Integrand size = 24, antiderivative size = 82 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {15 \text {arctanh}(\cos (c+d x))}{8 a d}+\frac {15 \sec (c+d x)}{8 a d}-\frac {5 \csc ^2(c+d x) \sec (c+d x)}{8 a d}-\frac {\csc ^4(c+d x) \sec (c+d x)}{4 a d} \]
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Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3254, 2702, 294, 327, 213} \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {15 \text {arctanh}(\cos (c+d x))}{8 a d}+\frac {15 \sec (c+d x)}{8 a d}-\frac {\csc ^4(c+d x) \sec (c+d x)}{4 a d}-\frac {5 \csc ^2(c+d x) \sec (c+d x)}{8 a d} \]
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Rule 213
Rule 294
Rule 327
Rule 2702
Rule 3254
Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc ^5(c+d x) \sec ^2(c+d x) \, dx}{a} \\ & = \frac {\text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\sec (c+d x)\right )}{a d} \\ & = -\frac {\csc ^4(c+d x) \sec (c+d x)}{4 a d}+\frac {5 \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{4 a d} \\ & = -\frac {5 \csc ^2(c+d x) \sec (c+d x)}{8 a d}-\frac {\csc ^4(c+d x) \sec (c+d x)}{4 a d}+\frac {15 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{8 a d} \\ & = \frac {15 \sec (c+d x)}{8 a d}-\frac {5 \csc ^2(c+d x) \sec (c+d x)}{8 a d}-\frac {\csc ^4(c+d x) \sec (c+d x)}{4 a d}+\frac {15 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{8 a d} \\ & = -\frac {15 \text {arctanh}(\cos (c+d x))}{8 a d}+\frac {15 \sec (c+d x)}{8 a d}-\frac {5 \csc ^2(c+d x) \sec (c+d x)}{8 a d}-\frac {\csc ^4(c+d x) \sec (c+d x)}{4 a d} \\ \end{align*}
Time = 2.74 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.61 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {14 \csc ^2\left (\frac {1}{2} (c+d x)\right )+\csc ^4\left (\frac {1}{2} (c+d x)\right )+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (78+\cos (c+d x) \left (-8 \left (8+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec ^4\left (\frac {1}{2} (c+d x)\right )\right )-14 \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}{64 a d} \]
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Time = 0.82 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06
method | result | size |
derivativedivides | \(\frac {\frac {1}{16 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {7}{16 \left (1+\cos \left (d x +c \right )\right )}-\frac {15 \ln \left (1+\cos \left (d x +c \right )\right )}{16}-\frac {1}{16 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {7}{16 \left (\cos \left (d x +c \right )-1\right )}+\frac {15 \ln \left (\cos \left (d x +c \right )-1\right )}{16}+\frac {1}{\cos \left (d x +c \right )}}{d a}\) | \(87\) |
default | \(\frac {\frac {1}{16 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {7}{16 \left (1+\cos \left (d x +c \right )\right )}-\frac {15 \ln \left (1+\cos \left (d x +c \right )\right )}{16}-\frac {1}{16 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {7}{16 \left (\cos \left (d x +c \right )-1\right )}+\frac {15 \ln \left (\cos \left (d x +c \right )-1\right )}{16}+\frac {1}{\cos \left (d x +c \right )}}{d a}\) | \(87\) |
parallelrisch | \(\frac {-160+120 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )+\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )+15 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(99\) |
norman | \(\frac {\frac {1}{64 a d}+\frac {15 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {15 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}\) | \(132\) |
risch | \(\frac {15 \,{\mathrm e}^{9 i \left (d x +c \right )}-40 \,{\mathrm e}^{7 i \left (d x +c \right )}+18 \,{\mathrm e}^{5 i \left (d x +c \right )}-40 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}}{4 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}\) | \(132\) |
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Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.65 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{4} - 50 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 16}{16 \, {\left (a d \cos \left (d x + c\right )^{5} - 2 \, a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )\right )}} \]
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\[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=- \frac {\int \frac {\csc ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \]
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Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 25 \, \cos \left (d x + c\right )^{2} + 8\right )}}{a \cos \left (d x + c\right )^{5} - 2 \, a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )} - \frac {15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (74) = 148\).
Time = 0.33 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.21 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {{\left (\frac {16 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {90 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} + \frac {60 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} - \frac {\frac {16 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {128}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{64 \, d} \]
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Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {15\,{\cos \left (c+d\,x\right )}^4}{8}-\frac {25\,{\cos \left (c+d\,x\right )}^2}{8}+1}{d\,\left (a\,{\cos \left (c+d\,x\right )}^5-2\,a\,{\cos \left (c+d\,x\right )}^3+a\,\cos \left (c+d\,x\right )\right )}-\frac {15\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{8\,a\,d} \]
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