Integrand size = 18, antiderivative size = 46 \[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=(2+2 i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-i e^{i x}\right )-\frac {e^x \cos (x)}{1-\sin (x)} \]
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Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4550, 4547, 4527, 2225, 2283, 2326} \[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=-4 i e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-i e^{i x}\right )+2 i e^x+\frac {e^x \cos (x)}{1-\sin (x)} \]
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Rule 2225
Rule 2283
Rule 2326
Rule 4527
Rule 4547
Rule 4550
Rubi steps \begin{align*} \text {integral}& = -\left (2 \int \frac {e^x \cos (x)}{1-\sin (x)} \, dx\right )+\int \frac {e^x (1+\cos (x))}{1-\sin (x)} \, dx \\ & = \frac {e^x \cos (x)}{1-\sin (x)}-2 \int e^x \tan \left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx \\ & = \frac {e^x \cos (x)}{1-\sin (x)}-2 i \int \left (-e^x+\frac {2 e^x}{1+e^{2 i \left (\frac {\pi }{4}+\frac {x}{2}\right )}}\right ) \, dx \\ & = \frac {e^x \cos (x)}{1-\sin (x)}+2 i \int e^x \, dx-4 i \int \frac {e^x}{1+e^{2 i \left (\frac {\pi }{4}+\frac {x}{2}\right )}} \, dx \\ & = 2 i e^x-4 i e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-i e^{i x}\right )+\frac {e^x \cos (x)}{1-\sin (x)} \\ \end{align*}
Time = 0.66 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.57 \[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\frac {1}{2} (-1+\cos (x)) \csc ^2\left (\frac {x}{2}\right ) \left (-\frac {e^x \left ((1-2 i)+(1+2 i) \cot \left (\frac {x}{2}\right )\right )}{-1+\cot \left (\frac {x}{2}\right )}+4 i \operatorname {Hypergeometric2F1}(-i,1,1-i,-i \cos (x)+\sin (x)) (\cosh (x)+\sinh (x))\right ) \]
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\[\int \frac {{\mathrm e}^{x} \left (1-\cos \left (x \right )\right )}{-\sin \left (x \right )+1}d x\]
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\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int { \frac {{\left (\cos \left (x\right ) - 1\right )} e^{x}}{\sin \left (x\right ) - 1} \,d x } \]
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\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int \frac {\left (\cos {\left (x \right )} - 1\right ) e^{x}}{\sin {\left (x \right )} - 1}\, dx \]
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\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int { \frac {{\left (\cos \left (x\right ) - 1\right )} e^{x}}{\sin \left (x\right ) - 1} \,d x } \]
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\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int { \frac {{\left (\cos \left (x\right ) - 1\right )} e^{x}}{\sin \left (x\right ) - 1} \,d x } \]
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Timed out. \[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (\cos \left (x\right )-1\right )}{\sin \left (x\right )-1} \,d x \]
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