Integrand size = 31, antiderivative size = 129 \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {2 \tan (2 a+2 b x)}{3 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b c} \]
[Out]
Time = 0.43 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4482, 3885, 4086, 3880, 213} \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}+\frac {2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}} \]
[In]
[Out]
Rule 213
Rule 3880
Rule 3885
Rule 4086
Rule 4482
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec ^3(2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx \\ & = \frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b c}+\frac {2 \int \frac {\sec (2 a+2 b x) \left (\frac {c}{2}+c \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{3 c} \\ & = \frac {2 \tan (2 a+2 b x)}{3 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)}{3 b c}+\int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx \\ & = \frac {2 \tan (2 a+2 b x)}{3 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)}{3 b c}-\frac {\text {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 {\text {arctanh}\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 {2 \tan (2 a+2 b x)}{3 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)}{3 b c} \\ \end{align*}
Time = 1.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\cos ^2(a+b x) \csc (2 (a+b x)) \left (2+2 \sec (2 (a+b x))+3 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{3 b c} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. \(606\) vs. \(2(112)=224\).
Time = 9.81 (sec) , antiderivative size = 607, normalized size of antiderivative = 4.71
method | result | size |
default | \(\frac {\sqrt {2}\, \sin \left (x b +a \right ) \left (12 \cos \left (x b +a \right )^{4} \ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right )-12 \cos \left (x b +a \right )^{4} \operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right )+8 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \cos \left (x b +a \right )^{4}+8 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \cos \left (x b +a \right )^{3}-12 \cos \left (x b +a \right )^{2} \ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right )+12 \cos \left (x b +a \right )^{2} \operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right )+3 \ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right )-3 \,\operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right )\right ) \sqrt {4}}{24 b \left (2 \cos \left (x b +a \right )^{2}-1\right )^{2} \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \left (1+\cos \left (x b +a \right )\right ) \left (-3+2 \sqrt {2}\right )^{2} \left (3+2 \sqrt {2}\right )^{2}}\) | \(607\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\left [\frac {\frac {3 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{3} - c \tan \left (b x + a\right )\right )} \log \left (\frac {\tan \left (b x + a\right )^{3} - \frac {2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt {c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right )}{\sqrt {c}} - 8 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{12 \, {\left (b c \tan \left (b x + a\right )^{3} - b c \tan \left (b x + a\right )\right )}}, -\frac {3 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{3} - c \tan \left (b x + a\right )\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {1}{c}}}{\tan \left (b x + a\right )}\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{6 \, {\left (b c \tan \left (b x + a\right )^{3} - b c \tan \left (b x + a\right )\right )}}\right ] \]
[In]
[Out]
Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )^{3}}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^3\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \]
[In]
[Out]