Integrand size = 21, antiderivative size = 77 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {a^2 x}{2}-b^2 x+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d} \]
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Time = 0.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3957, 2801, 2715, 8, 2672, 327, 212, 3554} \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a^2 x}{2}+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \tan (c+d x)}{d}-b^2 x \]
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Rule 8
Rule 212
Rule 327
Rule 2672
Rule 2715
Rule 2801
Rule 3554
Rule 3957
Rubi steps \begin{align*} \text {integral}& = \int (-b-a \cos (c+d x))^2 \tan ^2(c+d x) \, dx \\ & = \int \left (a^2 \sin ^2(c+d x)+2 a b \sin (c+d x) \tan (c+d x)+b^2 \tan ^2(c+d x)\right ) \, dx \\ & = a^2 \int \sin ^2(c+d x) \, dx+(2 a b) \int \sin (c+d x) \tan (c+d x) \, dx+b^2 \int \tan ^2(c+d x) \, dx \\ & = -\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}+\frac {1}{2} a^2 \int 1 \, dx-b^2 \int 1 \, dx+\frac {(2 a b) \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a^2 x}{2}-b^2 x-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}+\frac {(2 a b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a^2 x}{2}-b^2 x+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d} \\ \end{align*}
Time = 0.97 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.57 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=-\frac {-2 a^2 c+4 b^2 c-2 a^2 d x+4 b^2 d x+8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a b \sin (c+d x)+a^2 \sin (2 (c+d x))-4 b^2 \tan (c+d x)}{4 d} \]
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Time = 1.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(77\) |
default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(77\) |
parts | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}+\frac {2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
parallelrisch | \(\frac {-16 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+16 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-8 a b \sin \left (2 d x +2 c \right )-a^{2} \sin \left (3 d x +3 c \right )+4 d x \left (a^{2}-2 b^{2}\right ) \cos \left (d x +c \right )-\sin \left (d x +c \right ) \left (a^{2}-8 b^{2}\right )}{8 d \cos \left (d x +c \right )}\) | \(122\) |
risch | \(\frac {a^{2} x}{2}-b^{2} x +\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(146\) |
norman | \(\frac {\left (-\frac {a^{2}}{2}+b^{2}\right ) x +\left (-\frac {a^{2}}{2}+b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {a^{2}}{2}-b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {a^{2}}{2}-b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (a^{2}-4 a b -2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (a^{2}+4 a b -2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(232\) |
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Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {{\left (a^{2} - 2 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 2 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 4 \, a b \cos \left (d x + c\right ) - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
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\[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{2}{\left (c + d x \right )}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 4 \, {\left (d x + c - \tan \left (d x + c\right )\right )} b^{2} + 4 \, a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (73) = 146\).
Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.06 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
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Time = 14.66 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.86 \[ \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]
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