Optimal. Leaf size=106 \[ \frac{2 \tan ^5(c+d x)}{5 a^2 d}-\frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{\tan (c+d x)}{a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{4 \sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}-\frac{x}{a^2} \]
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Rubi [A] time = 0.282228, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2873, 2607, 30, 2606, 194, 3473, 8} \[ \frac{2 \tan ^5(c+d x)}{5 a^2 d}-\frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{\tan (c+d x)}{a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{4 \sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}-\frac{x}{a^2} \]
Antiderivative was successfully verified.
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Rule 2875
Rule 2873
Rule 2607
Rule 30
Rule 2606
Rule 194
Rule 3473
Rule 8
Rubi steps
\begin{align*} \int \frac{\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \sec ^2(c+d x) (a-a \sin (c+d x))^2 \tan ^4(c+d x) \, dx}{a^4}\\ &=\frac{\int \left (a^2 \sec ^2(c+d x) \tan ^4(c+d x)-2 a^2 \sec (c+d x) \tan ^5(c+d x)+a^2 \tan ^6(c+d x)\right ) \, dx}{a^4}\\ &=\frac{\int \sec ^2(c+d x) \tan ^4(c+d x) \, dx}{a^2}+\frac{\int \tan ^6(c+d x) \, dx}{a^2}-\frac{2 \int \sec (c+d x) \tan ^5(c+d x) \, dx}{a^2}\\ &=\frac{\tan ^5(c+d x)}{5 a^2 d}-\frac{\int \tan ^4(c+d x) \, dx}{a^2}+\frac{\operatorname{Subst}\left (\int x^4 \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac{2 \operatorname{Subst}\left (\int \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{2 \tan ^5(c+d x)}{5 a^2 d}+\frac{\int \tan ^2(c+d x) \, dx}{a^2}-\frac{2 \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac{2 \sec (c+d x)}{a^2 d}+\frac{4 \sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{\tan (c+d x)}{a^2 d}-\frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{2 \tan ^5(c+d x)}{5 a^2 d}-\frac{\int 1 \, dx}{a^2}\\ &=-\frac{x}{a^2}-\frac{2 \sec (c+d x)}{a^2 d}+\frac{4 \sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{\tan (c+d x)}{a^2 d}-\frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{2 \tan ^5(c+d x)}{5 a^2 d}\\ \end{align*}
Mathematica [A] time = 0.546513, size = 143, normalized size = 1.35 \[ -\frac{\sec (c+d x) \left (-10 \sin (c+d x)+60 c \sin (2 (c+d x))+60 d x \sin (2 (c+d x))-89 \sin (2 (c+d x))+26 \sin (3 (c+d x))+\frac{5}{4} (60 c+60 d x-89) \cos (c+d x)+44 \cos (2 (c+d x))-15 c \cos (3 (c+d x))-15 d x \cos (3 (c+d x))+\frac{89}{4} \cos (3 (c+d x))+20\right )}{60 a^2 d (\sin (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.089, size = 146, normalized size = 1.4 \begin{align*} -{\frac{1}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-{\frac{4}{5\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+2\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-{\frac{1}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{3}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{7}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.51984, size = 336, normalized size = 3.17 \begin{align*} -\frac{2 \,{\left (\frac{\frac{49 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{70 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{60 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 16}{a^{2} + \frac{4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.09352, size = 297, normalized size = 2.8 \begin{align*} -\frac{15 \, d x \cos \left (d x + c\right )^{3} - 30 \, d x \cos \left (d x + c\right ) - 22 \, \cos \left (d x + c\right )^{2} -{\left (30 \, d x \cos \left (d x + c\right ) + 26 \, \cos \left (d x + c\right )^{2} - 9\right )} \sin \left (d x + c\right ) + 6}{15 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\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 = 1.19629, size = 139, normalized size = 1.31 \begin{align*} -\frac{\frac{60 \,{\left (d x + c\right )}}{a^{2}} + \frac{15}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} + \frac{105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 510 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 920 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 610 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 143}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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