Integrand size = 19, antiderivative size = 51 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a x}{2}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \]
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Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3957, 2917, 2672, 327, 212, 2715, 8} \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=-\frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d} \]
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Rule 8
Rule 212
Rule 327
Rule 2672
Rule 2715
Rule 2917
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int (-b-a \cos (c+d x)) \sin (c+d x) \tan (c+d x) \, dx \\ & = a \int \sin ^2(c+d x) \, dx+b \int \sin (c+d x) \tan (c+d x) \, dx \\ & = -\frac {a \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{2} a \int 1 \, dx+\frac {b \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a x}{2}-\frac {b \sin (c+d x)}{d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a x}{2}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a (c+d x)}{2 d}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {a \sin (2 (c+d x))}{4 d} \]
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Time = 1.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(55\) |
default | \(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(55\) |
parts | \(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(57\) |
parallelrisch | \(\frac {2 a x d -4 \sin \left (d x +c \right ) b -a \sin \left (2 d x +2 c \right )+4 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}\) | \(63\) |
risch | \(\frac {a x}{2}+\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(90\) |
norman | \(\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {\left (a -2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {a x}{2}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}-\frac {\left (a +2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(126\) |
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Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a d x + b \log \left (\sin \left (d x + c\right ) + 1\right ) - b \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a \cos \left (d x + c\right ) + 2 \, b\right )} \sin \left (d x + c\right )}{2 \, d} \]
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\[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \sin ^{2}{\left (c + d x \right )}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a + 2 \, b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (47) = 94\).
Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.24 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {{\left (d x + c\right )} a + 2 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
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Time = 13.93 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.63 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {b\,\sin \left (c+d\,x\right )}{d} \]
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