Integrand size = 33, antiderivative size = 159 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {4 a^2 (4 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (A+2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (7 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+5 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
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Time = 0.36 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3054, 3047, 3100, 2827, 2716, 2719, 2720} \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 a^2 (A+2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {4 a^2 (4 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^2 (7 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+5 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
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Rule 2716
Rule 2719
Rule 2720
Rule 2827
Rule 3047
Rule 3054
Rule 3100
Rubi steps \begin{align*} \text {integral}& = \frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{2} a (7 A+5 B)+\frac {1}{2} a (A+5 B) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\frac {1}{2} a^2 (7 A+5 B)+\left (\frac {1}{2} a^2 (A+5 B)+\frac {1}{2} a^2 (7 A+5 B)\right ) \cos (c+d x)+\frac {1}{2} a^2 (A+5 B) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (7 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{15} \int \frac {\frac {3}{2} a^2 (4 A+5 B)+\frac {5}{2} a^2 (A+2 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (7 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{3} \left (2 a^2 (A+2 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (2 a^2 (4 A+5 B)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {4 a^2 (A+2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (7 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+5 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} \left (2 a^2 (4 A+5 B)\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {4 a^2 (4 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (A+2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (7 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+5 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.89 (sec) , antiderivative size = 883, normalized size of antiderivative = 5.55 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {(4 A+5 B) \csc (c) \sec (c)}{5 d}+\frac {A \sec (c) \sec ^3(c+d x) \sin (d x)}{10 d}+\frac {\sec (c) \sec ^2(c+d x) (3 A \sin (c)+10 A \sin (d x)+5 B \sin (d x))}{30 d}+\frac {\sec (c) \sec (c+d x) (10 A \sin (c)+5 B \sin (c)+24 A \sin (d x)+30 B \sin (d x))}{30 d}\right )-\frac {A (a+a \cos (c+d x))^2 \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 ) \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 \sqrt {1+\cot ^2(c)}}-\frac {2 B (a+a \cos (c+d x))^2 \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 ) \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 \sqrt {1+\cot ^2(c)}}+\frac {2 A (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\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}+\frac {B (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\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 )}{2 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(713\) vs. \(2(195)=390\).
Time = 10.22 (sec) , antiderivative size = 714, normalized size of antiderivative = 4.49
method | result | size |
default | \(\text {Expression too large to display}\) | \(714\) |
parts | \(\text {Expression too large to display}\) | \(801\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.50 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (6 \, {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \]
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Timed out. \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
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\[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
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Time = 2.14 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.44 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
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