Integrand size = 43, antiderivative size = 392 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=-\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a-b) (a+b)^2 d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \]
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Time = 1.68 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3134, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=-\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\sin (c+d x) \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}-\frac {\left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 d (a-b) (a+b)^2} \]
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Rule 2719
Rule 2720
Rule 2884
Rule 3081
Rule 3134
Rule 3138
Rubi steps \begin{align*} \text {integral}& = \frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (-5 A b^2+3 a b B+a^2 (2 A-3 C)\right )-a (A b-a B+b C) \cos (c+d x)+\frac {3}{2} \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )} \\ & = -\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {2 \int \frac {\frac {3}{4} \left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right )+\frac {1}{2} a \left (2 A b^2-3 a b B+a^2 (A+3 C)\right ) \cos (c+d x)-\frac {1}{4} b \left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {4 \int \frac {\frac {1}{8} \left (-15 A b^4-12 a^3 b B+9 a b^3 B+a^2 b^2 (16 A-3 C)+2 a^4 (A+3 C)\right )-\frac {1}{4} a \left (10 A b^3+3 a^3 B-6 a b^2 B-a^2 (7 A b-3 b C)\right ) \cos (c+d x)-\frac {3}{8} b \left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {4 \int \frac {\frac {1}{8} b \left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right )+\frac {1}{8} a b^2 \left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 b \left (a^2-b^2\right )}-\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}-\frac {\left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a-b) (a+b)^2 d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \\ \end{align*}
Time = 7.62 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \left (4 a^4 A+44 a^2 A b^2-45 A b^4-30 a^3 b B+27 a b^3 B+12 a^4 C-9 a^2 b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {\left (28 a^3 A b-40 a A b^3-12 a^4 B+24 a^2 b^2 B-12 a^3 b C\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}+\frac {2 \left (12 a^2 A b^2-15 A b^4-6 a^3 b B+9 a b^3 B-3 a^2 b^2 C\right ) \cos (2 (c+d x)) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {1-\cos ^2(c+d x)} \left (-1+2 \cos ^2(c+d x)\right )}}{12 a^3 (a-b) (a+b) d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {2 \sec (c+d x) (-2 A b \sin (c+d x)+a B \sin (c+d x))}{a^3}+\frac {A b^4 \sin (c+d x)-a b^3 B \sin (c+d x)+a^2 b^2 C \sin (c+d x)}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 A \sec (c+d x) \tan (c+d x)}{3 a^2}\right )}{d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(1010\) vs. \(2(460)=920\).
Time = 4.91 (sec) , antiderivative size = 1011, normalized size of antiderivative = 2.58
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Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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