Integrand size = 29, antiderivative size = 142 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 (3 a-5 b) (a-b) b \log (1+\sin (c+d x))}{16 d}-\frac {15 b^3 \sin (c+d x)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d} \]
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Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2916, 12, 1659, 788, 647, 31} \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 b (3 a-5 b) (a-b) \log (\sin (c+d x)+1)}{16 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}-\frac {15 b^3 \sin (c+d x)}{8 d} \]
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Rule 12
Rule 31
Rule 647
Rule 788
Rule 1659
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {x^3 (a+x)^3}{b^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^2 \text {Subst}\left (\int \frac {x^3 (a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {\text {Subst}\left (\int \frac {(a+x)^2 \left (-3 b^4-4 a b^2 x-4 b^2 x^2\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}+\frac {\text {Subst}\left (\int \frac {(a+x) \left (9 a b^4+15 b^4 x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^2 d} \\ & = -\frac {15 b^3 \sin (c+d x)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}-\frac {\text {Subst}\left (\int \frac {-9 a^2 b^4-15 b^6-24 a b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^2 d} \\ & = -\frac {15 b^3 \sin (c+d x)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}-\frac {(3 (3 a-5 b) (a-b) b) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {(3 b (a+b) (3 a+5 b)) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d} \\ & = -\frac {3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 (3 a-5 b) (a-b) b \log (1+\sin (c+d x))}{16 d}-\frac {15 b^3 \sin (c+d x)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {-3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))+3 (3 a-5 b) (a-b) b \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {3 (a+b)^2 (a+3 b)}{-1+\sin (c+d x)}-16 b^3 \sin (c+d x)+\frac {(a-b)^3}{(1+\sin (c+d x))^2}-\frac {3 (a-3 b) (a-b)^2}{1+\sin (c+d x)}}{16 d} \]
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Time = 0.84 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.58
method | result | size |
derivativedivides | \(\frac {\frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(225\) |
default | \(\frac {\frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(225\) |
parallelrisch | \(\frac {48 b^{2} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 b \left (a +\frac {5 b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 b \left (a -\frac {5 b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (-a^{3}-3 a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (a^{3}+9 a \,b^{2}\right ) \cos \left (4 d x +4 c \right )+15 \left (-a^{2} b -b^{3}\right ) \sin \left (3 d x +3 c \right )-2 b^{3} \sin \left (5 d x +5 c \right )+\left (9 a^{2} b -5 b^{3}\right ) \sin \left (d x +c \right )+3 a^{3}+3 a \,b^{2}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(273\) |
risch | \(3 i a \,b^{2} x +\frac {i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {6 i b^{2} a c}{d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (8 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+15 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-9 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+8 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+9 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-15 a^{2} b -9 b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}\) | \(413\) |
norman | \(\frac {\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (45 a^{2}+11 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (4 a^{3}+16 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (4 a^{3}+16 a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 b \left (3 a^{2}+5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {b \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 b \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (105 a^{2}+47 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (105 a^{2}+47 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 a \,b^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 b \left (3 a^{2}-8 a b +5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {3 b \left (3 a^{2}+8 a b +5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(466\) |
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Time = 0.30 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.24 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{3} + 12 \, a b^{2} - 8 \, {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b - 2 \, b^{3} + 3 \, {\left (5 \, a^{2} b + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
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Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\text {Timed out} \]
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Time = 0.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.22 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {16 \, b^{3} \sin \left (d x + c\right ) - 3 \, {\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{3} - 18 \, a b^{2} + 4 \, {\left (a^{3} + 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (9 \, a^{2} b + 7 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
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Time = 0.37 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.32 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {16 \, b^{3} \sin \left (d x + c\right ) - 3 \, {\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, a b^{2} \sin \left (d x + c\right )^{4} + 15 \, a^{2} b \sin \left (d x + c\right )^{3} + 9 \, b^{3} \sin \left (d x + c\right )^{3} + 4 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a b^{2} \sin \left (d x + c\right )^{2} - 9 \, a^{2} b \sin \left (d x + c\right ) - 7 \, b^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 11.94 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.51 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^3+18\,a\,b^2\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {15\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^3+18\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a^2\,b+10\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (6\,a^2\,b+10\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {9\,a^2\,b}{4}+\frac {15\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {33\,a^2\,b}{2}-\frac {9\,b^3}{2}\right )-6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,b^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {3\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+b\right )\,\left (3\,a+5\,b\right )}{8\,d}+\frac {3\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-b\right )\,\left (3\,a-5\,b\right )}{8\,d} \]
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