Integrand size = 27, antiderivative size = 74 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a \tan ^4(c+d x)}{4 d} \]
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Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2913, 2687, 30, 2691, 3855} \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {a \tan ^4(c+d x)}{4 d}+\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3 b \tan (c+d x) \sec (c+d x)}{8 d} \]
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Rule 30
Rule 2687
Rule 2691
Rule 2913
Rule 3855
Rubi steps \begin{align*} \text {integral}& = a \int \sec ^2(c+d x) \tan ^3(c+d x) \, dx+b \int \sec (c+d x) \tan ^4(c+d x) \, dx \\ & = \frac {b \sec (c+d x) \tan ^3(c+d x)}{4 d}-\frac {1}{4} (3 b) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\frac {a \text {Subst}\left (\int x^3 \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a \tan ^4(c+d x)}{4 d}+\frac {1}{8} (3 b) \int \sec (c+d x) \, dx \\ & = \frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a \tan ^4(c+d x)}{4 d} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {3 b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b \sec (c+d x) \tan ^3(c+d x)}{d}+\frac {a \tan ^4(c+d x)}{4 d} \]
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Time = 0.57 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.32
method | result | size |
derivativedivides | \(\frac {\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(98\) |
default | \(\frac {\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(98\) |
risch | \(\frac {i \left (8 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+5 b \,{\mathrm e}^{7 i \left (d x +c \right )}-3 b \,{\mathrm e}^{5 i \left (d x +c \right )}+8 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}\) | \(134\) |
parallelrisch | \(\frac {-6 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 a \cos \left (2 d x +2 c \right )+\cos \left (4 d x +4 c \right ) a +3 b \sin \left (d x +c \right )-5 b \sin \left (3 d x +3 c \right )+3 a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(152\) |
norman | \(\frac {-\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(187\) |
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Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 \, b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, a \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \]
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Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\text {Timed out} \]
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Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.20 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
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Time = 0.37 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 \, b \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, b \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 17.75 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.95 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \]
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