Optimal. Leaf size=178 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b c^{3/2}}+\frac {9 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]
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Rubi [A] time = 0.32, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {4397, 3817, 4022, 3920, 3774, 207, 3795} \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b c^{3/2}}+\frac {9 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 207
Rule 3774
Rule 3795
Rule 3817
Rule 3920
Rule 4022
Rule 4397
Rubi steps
\begin {align*} \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx &=\int \frac {\cos (2 a+2 b x)}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx\\ &=-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {\int \frac {\cos (2 a+2 b x) \left (3 c+\frac {3}{2} c \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{2 c^2}\\ &=-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {\int \frac {3 c^2+\frac {3}{2} c^2 \sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{2 c^3}\\ &=-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {3 \int \sqrt {-c+c \sec (2 a+2 b x)} \, dx}{2 c^2}-\frac {9 \int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{4 c}\\ &=-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {3 \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 c}+\frac {9 \operatorname {Subst}\left (\int \frac {1}{-2 c+x^2} \, dx,x,-\frac {c \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 b c}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b c^{3/2}}+\frac {9 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}}\\ \end {align*}
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Mathematica [A] time = 6.20, size = 342, normalized size = 1.92 \[ \frac {\tan ^2(a+b x) \tan ^2(2 (a+b x)) \left (\frac {1}{2} \sin (2 (a+b x))-\frac {1}{4} \cot (a+b x)-\frac {1}{8} \cot (a+b x) \csc ^2(a+b x)\right )}{b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}-\frac {3 \tan ^{\frac {3}{2}}(a+b x) \tan ^{\frac {3}{2}}(2 (a+b x)) \left (\frac {\tan ^{-1}\left (\sqrt {\tan ^2(a+b x)-1}\right ) \tan ^{\frac {3}{2}}(a+b x) \sqrt {\tan ^2(a+b x)-1} \sqrt {\tan (2 (a+b x))} \csc ^2(a+b x) \sec ^2(a+b x)}{\left (\tan ^2(a+b x)+1\right )^2}+\frac {\sqrt {2} \cos (2 (a+b x)) \tan ^{\frac {3}{2}}(a+b x) \sqrt {\tan (2 (a+b x))} \csc ^2(a+b x) \sec ^2(a+b x) \left (2 \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-2 \tan ^2(a+b x)}\right )-\sqrt {2} \tanh ^{-1}\left (\sqrt {1-\tan ^2(a+b x)}\right )\right )}{\sqrt {1-\tan ^2(a+b x)} \left (\tan ^2(a+b x)+1\right )}\right )}{8 b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.05, size = 528, normalized size = 2.97 \[ \left [\frac {9 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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 ) + 12 \, {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{16 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} + b c^{2} \tan \left (b x + a\right )^{3}\right )}}, \frac {9 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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 ) - 12 \, {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{8 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} + b c^{2} \tan \left (b x + a\right )^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 1.07, size = 1157, normalized size = 6.50 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (2 \, b x + 2 \, a\right )}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {\cos \left (2\,a+2\,b\,x\right )}{{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2}} \,d x \]
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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