Integrand size = 26, antiderivative size = 80 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {b^2 \log (b+a \cos (c+d x))}{a \left (a^2-b^2\right ) d} \]
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Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4482, 2916, 12, 1643} \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {b^2 \log (a \cos (c+d x)+b)}{a d \left (a^2-b^2\right )}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}+\frac {\log (\cos (c+d x)+1)}{2 d (a-b)} \]
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Rule 12
Rule 1643
Rule 2916
Rule 4482
Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos (c+d x) \cot (c+d x)}{b+a \cos (c+d x)} \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {x^2}{a^2 (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {x^2}{(b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{a d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {a}{2 (a+b) (a-x)}-\frac {a}{2 (a-b) (a+x)}+\frac {b^2}{(a-b) (a+b) (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{a d} \\ & = \frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {b^2 \log (b+a \cos (c+d x))}{a \left (a^2-b^2\right ) d} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {a (a+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-b^2 \log (b+a \cos (c+d x))+a (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a (a-b) (a+b) d} \]
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Time = 0.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}-\frac {b^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) a}}{d}\) | \(75\) |
default | \(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}-\frac {b^{2} \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right ) a}}{d}\) | \(75\) |
risch | \(\frac {i x}{a}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {2 i b^{2} x}{a \left (a^{2}-b^{2}\right )}+\frac {2 i b^{2} c}{a d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a d \left (a^{2}-b^{2}\right )}\) | \(193\) |
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Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d} \]
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\[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\cos {\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]
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Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {b^{2} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{3} - a b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {\log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a}}{d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (76) = 152\).
Time = 0.36 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.21 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {a \log \left ({\left | -a - b + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \right |}\right )}{a^{2} - b^{2}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (\frac {{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | a \right |} \right |}}{{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | a \right |} \right |}}\right )}{{\left (a^{2} - b^{2}\right )} {\left | a \right |}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]
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Time = 22.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}+\frac {b^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a\,b^2-a^3\right )} \]
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