Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))} \]
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Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2743, 12, 2739, 630, 31} \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}+\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d} \]
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Rule 12
Rule 31
Rule 630
Rule 2739
Rule 2743
Rubi steps \begin{align*} \text {integral}& = \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}+\frac {1}{16} \int -\frac {3}{3-5 \sin (c+d x)} \, dx \\ & = \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {3}{16} \int \frac {1}{3-5 \sin (c+d x)} \, dx \\ & = \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {3 \text {Subst}\left (\int \frac {1}{3-10 x+3 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \\ & = \frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))}-\frac {9 \text {Subst}\left (\int \frac {1}{-9+3 x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {9 \text {Subst}\left (\int \frac {1}{-1+3 x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{64 d} \\ & = \frac {3 \log \left (1-3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 \log \left (3-\tan \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 \cos (c+d x)}{16 d (3-5 \sin (c+d x))} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.44 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {9 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+20 \left (\frac {3}{\cos \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )}+\frac {1}{3 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{192 d} \]
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Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{64}-\frac {5}{48 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}}{d}\) | \(68\) |
default | \(\frac {-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{64}-\frac {5}{48 \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}}{d}\) | \(68\) |
risch | \(\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}-5 i}{8 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-5-6 i {\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {3 \ln \left (-\frac {4}{5}-\frac {3 i}{5}+{\mathrm e}^{i \left (d x +c \right )}\right )}{64 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {4}{5}-\frac {3 i}{5}\right )}{64 d}\) | \(86\) |
norman | \(\frac {\frac {5}{8 d}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d}}{3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )}{64 d}+\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 d}\) | \(87\) |
parallelrisch | \(\frac {-45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right ) \sin \left (d x +c \right )+45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3}\right ) \sin \left (d x +c \right )+27 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right )-27 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3}\right )+100 \sin \left (d x +c \right )-60 \cos \left (d x +c \right )-60}{192 d \left (-3+5 \sin \left (d x +c \right )\right )}\) | \(104\) |
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Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=-\frac {3 \, {\left (5 \, \sin \left (d x + c\right ) - 3\right )} \log \left (4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) - 3 \, {\left (5 \, \sin \left (d x + c\right ) - 3\right )} \log \left (-4 \, \cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) + 5\right ) + 40 \, \cos \left (d x + c\right )}{128 \, {\left (5 \, d \sin \left (d x + c\right ) - 3 \, d\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (78) = 156\).
Time = 0.75 (sec) , antiderivative size = 462, normalized size of antiderivative = 5.13 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\begin {cases} \frac {x}{\left (3 - 5 \sin {\left (2 \operatorname {atan}{\left (\frac {1}{3} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\left (\frac {1}{3} \right )} \\\frac {x}{\left (3 - 5 \sin {\left (2 \operatorname {atan}{\left (3 \right )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\left (3 \right )} \\\frac {x}{\left (3 - 5 \sin {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {90 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {27 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {27 \log {\left (3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {90 \log {\left (3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {27 \log {\left (3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} - \frac {200 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} + \frac {120}{576 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1920 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 576 d} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {\frac {40 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 3\right )}}{\frac {10 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 3} + 9 \, \log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right ) - 9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 3\right )}{192 \, d} \]
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Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=-\frac {\frac {40 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3\right )}}{3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3} - 9 \, \log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \right |}\right )}{192 \, d} \]
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Time = 6.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(3-5 \sin (c+d x))^2} \, dx=\frac {3\,\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {5}{4}\right )}{32\,d}-\frac {\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{72}-\frac {5}{24}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {10\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1\right )} \]
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