Integrand size = 19, antiderivative size = 89 \[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {3 a x}{8}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d} \]
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Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3957, 2917, 2672, 308, 212, 2715, 8} \[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=-\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d} \]
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Rule 8
Rule 212
Rule 308
Rule 2672
Rule 2715
Rule 2917
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int (-b-a \cos (c+d x)) \sin ^3(c+d x) \tan (c+d x) \, dx \\ & = a \int \sin ^4(c+d x) \, dx+b \int \sin ^3(c+d x) \tan (c+d x) \, dx \\ & = -\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{4} (3 a) \int \sin ^2(c+d x) \, dx+\frac {b \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} (3 a) \int 1 \, dx+\frac {b \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {3 a x}{8}-\frac {b \sin (c+d x)}{d}-\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {b \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {3 a x}{8}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 d} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {3 a (c+d x)}{8 d}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \sin (4 (c+d x))}{32 d} \]
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Time = 1.73 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(76\) |
default | \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(76\) |
parts | \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(78\) |
parallelrisch | \(\frac {36 a x d -120 \sin \left (d x +c \right ) b +3 a \sin \left (4 d x +4 c \right )-24 a \sin \left (2 d x +2 c \right )+96 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-96 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \sin \left (3 d x +3 c \right ) b}{96 d}\) | \(87\) |
risch | \(\frac {3 a x}{8}+\frac {5 i b \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {5 i b \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}+\frac {b \sin \left (3 d x +3 c \right )}{12 d}-\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(120\) |
norman | \(\frac {\frac {3 a x}{8}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {9 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {\left (3 a -8 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {\left (3 a +8 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (33 a -104 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {\left (33 a +104 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(208\) |
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Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {9 \, a d x + 12 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, b \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a \cos \left (d x + c\right )^{3} + 8 \, b \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 32 \, b\right )} \sin \left (d x + c\right )}{24 \, d} \]
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\[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \sin ^{4}{\left (c + d x \right )}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91 \[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a - 16 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} b}{96 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (81) = 162\).
Time = 0.30 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.93 \[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {9 \, {\left (d x + c\right )} a + 24 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 33 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 104 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 33 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 104 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
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Time = 14.54 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.00 \[ \int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx=\frac {3\,a\,\mathrm {atan}\left (\frac {27\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,\left (\frac {27\,a^3}{8}+24\,a\,b^2\right )}+\frac {24\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {27\,a^3}{8}+24\,a\,b^2}\right )}{4\,d}+\frac {2\,b\,\mathrm {atanh}\left (\frac {64\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,a^2\,b+64\,b^3}+\frac {9\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,a^2\,b+64\,b^3}\right )}{d}-\frac {\left (2\,b-\frac {3\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {26\,b}{3}-\frac {11\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {11\,a}{4}+\frac {26\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {3\,a}{4}+2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
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