Optimal. Leaf size=84 \[ \frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}} \]
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Rubi [A] time = 0.14, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {4397, 3805, 3774, 207} \[ \frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}} \]
Antiderivative was successfully verified.
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Rule 207
Rule 3774
Rule 3805
Rule 4397
Rubi steps
\begin {align*} \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx &=\int \cos (2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {1}{2} \int \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c \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 )}{2 b}\\ &=\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {-c+c \sec (2 a+2 b x)}}\\ \end {align*}
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Mathematica [A] time = 0.24, size = 92, normalized size = 1.10 \[ -\frac {\csc (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \left (\cos (a+b x)+\cos (3 (a+b x))-\sqrt {2} \sqrt {\cos (2 (a+b x))} \tanh ^{-1}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right )\right )}{4 b} \]
Antiderivative was successfully verified.
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fricas [B] time = 2.28, size = 351, normalized size = 4.18 \[ \left [\frac {{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{8 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, -\frac {{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{4 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.14, size = 391, normalized size = 4.65 \[ \frac {\sqrt {\frac {c \left (1-\left (\cos ^{2}\left (b x +a \right )\right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \sin \left (b x +a \right ) \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \sqrt {4}}{2 b \left (-1+\cos \left (b x +a \right )\right )}-\frac {\sqrt {2}\, \sin \left (b x +a \right ) \sqrt {\frac {c \left (1-\left (\cos ^{2}\left (b x +a \right )\right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}}\, \left (\sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \cos \left (b x +a \right )+\sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )-4 \left (\cos ^{3}\left (b x +a \right )\right )+2 \cos \left (b x +a \right )\right ) \sqrt {4}}{8 b \left (-1+\cos ^{2}\left (b x +a \right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.61, size = 1049, normalized size = 12.49 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (2\,a+2\,b\,x\right )\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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