Integrand size = 27, antiderivative size = 150 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (3 a^2-16 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {b^2 \sin (c+d x)}{d}-\frac {\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \]
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Time = 0.20 (sec) , antiderivative size = 150, 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, 1824, 647, 31} \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (3 a^2-16 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {\sec ^2(c+d x) (5 a \sin (c+d x)+7 b) (a+b \sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {b^2 \sin (c+d x)}{d} \]
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Rule 12
Rule 31
Rule 647
Rule 1659
Rule 1824
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {x^4 (a+x)^2}{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)^2}{\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))^2 \tan (c+d x)}{4 d}+\frac {\text {Subst}\left (\int \frac {(a+x) \left (-a b^4-3 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) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\text {Subst}\left (\int \frac {b^4 \left (3 a^2+7 b^2\right )+16 a b^4 x+8 b^4 x^2}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d} \\ & = -\frac {\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\text {Subst}\left (\int \left (-8 b^4+\frac {3 b^4 \left (a^2+5 b^2\right )+16 a b^4 x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^3 d} \\ & = -\frac {b^2 \sin (c+d x)}{d}-\frac {\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\text {Subst}\left (\int \frac {3 b^4 \left (a^2+5 b^2\right )+16 a b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d} \\ & = -\frac {b^2 \sin (c+d x)}{d}-\frac {\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\left (3 a^2+16 a b+15 b^2\right ) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (8 a b^4-\frac {3}{2} b^3 \left (a^2+5 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d} \\ & = -\frac {\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (16 a b-3 \left (a^2+5 b^2\right )\right ) \log (1+\sin (c+d x))}{16 d}-\frac {b^2 \sin (c+d x)}{d}-\frac {\sec ^2(c+d x) (7 b+5 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \\ \end{align*}
Time = 0.73 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {-\left (\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))\right )+\left (3 a^2-16 a b+15 b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^2}{(-1+\sin (c+d x))^2}+\frac {(a+b) (5 a+9 b)}{-1+\sin (c+d x)}-16 b^2 \sin (c+d x)-\frac {(a-b)^2}{(1+\sin (c+d x))^2}+\frac {(5 a-9 b) (a-b)}{1+\sin (c+d x)}}{16 d} \]
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Time = 0.83 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.33
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(200\) |
default | \(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(200\) |
parallelrisch | \(\frac {64 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {16}{3} a b +5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {16}{3} a b +5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-16 a b \cos \left (2 d x +2 c \right )+12 \cos \left (4 d x +4 c \right ) a b +\left (-10 a^{2}-30 b^{2}\right ) \sin \left (3 d x +3 c \right )-4 b^{2} \sin \left (5 d x +5 c \right )+\left (6 a^{2}-10 b^{2}\right ) \sin \left (d x +c \right )+4 a b}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(250\) |
norman | \(\frac {\frac {8 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 \left (a^{2}+5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {5 \left (a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 \left (a^{2}+5 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {3 \left (a^{2}+5 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (15 a^{2}+11 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (15 a^{2}+11 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {4 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {24 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (3 a^{2}-16 a b +15 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (3 a^{2}+16 a b +15 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {2 a b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(361\) |
risch | \(2 i x a b +\frac {i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}+\frac {4 i a b c}{d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (5 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+9 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+32 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+32 i a b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 a^{2}-9 b^{2}+32 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}\) | \(361\) |
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Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 32 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b - 2 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{2} + 9 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
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Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {16 \, b^{2} \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (16 \, a b \sin \left (d x + c\right )^{2} + {\left (5 \, a^{2} + 9 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 12 \, a b - {\left (3 \, a^{2} + 7 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
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Time = 0.51 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {16 \, b^{2} \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, a b \sin \left (d x + c\right )^{4} + 5 \, a^{2} \sin \left (d x + c\right )^{3} + 9 \, b^{2} \sin \left (d x + c\right )^{3} - 8 \, a b \sin \left (d x + c\right )^{2} - 3 \, a^{2} \sin \left (d x + c\right ) - 7 \, b^{2} \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 12.40 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.21 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a^2}{8}-2\,a\,b+\frac {15\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3\,a^2}{8}+2\,a\,b+\frac {15\,b^2}{8}\right )}{d}+\frac {\left (-\frac {3\,a^2}{4}-\frac {15\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (2\,a^2+10\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {11\,a^2}{2}-\frac {9\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (2\,a^2+10\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {3\,a^2}{4}-\frac {15\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]
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