Integrand size = 35, antiderivative size = 156 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {16 a^2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (15 A+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
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Time = 0.53 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4199, 3123, 3054, 3047, 3100, 2827, 2720, 2719} \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (15 A+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}-\frac {16 a^2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
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Rule 2719
Rule 2720
Rule 2827
Rule 3047
Rule 3054
Rule 3100
Rule 3123
Rule 4199
Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+a \cos (c+d x))^2 \left (C+A \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^2 \left (2 a C+\frac {1}{2} a (5 A-C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a} \\ & = \frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (15 A+17 C)+\frac {1}{4} a^2 (15 A-7 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} a^3 (15 A+17 C)+\left (\frac {1}{4} a^3 (15 A-7 C)+\frac {1}{4} a^3 (15 A+17 C)\right ) \cos (c+d x)+\frac {1}{4} a^3 (15 A-7 C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {2 a^2 (15 A+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 \int \frac {\frac {5}{4} a^3 (3 A+C)-3 a^3 C \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a} \\ & = \frac {2 a^2 (15 A+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {1}{5} \left (8 a^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {16 a^2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (15 A+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 C \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 11.10 (sec) , antiderivative size = 800, normalized size of antiderivative = 5.13 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\cos ^{\frac {9}{2}}(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {(-5 A-16 C+5 A \cos (2 c)) \csc (c) \sec (c)}{10 d}+\frac {C \sec (c) \sec ^3(c+d x) \sin (d x)}{5 d}+\frac {\sec (c) \sec ^2(c+d x) (3 C \sin (c)+10 C \sin (d x))}{15 d}+\frac {\sec (c) \sec (c+d x) (10 C \sin (c)+15 A \sin (d x)+24 C \sin (d x))}{15 d}\right )}{A+2 C+A \cos (2 c+2 d x)}-\frac {2 A \cos ^4(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (A+2 C+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {2 C \cos ^4(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (A+2 C+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}+\frac {4 C \cos ^4(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (A+2 C+A \cos (2 c+2 d x))} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(755\) vs. \(2(192)=384\).
Time = 2.60 (sec) , antiderivative size = 756, normalized size of antiderivative = 4.85
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.43 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {2} C a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {2} C a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, {\left (5 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 10 \, C a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \]
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\[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int A \sqrt {\cos {\left (c + d x \right )}}\, dx + \int 2 A \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx + \int A \sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \sqrt {\cos {\left (c + d x \right )}} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sqrt {\cos {\left (c + d x \right )}} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
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\[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )} \,d x } \]
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\[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )} \,d x } \]
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Time = 20.19 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.29 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {6\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,C\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,C\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
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