Optimal. Leaf size=152 \[ -\frac{2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2-6 b^2\right )}{2 a^4}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{a d (a \cos (c+d x)+b)} \]
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Rubi [A] time = 0.570967, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2889, 3048, 3050, 3023, 2735, 2659, 208} \[ -\frac{2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2-6 b^2\right )}{2 a^4}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{a d (a \cos (c+d x)+b)} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2889
Rule 3048
Rule 3050
Rule 3023
Rule 2735
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sin ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\int \frac{\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{(-b-a \cos (c+d x))^2} \, dx\\ &=\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (2 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{-b-a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac{\int \frac{3 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \cos (c+d x)-6 b \left (a^2-b^2\right ) \cos ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}-\frac{\int \frac{-3 a b \left (a^2-b^2\right )+\left (a^2-6 b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{-b-a \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2-6 b^2\right ) x}{2 a^4}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac{\left (b \left (2 a^2-3 b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^4}\\ &=\frac{\left (a^2-6 b^2\right ) x}{2 a^4}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac{\left (2 b \left (2 a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=\frac{\left (a^2-6 b^2\right ) x}{2 a^4}-\frac{2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 \sqrt{a-b} \sqrt{a+b} d}+\frac{3 b \sin (c+d x)}{a^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.1514, size = 178, normalized size = 1.17 \[ \frac{\frac{-a \left (a^2-24 b^2\right ) \sin (c+d x)+4 a \left (a^2-6 b^2\right ) (c+d x) \cos (c+d x)+6 a^2 b \sin (2 (c+d x))+4 a^2 b c+4 a^2 b d x+a^3 (-\sin (3 (c+d x)))-24 b^3 c-24 b^3 d x}{a \cos (c+d x)+b}+\frac{16 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{8 a^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.072, size = 325, normalized size = 2.1 \begin{align*}{\frac{1}{d{a}^{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{b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+4\,{\frac{b\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{d{a}^{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}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){b}^{2}}{d{a}^{4}}}+{\frac{1}{d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-4\,{\frac{b}{d{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+6\,{\frac{{b}^{3}}{d{a}^{4}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \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 = 2.03398, size = 1204, normalized size = 7.92 \begin{align*} \left [\frac{{\left (a^{5} - 7 \, a^{3} b^{2} + 6 \, a b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (a^{4} b - 7 \, a^{2} b^{3} + 6 \, b^{5}\right )} d x -{\left (2 \, a^{2} b^{2} - 3 \, b^{4} +{\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) +{\left (6 \, a^{3} b^{2} - 6 \, a b^{4} -{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - a^{5} b^{2}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - a^{4} b^{3}\right )} d\right )}}, \frac{{\left (a^{5} - 7 \, a^{3} b^{2} + 6 \, a b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (a^{4} b - 7 \, a^{2} b^{3} + 6 \, b^{5}\right )} d x - 2 \,{\left (2 \, a^{2} b^{2} - 3 \, b^{4} +{\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) +{\left (6 \, a^{3} b^{2} - 6 \, a b^{4} -{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - a^{5} b^{2}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - a^{4} b^{3}\right )} d\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.30816, size = 324, normalized size = 2.13 \begin{align*} -\frac{\frac{4 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (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 )^{2} - a - b\right )} a^{3}} - \frac{{\left (a^{2} - 6 \, b^{2}\right )}{\left (d x + c\right )}}{a^{4}} + \frac{4 \,{\left (2 \, a^{2} b - 3 \, b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{\sqrt{-a^{2} + b^{2}} a^{4}} - \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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