Integrand size = 33, antiderivative size = 177 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {(4 A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {(5 A-2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(5 A-2 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
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Time = 0.40 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4105, 3872, 3856, 2719, 2720} \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=-\frac {(5 A-2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a^2 d (\sec (c+d x)+1)}-\frac {(5 A-2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(4 A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2} \]
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Rule 2719
Rule 2720
Rule 3856
Rule 3872
Rule 4105
Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (7 A-B)-\frac {3}{2} a (A-B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{3 a^2} \\ & = -\frac {(5 A-2 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {3}{2} a^2 (4 A-B)-\frac {1}{2} a^2 (5 A-2 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^4} \\ & = -\frac {(5 A-2 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(5 A-2 B) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}+\frac {(4 A-B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(5 A-2 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\left ((5 A-2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}+\frac {\left ((4 A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2} \\ & = \frac {(4 A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {(5 A-2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(5 A-2 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.26 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.86 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-8 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+2 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-\frac {\left (5 (4 A-B) \cos \left (\frac {1}{2} (c-d x)\right )+4 (4 A-B) \cos \left (\frac {1}{2} (3 c+d x)\right )+9 A \cos \left (\frac {1}{2} (c+3 d x)\right )-3 B \cos \left (\frac {1}{2} (c+3 d x)\right )+3 A \cos \left (\frac {1}{2} (5 c+3 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}-20 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+8 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}\right ) \sec (c+d x) (A+B \sec (c+d x))}{3 a^2 d (B+A \cos (c+d x)) (1+\sec (c+d x))^2} \]
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Time = 10.50 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.38
method | result | size |
default | \(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+10 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-38 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+20 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+15 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A +B \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(421\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.06 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} {\left (5 i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-5 i \, A + 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-5 i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (5 i \, A - 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-4 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-4 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-4 i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (4 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (4 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (4 i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
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\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{2}} \]
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\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
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