Optimal. Leaf size=180 \[ -\frac {11 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}+\frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{12 b c^2}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}+\frac {13 \tan (2 a+2 b x)}{6 b c \sqrt {c \sec (2 a+2 b x)-c}} \]
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Rubi [A] time = 0.51, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4397, 3816, 4010, 4001, 3795, 207} \[ \frac {7 \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{12 b c^2}-\frac {11 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}+\frac {13 \tan (2 a+2 b x)}{6 b c \sqrt {c \sec (2 a+2 b x)-c}} \]
Antiderivative was successfully verified.
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Rule 207
Rule 3795
Rule 3816
Rule 4001
Rule 4010
Rule 4397
Rubi steps
\begin {align*} \int \frac {\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx &=\int \frac {\sec ^4(2 a+2 b x)}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx\\ &=-\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {\int \frac {\sec ^2(2 a+2 b x) \left (2 c+\frac {7}{2} c \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{2 c^2}\\ &=-\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {7 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}+\frac {\int \frac {\sec (2 a+2 b x) \left (\frac {7 c^2}{4}+\frac {13}{2} c^2 \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{3 c^3}\\ &=-\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {13 \tan (2 a+2 b x)}{6 b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {7 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}+\frac {11 \int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{4 c}\\ &=-\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {13 \tan (2 a+2 b x)}{6 b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {7 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}-\frac {11 \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 {11 \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 {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {13 \tan (2 a+2 b x)}{6 b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {7 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}\\ \end {align*}
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Mathematica [A] time = 1.30, size = 100, normalized size = 0.56 \[ -\frac {\cot (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \left (\csc ^2(a+b x) ((19 \cos (4 (a+b x))+11) \sec (2 (a+b x))-24)-66 \tan ^{-1}\left (\sqrt {\tan ^2(a+b x)-1}\right ) \sqrt {\tan ^2(a+b x)-1}\right )}{48 b c^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.15, size = 350, normalized size = 1.94 \[ \left [\frac {33 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} - \tan \left (b x + a\right )^{3}\right )} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) + 2 \, \sqrt {2} {\left (27 \, \tan \left (b x + a\right )^{4} - 46 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{48 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} - b c^{2} \tan \left (b x + a\right )^{3}\right )}}, -\frac {33 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} - \tan \left (b x + a\right )^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) - \sqrt {2} {\left (27 \, \tan \left (b x + a\right )^{4} - 46 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{24 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} - b c^{2} \tan \left (b x + a\right )^{3}\right )}}\right ] \]
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 = 1.13, size = 1211, normalized size = 6.73 \[ \text {result too large to display} \]
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 \[ \int \frac {\sec \left (2 \, b x + 2 \, a\right )^{4}}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^4\,{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2}} \,d x \]
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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