Integrand size = 19, antiderivative size = 52 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\log (\sin (c+d x))}{a^2 d}-\frac {\log (1+\sin (c+d x))}{a^2 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \]
[Out]
Time = 0.04 (sec) , antiderivative size = 52, 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 = {2786, 46} \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\log (\sin (c+d x))}{a^2 d}-\frac {\log (\sin (c+d x)+1)}{a^2 d} \]
[In]
[Out]
Rule 46
Rule 2786
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a^2 x}-\frac {1}{a (a+x)^2}-\frac {1}{a^2 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\log (\sin (c+d x))}{a^2 d}-\frac {\log (1+\sin (c+d x))}{a^2 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.69 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\log (\sin (c+d x))-\log (1+\sin (c+d x))+\frac {1}{1+\sin (c+d x)}}{a^2 d} \]
[In]
[Out]
Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.56
method | result | size |
derivativedivides | \(-\frac {\frac {1}{\csc \left (d x +c \right )+1}+\ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\) | \(29\) |
default | \(-\frac {\frac {1}{\csc \left (d x +c \right )+1}+\ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\) | \(29\) |
risch | \(\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{2}}\) | \(74\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.13 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 1}{a^{2} d \sin \left (d x + c\right ) + a^{2} d} \]
[In]
[Out]
\[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {1}{a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {\log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{d} \]
[In]
[Out]
none
Time = 0.34 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {a {\left (\frac {\log \left ({\left | -\frac {a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{3}} + \frac {1}{{\left (a \sin \left (d x + c\right ) + a\right )} a^{2}}\right )}}{d} \]
[In]
[Out]
Time = 9.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.67 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\right )} \]
[In]
[Out]