Integrand size = 37, antiderivative size = 489 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (8 A b^4+3 a^4 (A-C)-a^2 b^2 (15 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 \left (6 a A b^2+8 A b^3-3 a^3 (A-C)-a^2 b (9 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 \left (2 A b^4-a^4 C-a^2 b^2 (4 A+C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]
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Time = 1.42 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4306, 3135, 3134, 3077, 2895, 3073} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (3 a^4 (A-C)-a^2 b^2 (15 A+C)+8 A b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^4 d \sqrt {a+b} \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}-\frac {4 \left (a^4 (-C)-a^2 b^2 (4 A+C)+2 A b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-3 a^3 (A-C)-a^2 b (9 A+C)+6 a A b^2+8 A b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{3 a^3 d \sqrt {a+b} \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}} \]
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Rule 2895
Rule 3073
Rule 3077
Rule 3134
Rule 3135
Rule 4306
Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx \\ & = \frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (-4 A b^2+a^2 (3 A-C)\right )-\frac {3}{2} a b (A+C) \cos (c+d x)+\left (A b^2+a^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 \left (2 A b^4-a^4 C-a^2 b^2 (4 A+C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (8 A b^4+3 a^4 (A-C)-a^2 b^2 (15 A+C)\right )+\frac {1}{2} a b \left (A b^2-a^2 (3 A+2 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 \left (2 A b^4-a^4 C-a^2 b^2 (4 A+C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left ((a-b) \left (6 a A b^2+8 A b^3-3 a^3 (A-C)-a^2 b (9 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}-\frac {\left (\left (-8 A b^4-3 a^4 (A-C)+a^2 b^2 (15 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (8 A b^4+3 a^4 (A-C)-a^2 b^2 (15 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 \left (6 a A b^2+8 A b^3-3 a^3 (A-C)-a^2 b (9 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 \left (2 A b^4-a^4 C-a^2 b^2 (4 A+C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \\ \end{align*}
Time = 19.27 (sec) , antiderivative size = 596, normalized size of antiderivative = 1.22 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \left (3 a^4 A-15 a^2 A b^2+8 A b^4-3 a^4 C-a^2 b^2 C\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2}+\frac {2 \left (A b^2 \sin (c+d x)+a^2 C \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {4 \left (4 a^2 A b^2 \sin (c+d x)-2 A b^4 \sin (c+d x)+a^4 C \sin (c+d x)+a^2 b^2 C \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}+\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-2 (a+b) \left (8 A b^4+3 a^4 (A-C)-a^2 b^2 (15 A+C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+2 a (a+b) \left (-6 a A b^2+8 A b^3+3 a^3 (A-C)-a^2 b (9 A+C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-\left (8 A b^4+3 a^4 (A-C)-a^2 b^2 (15 A+C)\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(6141\) vs. \(2(449)=898\).
Time = 16.09 (sec) , antiderivative size = 6142, normalized size of antiderivative = 12.56
method | result | size |
default | \(\text {Expression too large to display}\) | \(6142\) |
parts | \(\text {Expression too large to display}\) | \(6291\) |
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\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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