Integrand size = 25, antiderivative size = 155 \[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (1+\cos (c+d x))}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \csc (c+d x) \operatorname {EllipticPi}\left (\frac {b}{a+b},\arcsin \left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right ),-\frac {a-b}{a+b}\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b} d} \]
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Time = 0.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4307, 2890} \[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=-\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\sec (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (\cos (c+d x)+1)}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \operatorname {EllipticPi}\left (\frac {b}{a+b},\arcsin \left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right ),-\frac {a-b}{a+b}\right )}{d \sqrt {a+b}} \]
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Rule 2890
Rule 4307
Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 \sqrt {\cos (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (1+\cos (c+d x))}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \csc (c+d x) \operatorname {EllipticPi}\left (\frac {b}{a+b},\arcsin \left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right ),-\frac {a-b}{a+b}\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b} d} \\ \end{align*}
Time = 1.51 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.94 \[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=\frac {2 \sqrt {a+b \cos (c+d x)} \left ((a-b) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+2 b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )\right ) \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)}}{(a+b) d \sqrt {\frac {(a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}} \]
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Time = 8.20 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.17
method | result | size |
default | \(-\frac {2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \left (F\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) a -F\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) b +2 b \Pi \left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {-\frac {a -b}{a +b}}\right )\right ) \left (\sqrt {\sec }\left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )}{d \sqrt {a +\cos \left (d x +c \right ) b}}\) | \(181\) |
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\[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )} \,d x } \]
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\[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=\int \sqrt {a + b \cos {\left (c + d x \right )}} \sqrt {\sec {\left (c + d x \right )}}\, dx \]
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\[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )} \,d x } \]
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\[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )} \,d x } \]
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Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=\int \sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {a+b\,\cos \left (c+d\,x\right )} \,d x \]
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