Integrand size = 33, antiderivative size = 187 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a b (3 A+4 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]
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Time = 0.49 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3127, 3110, 3100, 2827, 3853, 3855, 3852, 8} \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {a b (3 A+4 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a b (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{10 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
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Rule 8
Rule 2827
Rule 3100
Rule 3110
Rule 3127
Rule 3852
Rule 3853
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x)) \left (2 A b+a (4 A+5 C) \cos (c+d x)+b (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{20} \int \left (-4 \left (2 A b^2+a^2 (4 A+5 C)\right )-10 a b (3 A+4 C) \cos (c+d x)-4 b^2 (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-30 a b (3 A+4 C)-4 \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} (a b (3 A+4 C)) \int \sec ^3(c+d x) \, dx-\frac {1}{15} \left (-5 b^2 (2 A+3 C)-2 a^2 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (a b (3 A+4 C)) \int \sec (c+d x) \, dx-\frac {\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d} \\ & = \frac {a b (3 A+4 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \\ \end{align*}
Time = 1.18 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.61 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 a b (3 A+4 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (60 \left (a^2+b^2\right ) (A+C)+15 a b (3 A+4 C) \sec (c+d x)+30 a A b \sec ^3(c+d x)+20 \left (A b^2+a^2 (2 A+C)\right ) \tan ^2(c+d x)+12 a^2 A \tan ^4(c+d x)\right )}{60 d} \]
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Time = 10.86 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.95
method | result | size |
parts | \(-\frac {A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (A \,b^{2}+a^{2} C \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{2} C \tan \left (d x +c \right )}{d}+\frac {2 A a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {2 C a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(178\) |
derivativedivides | \(\frac {-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 A a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 C a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right ) b^{2}}{d}\) | \(184\) |
default | \(\frac {-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 A a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 C a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right ) b^{2}}{d}\) | \(184\) |
parallelrisch | \(\frac {-45 b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {4 C}{3}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {4 C}{3}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (160 A +200 C \right ) a^{2}+200 b^{2} \left (A +\frac {9 C}{10}\right )\right ) \sin \left (3 d x +3 c \right )+\left (\left (32 A +40 C \right ) a^{2}+40 b^{2} \left (A +\frac {3 C}{2}\right )\right ) \sin \left (5 d x +5 c \right )+420 b \left (A +\frac {4 C}{7}\right ) a \sin \left (2 d x +2 c \right )+90 b \left (A +\frac {4 C}{3}\right ) a \sin \left (4 d x +4 c \right )+320 \left (\frac {\left (A +\frac {3 C}{4}\right ) b^{2}}{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) \sin \left (d x +c \right )}{60 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(266\) |
risch | \(-\frac {i \left (45 A a b \,{\mathrm e}^{9 i \left (d x +c \right )}+60 C a b \,{\mathrm e}^{9 i \left (d x +c \right )}-60 C \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+210 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+120 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-120 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 C \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-240 C \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-320 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-360 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-210 A a b \,{\mathrm e}^{3 i \left (d x +c \right )}-120 C a b \,{\mathrm e}^{3 i \left (d x +c \right )}-160 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-240 b^{2} C \,{\mathrm e}^{2 i \left (d x +c \right )}-45 A a b \,{\mathrm e}^{i \left (d x +c \right )}-60 C a b \,{\mathrm e}^{i \left (d x +c \right )}-32 A \,a^{2}-40 A \,b^{2}-40 a^{2} C -60 b^{2} C \right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 A a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {3 A a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(421\) |
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Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, {\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, A a b \cos \left (d x + c\right ) + 12 \, A a^{2} + 4 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]
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Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 15 \, A a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C b^{2} \tan \left (d x + c\right )}{120 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (175) = 350\).
Time = 0.34 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.84 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, A a b + 4 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, A a b + 4 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 80 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 240 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 232 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 360 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]
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Time = 5.86 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.72 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,C\right )}{2\,d}-\frac {\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2-\frac {5\,A\,a\,b}{2}-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (A\,a\,b-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-8\,C\,b^2-\frac {8\,A\,a^2}{3}+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^2}{15}+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+12\,C\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^2}{3}-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-8\,C\,b^2-A\,a\,b-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2+\frac {5\,A\,a\,b}{2}+2\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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