Integrand size = 43, antiderivative size = 267 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 (4 A+5 B+4 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \]
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Time = 0.98 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {4197, 3122, 3054, 3055, 3047, 3102, 2827, 2720, 2719} \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {8 a^4 (4 A+5 B+4 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}-\frac {4 (6 A+25 B+34 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{15 d}+\frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (5 A+15 B+19 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a (5 B+8 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
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Rule 2719
Rule 2720
Rule 2827
Rule 3047
Rule 3054
Rule 3055
Rule 3102
Rule 3122
Rule 4197
Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+a \cos (c+d x))^4 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^4 \left (\frac {1}{2} a (5 B+8 C)+\frac {5}{2} a (A-C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a} \\ & = \frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x))^3 \left (\frac {3}{4} a^2 (5 A+15 B+19 C)+\frac {5}{4} a^2 (3 A-5 B-11 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {8 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {5}{4} a^3 (9 A+20 B+23 C)-\frac {5}{4} a^3 (6 A+25 B+34 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{15 a} \\ & = \frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {16 \int \frac {(a+a \cos (c+d x)) \left (\frac {15}{8} a^4 (13 A+25 B+27 C)+\frac {15}{8} a^4 (A-25 B-41 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{75 a} \\ & = \frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {16 \int \frac {\frac {15}{8} a^5 (13 A+25 B+27 C)+\left (\frac {15}{8} a^5 (A-25 B-41 C)+\frac {15}{8} a^5 (13 A+25 B+27 C)\right ) \cos (c+d x)+\frac {15}{8} a^5 (A-25 B-41 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{75 a} \\ & = \frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {32 \int \frac {\frac {75}{8} a^5 (4 A+5 B+4 C)+\frac {315}{8} a^5 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{225 a} \\ & = \frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (28 a^4 (A-C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (4 a^4 (4 A+5 B+4 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {56 a^4 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a^4 (4 A+5 B+4 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^4 (A-25 B-41 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (5 B+8 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 A+15 B+19 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {4 (6 A+25 B+34 C) \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 13.58 (sec) , antiderivative size = 1449, normalized size of antiderivative = 5.43 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\cos ^{\frac {13}{2}}(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {(23 A-20 B-61 C+33 A \cos (2 c)+20 B \cos (2 c)+5 C \cos (2 c)) \csc (c) \sec (c)}{40 d}+\frac {(4 A+B) \cos (d x) \sin (c)}{12 d}+\frac {A \cos (2 d x) \sin (2 c)}{40 d}+\frac {(4 A+B) \cos (c) \sin (d x)}{12 d}+\frac {C \sec (c) \sec ^3(c+d x) \sin (d x)}{20 d}+\frac {\sec (c) \sec ^2(c+d x) (3 C \sin (c)+5 B \sin (d x)+20 C \sin (d x))}{60 d}+\frac {\sec (c) \sec (c+d x) (5 B \sin (c)+20 C \sin (c)+15 A \sin (d x)+60 B \sin (d x)+99 C \sin (d x))}{60 d}+\frac {A \cos (2 c) \sin (2 d x)}{40 d}\right )}{A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)}-\frac {4 A \cos ^6(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 ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+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+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {5 B \cos ^6(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 ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+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+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {4 C \cos ^6(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 ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+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+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {7 A \cos ^6(c+d x) \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+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 )}{10 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {7 C \cos ^6(c+d x) \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+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 )}{10 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(1213\) vs. \(2(295)=590\).
Time = 647.27 (sec) , antiderivative size = 1214, normalized size of antiderivative = 4.55
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.05 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 10 i \, \sqrt {2} {\left (4 \, A + 5 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 42 i \, \sqrt {2} {\left (A - C\right )} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 5 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 20 \, B + 33 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \]
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Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
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Time = 22.20 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.97 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (12\,B\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+19\,B\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+B\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{3\,d}+\frac {2\,\left (C\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,C\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {2\,\left (\frac {34\,C\,a^4\,\sin \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d}+\frac {4\,A\,a^4\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {12\,A\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^4\,\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}}-\frac {2\,A\,a^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,B\,a^4\,\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}}+\frac {2\,B\,a^4\,\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 )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,C\,a^4\,\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 )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {8\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {7}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
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