Optimal. Leaf size=182 \[ \frac{11 c^2 \sin (2 a+2 b x)}{16 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{c^2 \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{11 c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{24 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{11 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{16 b} \]
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Rubi [A] time = 0.31209, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4397, 3813, 21, 3805, 3774, 207} \[ \frac{11 c^2 \sin (2 a+2 b x)}{16 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{c^2 \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{11 c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{24 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{11 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{16 b} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 3813
Rule 21
Rule 3805
Rule 3774
Rule 207
Rubi steps
\begin{align*} \int \cos ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \cos ^3(2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac{c^2 \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{1}{3} c \int \frac{\cos ^2(2 a+2 b x) \left (\frac{11 c}{2}-\frac{11}{2} c \sec (2 a+2 b x)\right )}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac{c^2 \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{1}{6} (11 c) \int \cos ^2(2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac{11 c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{24 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{1}{8} (11 c) \int \cos (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac{11 c^2 \sin (2 a+2 b x)}{16 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{11 c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{24 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{1}{16} (11 c) \int \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac{11 c^2 \sin (2 a+2 b x)}{16 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{11 c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{24 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{\left (11 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{-c+x^2} \, dx,x,-\frac{c \tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{16 b}\\ &=-\frac{11 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{16 b}+\frac{11 c^2 \sin (2 a+2 b x)}{16 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{11 c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{24 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt{-c+c \sec (2 a+2 b x)}}\\ \end{align*}
Mathematica [A] time = 0.249013, size = 117, normalized size = 0.64 \[ \frac{c \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \left (-42 \sin (2 (a+b x))+14 \sin (4 (a+b x))-4 \sin (6 (a+b x))+38 \cot (a+b x)-33 \sqrt{2} \sqrt{\cos (2 (a+b x))} \csc (a+b x) \tanh ^{-1}\left (\frac{\sqrt{2} \cos (a+b x)}{\sqrt{\cos (2 (a+b x))}}\right )\right )}{96 b} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.457, size = 1078, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [A] time = 2.56592, size = 1310, normalized size = 7.2 \begin{align*} \left [\frac{33 \,{\left (c \tan \left (b x + a\right )^{7} + 3 \, c \tan \left (b x + a\right )^{5} + 3 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt{c} \log \left (-\frac{c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} - 4 \, \sqrt{2}{\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt{c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) - 4 \, \sqrt{2}{\left (63 \, c \tan \left (b x + a\right )^{6} - 13 \, c \tan \left (b x + a\right )^{4} - 31 \, c \tan \left (b x + a\right )^{2} - 19 \, c\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{192 \,{\left (b \tan \left (b x + a\right )^{7} + 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, \frac{33 \,{\left (c \tan \left (b x + a\right )^{7} + 3 \, c \tan \left (b x + a\right )^{5} + 3 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt{-c} \arctan \left (\frac{2 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt{-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt{2}{\left (63 \, c \tan \left (b x + a\right )^{6} - 13 \, c \tan \left (b x + a\right )^{4} - 31 \, c \tan \left (b x + a\right )^{2} - 19 \, c\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{96 \,{\left (b \tan \left (b x + a\right )^{7} + 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\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 [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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