Optimal. Leaf size=138 \[ \frac{\sin (2 a+2 b x)}{2 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{2 b \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]
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Rubi [A] time = 0.278831, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4397, 3823, 3904, 3887, 481, 206} \[ \frac{\sin (2 a+2 b x)}{2 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{c \sec (2 a+2 b x)-c}}\right )}{2 b \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 3823
Rule 3904
Rule 3887
Rule 481
Rule 206
Rubi steps
\begin{align*} \int \frac{\cos (2 (a+b x))}{\sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \, dx &=\int \frac{\cos (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac{\sin (2 a+2 b x)}{2 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{\int \frac{-c-c \sec (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx}{2 c}\\ &=\frac{\sin (2 a+2 b x)}{2 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{1}{2} c \int \frac{\tan ^2(2 a+2 b x)}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx\\ &=\frac{\sin (2 a+2 b x)}{2 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{c \operatorname{Subst}\left (\int \frac{x^2}{\left (1-c x^2\right ) \left (2-c x^2\right )} \, dx,x,-\frac{\tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{2 b}\\ &=\frac{\sin (2 a+2 b x)}{2 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,-\frac{\tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{1}{2-c x^2} \, dx,x,-\frac{\tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{b}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{2 b \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{-c+c \sec (2 a+2 b x)}}\right )}{\sqrt{2} b \sqrt{c}}+\frac{\sin (2 a+2 b x)}{2 b \sqrt{-c+c \sec (2 a+2 b x)}}\\ \end{align*}
Mathematica [A] time = 2.47319, size = 166, normalized size = 1.2 \[ \frac{\tan (a+b x) \left (\sqrt{2} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-2 \tan ^2(a+b x)}\right )-\tanh ^{-1}\left (\sqrt{1-\tan ^2(a+b x)}\right )+\sqrt{2} \cos ^2(a+b x) \sqrt{\frac{1}{\sec (2 (a+b x))+1}} \left (\tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right ) \sqrt{\tan ^2(a+b x)-1} \sec (2 (a+b x))+2\right )\right )}{2 b \sqrt{1-\tan ^2(a+b x)} \sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.466, size = 1030, normalized size = 7.5 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (2 \, b x + 2 \, a\right )}{\sqrt{c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.2922, size = 1274, normalized size = 9.23 \begin{align*} \left [\frac{\sqrt{2}{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt{c} \log \left (\frac{c \tan \left (b x + a\right )^{3} - 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} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) +{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt{c} \log \left (\frac{c \tan \left (b x + a\right )^{3} + 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} - 3 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) - 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 )}}{4 \,{\left (b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}, -\frac{\sqrt{2}{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt{-c} \arctan \left (\frac{\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 )}\right ) -{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt{-c} \arctan \left (\frac{\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}}{2 \, c \tan \left (b x + a\right )}\right ) + \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 )}}{2 \,{\left (b c \tan \left (b x + a\right )^{3} + b c \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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