Integrand size = 27, antiderivative size = 115 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {15 a \csc (c+d x)}{8 d}+\frac {b \log (\tan (c+d x))}{d}+\frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {b \tan ^2(c+d x)}{d}+\frac {b \tan ^4(c+d x)}{4 d} \]
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Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2913, 2701, 294, 327, 213, 2700, 272, 45} \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {15 a \csc (c+d x)}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \tan ^4(c+d x)}{4 d}+\frac {b \tan ^2(c+d x)}{d}+\frac {b \log (\tan (c+d x))}{d} \]
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Rule 45
Rule 213
Rule 272
Rule 294
Rule 327
Rule 2700
Rule 2701
Rule 2913
Rubi steps \begin{align*} \text {integral}& = a \int \csc ^2(c+d x) \sec ^5(c+d x) \, dx+b \int \csc (c+d x) \sec ^5(c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (c+d x)\right )}{d}+\frac {b \text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}-\frac {(5 a) \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{4 d}+\frac {b \text {Subst}\left (\int \frac {(1+x)^2}{x} \, dx,x,\tan ^2(c+d x)\right )}{2 d} \\ & = \frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}-\frac {(15 a) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}+\frac {b \text {Subst}\left (\int \left (2+\frac {1}{x}+x\right ) \, dx,x,\tan ^2(c+d x)\right )}{2 d} \\ & = -\frac {15 a \csc (c+d x)}{8 d}+\frac {b \log (\tan (c+d x))}{d}+\frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {b \tan ^2(c+d x)}{d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {(15 a) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d} \\ & = \frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {15 a \csc (c+d x)}{8 d}+\frac {b \log (\tan (c+d x))}{d}+\frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {b \tan ^2(c+d x)}{d}+\frac {b \tan ^4(c+d x)}{4 d} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(c+d x)\right )}{d}-\frac {b \log (\cos (c+d x))}{d}+\frac {b \log (\sin (c+d x))}{d}+\frac {b \sec ^2(c+d x)}{2 d}+\frac {b \sec ^4(c+d x)}{4 d} \]
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Time = 0.94 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(101\) |
default | \(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(101\) |
parallelrisch | \(\frac {-15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {8 b}{15}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {8 b}{15}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b \left (\frac {3 \cos \left (4 d x +4 c \right )}{4}-\frac {7}{4}+\cos \left (2 d x +2 c \right )\right )}{2 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(216\) |
risch | \(-\frac {i \left (15 a \,{\mathrm e}^{9 i \left (d x +c \right )}+40 a \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i b \,{\mathrm e}^{8 i \left (d x +c \right )}+18 a \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i b \,{\mathrm e}^{6 i \left (d x +c \right )}+40 a \,{\mathrm e}^{3 i \left (d x +c \right )}-24 i b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 a \,{\mathrm e}^{i \left (d x +c \right )}-8 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d}+\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(240\) |
norman | \(\frac {\frac {4 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a}{2 d}+\frac {13 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {5 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {13 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (15 a -8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (15 a +8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(250\) |
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Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.38 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {16 \, b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (15 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (15 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 30 \, a \cos \left (d x + c\right )^{4} + 10 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]
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Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.10 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (15 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (15 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (15 \, a \sin \left (d x + c\right )^{4} + 4 \, b \sin \left (d x + c\right )^{3} - 25 \, a \sin \left (d x + c\right )^{2} - 6 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]
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Time = 0.37 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.17 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (15 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (15 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {16 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (6 \, b \sin \left (d x + c\right )^{4} - 7 \, a \sin \left (d x + c\right )^{3} - 16 \, b \sin \left (d x + c\right )^{2} + 9 \, a \sin \left (d x + c\right ) + 12 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {15\,a}{16}-\frac {b}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {15\,a}{16}+\frac {b}{2}\right )}{d}+\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {15\,a\,{\sin \left (c+d\,x\right )}^4}{8}+\frac {b\,{\sin \left (c+d\,x\right )}^3}{2}-\frac {25\,a\,{\sin \left (c+d\,x\right )}^2}{8}-\frac {3\,b\,\sin \left (c+d\,x\right )}{4}+a}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )} \]
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