Optimal. Leaf size=175 \[ \frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{15 b c}+\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b \sqrt {c}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.60, antiderivative size = 175, 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, 3822, 4010, 4001, 3795, 207} \[ \frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{15 b c}+\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b \sqrt {c}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 207
Rule 3795
Rule 3822
Rule 4001
Rule 4010
Rule 4397
Rubi steps
\begin {align*} \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx &=\int \frac {\sec ^4(2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\int \frac {\sec ^2(2 a+2 b x) (4 c+c \sec (2 a+2 b x))}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{5 c}\\ &=\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}+\frac {2 \int \frac {\sec (2 a+2 b x) \left (\frac {c^2}{2}+7 c^2 \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{15 c^2}\\ &=\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}+\int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}-\frac {\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 )}{b}\\ &=-\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 {14 \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.67, size = 112, normalized size = 0.64 \[ \frac {\sin (a+b x) \cos (a+b x) \sec ^3(2 (a+b x)) \left (4 \cos (2 (a+b x))+26 \cos (4 (a+b x))+30 \cos ^2(2 (a+b x)) \tan ^{-1}\left (\sqrt {\tan ^2(a+b x)-1}\right ) \sqrt {\tan ^2(a+b x)-1}+38\right )}{30 b \sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.51, size = 380, normalized size = 2.17 \[ \left [\frac {4 \, \sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 20 \, \tan \left (b x + a\right )^{2} + 17\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} + \frac {15 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{5} - 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \log \left (\frac {\tan \left (b x + a\right )^{3} - \frac {2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt {c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right )}{\sqrt {c}}}{60 \, {\left (b c \tan \left (b x + a\right )^{5} - 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}, -\frac {15 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{5} - 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {1}{c}}}{\tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 20 \, \tan \left (b x + a\right )^{2} + 17\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{30 \, {\left (b c \tan \left (b x + a\right )^{5} - 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 1.36, size = 984, normalized size = 5.62 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (2 \, b x + 2 \, a\right )^{4}}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^4\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________