Integrand size = 19, antiderivative size = 74 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {a \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \]
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Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2800, 815} \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}-\frac {\log (\sin (c+d x)+1)}{2 d (a-b)} \]
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Rule 815
Rule 2800
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{2 (a+b) (b-x)}+\frac {a}{(a-b) (a+b) (a+x)}-\frac {1}{2 (a-b) (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {a \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {(-a+b) \log (1-\sin (c+d x))-(a+b) \log (1+\sin (c+d x))+2 a \log (a+b \sin (c+d x))}{2 (a-b) (a+b) d} \]
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Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}}{d}\) | \(71\) |
default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}}{d}\) | \(71\) |
risch | \(\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {2 i a x}{a^{2}-b^{2}}-\frac {2 i a c}{d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{2}-b^{2}\right )}\) | \(175\) |
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Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a - b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]
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\[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]
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Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]
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Time = 14.55 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.23 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^2-b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )} \]
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