Integrand size = 29, antiderivative size = 235 \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}-\frac {2 a^4 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d} \]
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Time = 0.61 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2968, 3129, 3128, 3102, 2814, 2739, 632, 210} \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac {2 a^4 \sqrt {a^2-b^2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {a x \left (8 a^4-4 a^2 b^2-b^4\right )}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}+\frac {\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]
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Rule 210
Rule 632
Rule 2739
Rule 2814
Rule 2968
Rule 3102
Rule 3128
Rule 3129
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin ^4(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx \\ & = \frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {\sin ^3(c+d x) \left (-4 a+b \sin (c+d x)+5 a \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 b} \\ & = -\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {\sin ^2(c+d x) \left (15 a^2-a b \sin (c+d x)-4 \left (5 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 b^2} \\ & = \frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {\sin (c+d x) \left (-8 a \left (5 a^2-b^2\right )+b \left (5 a^2+8 b^2\right ) \sin (c+d x)+15 a \left (4 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 b^3} \\ & = -\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {15 a^2 \left (4 a^2-b^2\right )-a b \left (20 a^2-b^2\right ) \sin (c+d x)-8 \left (15 a^4-5 a^2 b^2-2 b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^4} \\ & = \frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {15 a^2 b \left (4 a^2-b^2\right )+15 a \left (8 a^4-4 a^2 b^2-b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^5} \\ & = \frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\left (a^4 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^6} \\ & = \frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\left (2 a^4 \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^6 d} \\ & = \frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\left (4 a^4 \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^6 d} \\ & = \frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}-\frac {2 a^4 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d} \\ \end{align*}
Time = 1.64 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-960 a^4 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )-60 b \left (-8 a^4+2 a^2 b^2+b^4\right ) \cos (c+d x)-10 \left (4 a^2 b^3+b^5\right ) \cos (3 (c+d x))+6 b^5 \cos (5 (c+d x))+15 a \left (4 \left (8 a^4-4 a^2 b^2-b^4\right ) (c+d x)-8 a^2 b^2 \sin (2 (c+d x))+b^4 \sin (4 (c+d x))\right )}{480 b^6 d} \]
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Result contains complex when optimal does not.
Time = 0.77 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.44
method | result | size |
risch | \(\frac {a^{5} x}{b^{6}}-\frac {a^{3} x}{2 b^{4}}-\frac {a x}{8 b^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )} a^{4}}{2 b^{5} d}-\frac {{\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{16 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{4}}{2 b^{5} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{16 b d}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{6}}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{6}}+\frac {\cos \left (5 d x +5 c \right )}{80 b d}+\frac {a \sin \left (4 d x +4 c \right )}{32 b^{2} d}-\frac {\cos \left (3 d x +3 c \right ) a^{2}}{12 b^{3} d}-\frac {\cos \left (3 d x +3 c \right )}{48 b d}-\frac {a^{3} \sin \left (2 d x +2 c \right )}{4 d \,b^{4}}\) | \(339\) |
derivativedivides | \(\frac {\frac {\frac {2 \left (\left (\frac {1}{2} a^{3} b^{2}-\frac {1}{8} a \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{4} b -a^{2} b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{3} b^{2}+\frac {3}{4} a \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -2 a^{2} b^{3}-2 b^{5}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 a^{4} b +\frac {2}{3} b^{5}-\frac {4}{3} a^{2} b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{3} b^{2}-\frac {3}{4} a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -\frac {2}{3} a^{2} b^{3}-\frac {2}{3} b^{5}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{3} b^{2}+\frac {1}{8} a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{4} b -\frac {a^{2} b^{3}}{3}-\frac {2 b^{5}}{15}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a \left (8 a^{4}-4 a^{2} b^{2}-b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{6}}-\frac {2 a^{4} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{6}}}{d}\) | \(355\) |
default | \(\frac {\frac {\frac {2 \left (\left (\frac {1}{2} a^{3} b^{2}-\frac {1}{8} a \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{4} b -a^{2} b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{3} b^{2}+\frac {3}{4} a \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -2 a^{2} b^{3}-2 b^{5}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 a^{4} b +\frac {2}{3} b^{5}-\frac {4}{3} a^{2} b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{3} b^{2}-\frac {3}{4} a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -\frac {2}{3} a^{2} b^{3}-\frac {2}{3} b^{5}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{3} b^{2}+\frac {1}{8} a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{4} b -\frac {a^{2} b^{3}}{3}-\frac {2 b^{5}}{15}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a \left (8 a^{4}-4 a^{2} b^{2}-b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{6}}-\frac {2 a^{4} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{6}}}{d}\) | \(355\) |
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Time = 0.45 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.82 \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 60 \, \sqrt {-a^{2} + b^{2}} a^{4} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 40 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}, \frac {24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 120 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 40 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}\right ] \]
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Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 467 vs. \(2 (218) = 436\).
Time = 0.35 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.99 \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} {\left (d x + c\right )}}{b^{6}} - \frac {240 \, {\left (a^{6} - a^{4} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{6}} + \frac {2 \, {\left (60 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 15 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 120 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 120 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 90 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 720 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 160 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, a^{4} - 40 \, a^{2} b^{2} - 16 \, b^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} b^{5}}}{120 \, d} \]
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Time = 13.74 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^4\,\cos \left (c+d\,x\right )}{b^5\,d}-\frac {\frac {\cos \left (c+d\,x\right )}{8}+\frac {\cos \left (3\,c+3\,d\,x\right )}{48}-\frac {\cos \left (5\,c+5\,d\,x\right )}{80}}{b\,d}-\frac {\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {a\,\sin \left (4\,c+4\,d\,x\right )}{32}}{b^2\,d}-\frac {\frac {a^2\,\cos \left (c+d\,x\right )}{4}+\frac {a^2\,\cos \left (3\,c+3\,d\,x\right )}{12}}{b^3\,d}-\frac {a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^4\,d}+\frac {2\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^6\,d}-\frac {2\,a^4\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}}{b^6\,d} \]
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