Integrand size = 21, antiderivative size = 78 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-a^2 x+\frac {b^2 x}{2}-\frac {2 a b \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a b \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{2 d} \]
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Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2801, 2715, 8, 2672, 327, 212, 3554} \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a^2 \cot (c+d x)}{d}+a^2 (-x)-\frac {2 a b \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a b \cos (c+d x)}{d}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {b^2 x}{2} \]
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Rule 8
Rule 212
Rule 327
Rule 2672
Rule 2715
Rule 2801
Rule 3554
Rubi steps \begin{align*} \text {integral}& = \int \left (b^2 \cos ^2(c+d x)+2 a b \cos (c+d x) \cot (c+d x)+a^2 \cot ^2(c+d x)\right ) \, dx \\ & = a^2 \int \cot ^2(c+d x) \, dx+(2 a b) \int \cos (c+d x) \cot (c+d x) \, dx+b^2 \int \cos ^2(c+d x) \, dx \\ & = -\frac {a^2 \cot (c+d x)}{d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{2 d}-a^2 \int 1 \, dx+\frac {1}{2} b^2 \int 1 \, dx-\frac {(2 a b) \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -a^2 x+\frac {b^2 x}{2}+\frac {2 a b \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {(2 a b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -a^2 x+\frac {b^2 x}{2}-\frac {2 a b \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a b \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-4 a^2 c+2 b^2 c-4 a^2 d x+2 b^2 d x+8 a b \cos (c+d x)-2 a^2 \cot \left (\frac {1}{2} (c+d x)\right )-8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^2 \sin (2 (c+d x))+2 a^2 \tan \left (\frac {1}{2} (c+d x)\right )}{4 d} \]
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Time = 0.35 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 a b \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(79\) |
default | \(\frac {a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 a b \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(79\) |
parallelrisch | \(\frac {2 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}-4 a^{2} d x +2 b^{2} d x +8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b -4 a^{2} \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+8 a b \cos \left (d x +c \right )+b^{2} \sin \left (2 d x +2 c \right )+8 a b}{4 d}\) | \(99\) |
risch | \(-a^{2} x +\frac {b^{2} x}{2}-\frac {i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\) | \(140\) |
norman | \(\frac {\left (-a^{2}+\frac {b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-a^{2}+\frac {b^{2}}{2}\right ) x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 a^{2}+b^{2}\right ) x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a^{2}}{2 d}+\frac {a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {\left (a^{2}-2 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (a^{2}-2 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {4 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(228\) |
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Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.51 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {b^{2} \cos \left (d x + c\right )^{3} + 2 \, a b \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a b \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right ) + {\left ({\left (2 \, a^{2} - b^{2}\right )} d x - 4 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \]
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\[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cos ^{2}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {4 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{2} - {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2} - 4 \, a b {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]
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Time = 0.40 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.90 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - {\left (2 \, a^{2} - b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
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Time = 10.18 (sec) , antiderivative size = 277, normalized size of antiderivative = 3.55 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {b^2\,\mathrm {atan}\left (\frac {-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )-2\,a^2\,\mathrm {atan}\left (\frac {-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )+2\,a\,b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\cos \left (c+d\,x\right )-\frac {b^2\,\cos \left (c+d\,x\right )}{8}+\frac {b^2\,\cos \left (3\,c+3\,d\,x\right )}{8}-a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\sin \left (c+d\,x\right )} \]
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