Optimal. Leaf size=165 \[ \frac{a \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{2 a^2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac{a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{a^2 \sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.237724, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2902, 3767, 2606, 30, 2696, 12, 2660, 618, 204} \[ \frac{a \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{2 a^2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac{a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{a^2 \sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 2902
Rule 3767
Rule 2606
Rule 30
Rule 2696
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{a \int \sec ^4(c+d x) \, dx}{a^2-b^2}-\frac{a^2 \int \frac{\sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac{b \int \sec ^3(c+d x) \tan (c+d x) \, dx}{a^2-b^2}\\ &=\frac{a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac{a^2 \int \frac{b^2}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{a \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=-\frac{b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac{a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 b^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac{b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac{a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2 b^2\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 )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac{b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac{a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac{\left (4 a^2 b^2\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 )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{2 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 )}{\left (a^2-b^2\right )^{5/2} d}-\frac{b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac{a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 1.2621, size = 200, normalized size = 1.21 \[ \frac{\frac{48 a^2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{\sec ^3(c+d x) \left (3 b \left (5 a^2+b^2\right ) \cos (c+d x)-12 a^2 b \cos (2 (c+d x))+5 a^2 b \cos (3 (c+d x))-4 a^2 b-6 a^3 \sin (c+d x)+2 a^3 \sin (3 (c+d x))+12 a b^2 \sin (c+d x)+4 a b^2 \sin (3 (c+d x))+b^3 \cos (3 (c+d x))-8 b^3\right )}{(a-b)^2 (a+b)^2}}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.102, size = 224, normalized size = 1.4 \begin{align*} -{\frac{8}{3\,d \left ( 8\,a+8\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-4\,{\frac{1}{d \left ( 8\,a+8\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}-{\frac{b}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{{a}^{2}{b}^{2}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{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 ) }-{\frac{8}{3\,d \left ( 8\,a-8\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+4\,{\frac{1}{d \left ( 8\,a-8\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+{\frac{b}{2\,d \left ( a-b \right ) ^{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 [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 = 1.7465, size = 1033, normalized size = 6.26 \begin{align*} \left [-\frac{3 \, \sqrt{-a^{2} + b^{2}} a^{2} b^{2} \cos \left (d x + c\right )^{3} \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 ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} -{\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac{3 \, \sqrt{a^{2} - b^{2}} 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 ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} -{\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\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 = 1.21032, size = 309, normalized size = 1.87 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\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 )} a^{2} b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 4 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{2} b - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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