Integrand size = 21, antiderivative size = 50 \[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\cot (c+d x)}{a d}-\frac {b \log (\tan (c+d x))}{a^2 d}+\frac {b \log (a+b \tan (c+d x))}{a^2 d} \]
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Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3597, 46} \[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {b \log (\tan (c+d x))}{a^2 d}+\frac {b \log (a+b \tan (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a d} \]
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Rule 46
Rule 3597
Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {1}{x^2 (a+x)} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \left (\frac {1}{a x^2}-\frac {1}{a^2 x}+\frac {1}{a^2 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d} \\ & = -\frac {\cot (c+d x)}{a d}-\frac {b \log (\tan (c+d x))}{a^2 d}+\frac {b \log (a+b \tan (c+d x))}{a^2 d} \\ \end{align*}
Time = 0.53 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-a \cot (c+d x)+b (-\log (\sin (c+d x))+\log (a \cos (c+d x)+b \sin (c+d x)))}{a^2 d} \]
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Time = 0.79 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {1}{a \tan \left (d x +c \right )}-\frac {b \ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}}}{d}\) | \(48\) |
default | \(\frac {-\frac {1}{a \tan \left (d x +c \right )}-\frac {b \ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}}}{d}\) | \(48\) |
risch | \(-\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{2} d}\) | \(82\) |
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Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.90 \[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {b \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) \sin \left (d x + c\right ) - b \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right )}{2 \, a^{2} d \sin \left (d x + c\right )} \]
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\[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2}} - \frac {b \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac {1}{a \tan \left (d x + c\right )}}{d} \]
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Time = 0.42 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.20 \[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {b \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {b \tan \left (d x + c\right ) - a}{a^{2} \tan \left (d x + c\right )}}{d} \]
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Time = 4.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78 \[ \int \frac {\csc ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {2\,b\,\mathrm {tan}\left (c+d\,x\right )}{a}+1\right )}{a^2\,d}-\frac {\mathrm {cot}\left (c+d\,x\right )}{a\,d} \]
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