Integrand size = 27, antiderivative size = 177 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \]
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Time = 0.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2916, 12, 1659, 1643, 647, 31} \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {\sec ^2(c+d x) (5 a \sin (c+d x)+8 b) (a+b \sin (c+d x))^2}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {b^3 \sin ^2(c+d x)}{2 d} \]
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Rule 12
Rule 31
Rule 647
Rule 1643
Rule 1659
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {x^4 (a+x)^3}{b^4 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \frac {x^4 (a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\text {Subst}\left (\int \frac {(a+x)^2 \left (-a b^4-4 b^4 x-4 a b^2 x^2-4 b^2 x^3\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b d} \\ & = -\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\text {Subst}\left (\int \frac {(a+x) \left (b^4 \left (3 a^2+16 b^2\right )+21 a b^4 x+8 b^4 x^2\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d} \\ & = -\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\text {Subst}\left (\int \left (-29 a b^4-8 b^4 x+\frac {3 \left (a b^4 \left (a^2+15 b^2\right )+8 b^4 \left (a^2+b^2\right ) x\right )}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^3 d} \\ & = -\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {3 \text {Subst}\left (\int \frac {a b^4 \left (a^2+15 b^2\right )+8 b^4 \left (a^2+b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d} \\ & = -\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (3 (a-b) \left (a^2-7 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (3 (a+b) \left (a^2+7 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d} \\ & = -\frac {3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.98 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {-3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))+3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {(a+b)^2 (5 a+11 b)}{-1+\sin (c+d x)}-48 a b^2 \sin (c+d x)-8 b^3 \sin ^2(c+d x)-\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(5 a-11 b) (a-b)^2}{1+\sin (c+d x)}}{16 d} \]
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Time = 0.98 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.60
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(284\) |
default | \(\frac {a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(284\) |
parallelrisch | \(\frac {96 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (a^{2}+7 a b +8 b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-7 a b +8 b^{2}\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-24 a^{2} b -9 b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (18 a^{2} b +12 b^{3}\right ) \cos \left (4 d x +4 c \right )+\left (-10 a^{3}-90 a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+b^{3} \cos \left (6 d x +6 c \right )-12 \sin \left (5 d x +5 c \right ) a \,b^{2}+\left (6 a^{3}-30 a \,b^{2}\right ) \sin \left (d x +c \right )+6 a^{2} b -4 b^{3}}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(305\) |
norman | \(\frac {\frac {\left (48 a^{2} b +16 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (48 a^{2} b +16 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (2 a^{2} b +2 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (2 a^{2} b +2 b^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (6 a^{2} b +6 b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (6 a^{2} b +6 b^{3}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a \left (a^{2}+15 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (a^{2}+15 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (a^{2}+15 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 a \left (a^{2}+15 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {3 a \left (5 a^{2}+11 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (35 a^{2}+141 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (35 a^{2}+141 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 \left (a^{3}-8 a^{2} b +15 a \,b^{2}-8 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {3 \left (a^{3}+8 a^{2} b +15 a \,b^{2}+8 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 b \left (a^{2}+b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(489\) |
risch | \(\frac {3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {6 i b^{3} c}{d}+\frac {b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {6 i a^{2} b c}{d}+3 i a^{2} b x +\frac {b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+3 i b^{3} x -\frac {3 i a \,b^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {{\mathrm e}^{i \left (d x +c \right )} \left (-5 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-27 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+48 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+24 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+48 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+32 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+5 i a^{3}+27 i a \,b^{2}+48 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+24 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{d}+\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{d}-\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{d}\) | \(531\) |
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Time = 0.30 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.18 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {8 \, b^{3} \cos \left (d x + c\right )^{6} - 4 \, b^{3} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, a^{2} b + 4 \, b^{3} - 24 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (24 \, a b^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a b^{2} + {\left (5 \, a^{3} + 27 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
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Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.07 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (5 \, a^{3} + 27 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 18 \, a^{2} b - 10 \, b^{3} + 12 \, {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{2} - 3 \, {\left (a^{3} + 7 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
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Time = 0.39 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.25 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, a^{2} b \sin \left (d x + c\right )^{4} + 18 \, b^{3} \sin \left (d x + c\right )^{4} + 5 \, a^{3} \sin \left (d x + c\right )^{3} + 27 \, a b^{2} \sin \left (d x + c\right )^{3} - 12 \, a^{2} b \sin \left (d x + c\right )^{2} - 24 \, b^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - 21 \, a b^{2} \sin \left (d x + c\right ) + 8 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 12.03 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.54 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (3\,a^2\,b+3\,b^3\right )}{d}-\frac {\left (-\frac {3\,a^3}{4}-\frac {45\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-6\,a^2\,b-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {5\,a^3}{4}+\frac {75\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (12\,a^2\,b+12\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (\frac {15\,a^3}{2}+\frac {33\,a\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (36\,a^2\,b+4\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {15\,a^3}{2}+\frac {33\,a\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (12\,a^2\,b+12\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {5\,a^3}{4}+\frac {75\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-6\,a^2\,b-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {3\,a^3}{4}-\frac {45\,a\,b^2}{4}\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 )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+b\right )\,\left (a^2+7\,a\,b+8\,b^2\right )}{8\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-b\right )\,\left (a^2-7\,a\,b+8\,b^2\right )}{8\,d} \]
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