Integrand size = 13, antiderivative size = 139 \[ \int \cot (c+d x) \sin (a+b x) \, dx=-\frac {i e^{-i (a+b x)}}{2 b}-\frac {i e^{i (a+b x)}}{2 b}+\frac {i e^{-i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{2 d},1-\frac {b}{2 d},e^{2 i (c+d x)}\right )}{b}+\frac {i e^{i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{2 d},1+\frac {b}{2 d},e^{2 i (c+d x)}\right )}{b} \]
[Out]
Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4655, 2225, 2283} \[ \int \cot (c+d x) \sin (a+b x) \, dx=\frac {i e^{-i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{2 d},1-\frac {b}{2 d},e^{2 i (c+d x)}\right )}{b}+\frac {i e^{i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{2 d},\frac {b}{2 d}+1,e^{2 i (c+d x)}\right )}{b}-\frac {i e^{-i (a+b x)}}{2 b}-\frac {i e^{i (a+b x)}}{2 b} \]
[In]
[Out]
Rule 2225
Rule 2283
Rule 4655
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2} e^{-i (a+b x)}+\frac {1}{2} e^{i (a+b x)}+\frac {e^{-i (a+b x)}}{1-e^{2 i (c+d x)}}-\frac {e^{i (a+b x)}}{1-e^{2 i (c+d x)}}\right ) \, dx \\ & = -\left (\frac {1}{2} \int e^{-i (a+b x)} \, dx\right )+\frac {1}{2} \int e^{i (a+b x)} \, dx+\int \frac {e^{-i (a+b x)}}{1-e^{2 i (c+d x)}} \, dx-\int \frac {e^{i (a+b x)}}{1-e^{2 i (c+d x)}} \, dx \\ & = -\frac {i e^{-i (a+b x)}}{2 b}-\frac {i e^{i (a+b x)}}{2 b}+\frac {i e^{-i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{2 d},1-\frac {b}{2 d},e^{2 i (c+d x)}\right )}{b}+\frac {i e^{i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{2 d},1+\frac {b}{2 d},e^{2 i (c+d x)}\right )}{b} \\ \end{align*}
Time = 5.37 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.87 \[ \int \cot (c+d x) \sin (a+b x) \, dx=\frac {-\cos (a) \cos (b x) \cot (c)-\frac {i e^{-i (a-2 c+b x)} \left (b e^{2 i d x} \operatorname {Hypergeometric2F1}\left (1,1-\frac {b}{2 d},2-\frac {b}{2 d},e^{2 i (c+d x)}\right )-(b-2 d) \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{2 d},1-\frac {b}{2 d},e^{2 i (c+d x)}\right )\right )}{(b-2 d) \left (-1+e^{2 i c}\right )}-\frac {i e^{i (a+2 c+b x)} \left (b e^{2 i d x} \operatorname {Hypergeometric2F1}\left (1,1+\frac {b}{2 d},2+\frac {b}{2 d},e^{2 i (c+d x)}\right )-(b+2 d) \operatorname {Hypergeometric2F1}\left (1,\frac {b}{2 d},1+\frac {b}{2 d},e^{2 i (c+d x)}\right )\right )}{(b+2 d) \left (-1+e^{2 i c}\right )}+\cot (c) \sin (a) \sin (b x)}{b} \]
[In]
[Out]
\[\int \cot \left (d x +c \right ) \sin \left (x b +a \right )d x\]
[In]
[Out]
\[ \int \cot (c+d x) \sin (a+b x) \, dx=\int { \cot \left (d x + c\right ) \sin \left (b x + a\right ) \,d x } \]
[In]
[Out]
\[ \int \cot (c+d x) \sin (a+b x) \, dx=\int \sin {\left (a + b x \right )} \cot {\left (c + d x \right )}\, dx \]
[In]
[Out]
\[ \int \cot (c+d x) \sin (a+b x) \, dx=\int { \cot \left (d x + c\right ) \sin \left (b x + a\right ) \,d x } \]
[In]
[Out]
\[ \int \cot (c+d x) \sin (a+b x) \, dx=\int { \cot \left (d x + c\right ) \sin \left (b x + a\right ) \,d x } \]
[In]
[Out]
Timed out. \[ \int \cot (c+d x) \sin (a+b x) \, dx=\int \mathrm {cot}\left (c+d\,x\right )\,\sin \left (a+b\,x\right ) \,d x \]
[In]
[Out]