf size=148 \[ \frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d}+\frac{a x \left (2 a^2-b^2\right )}{2 b^4}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{3 b d} \]
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Rubi [A] time = 0.458729, antiderivative size = 148, 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, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d}+\frac{a x \left (2 a^2-b^2\right )}{2 b^4}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{3 b d} \]
Antiderivative was successfully verified.
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Rule 2889
Rule 3050
Rule 3049
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)} \, dx &=\int \frac{\sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac{\int \frac{\sin (c+d x) \left (-2 a+b \sin (c+d x)+3 a \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 b}\\ &=-\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac{\int \frac{3 a^2-a b \sin (c+d x)-2 \left (3 a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^2}\\ &=\frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac{\int \frac{3 a^2 b+3 a \left (2 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^3}\\ &=\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{3 b d}-\frac{\left (a^2 \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^4}\\ &=\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{3 b d}-\frac{\left (2 a^2 \left (a^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{a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac{\left (4 a^2 \left (a^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{a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac{2 a^2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^4 d}+\frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{3 b d}\\ \end{align*}
Mathematica [A] time = 0.25919, size = 130, normalized size = 0.88 \[ -\frac{3 b \left (b^2-4 a^2\right ) \cos (c+d x)+24 a^2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-12 a^3 c-12 a^3 d x+3 a b^2 \sin (2 (c+d x))+6 a b^2 c+6 a b^2 d x+b^3 \cos (3 (c+d x))}{12 b^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.082, size = 318, normalized size = 2.2 \begin{align*}{\frac{a}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}{a}^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+4\,{\frac{{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{a}{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 ) ^{-3}}+2\,{\frac{{a}^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{2}{3\,bd} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{3}}{d{b}^{4}}}-{\frac{a}{d{b}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{a}^{2}\sqrt{{a}^{2}-{b}^{2}}}{d{b}^{4}}\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 = 1.60131, size = 736, normalized size = 4.97 \begin{align*} \left [-\frac{2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 3 \, \sqrt{-a^{2} + b^{2}} a^{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 ) - 3 \,{\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}, -\frac{2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 6 \, \sqrt{a^{2} - b^{2}} a^{2} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 3 \,{\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}\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 = 1.19669, size = 279, normalized size = 1.89 \begin{align*} \frac{\frac{3 \,{\left (2 \, a^{3} - a b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{12 \,{\left (a^{4} - a^{2} 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{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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