Integrand size = 14, antiderivative size = 43 \[ \int \frac {e^x (1+\cos (x))}{1+\sin (x)} \, dx=(-2-2 i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,i e^{i x}\right )+\frac {e^x \cos (x)}{1+\sin (x)} \]
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Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4550, 4547, 4527, 2225, 2283, 2326} \[ \int \frac {e^x (1+\cos (x))}{1+\sin (x)} \, dx=4 i e^x \operatorname {Hypergeometric2F1}\left (i,1,1+i,-i e^{-i x}\right )-2 i e^x-\frac {e^x \cos (x)}{\sin (x)+1} \]
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Rule 2225
Rule 2283
Rule 2326
Rule 4527
Rule 4547
Rule 4550
Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {e^x \cos (x)}{1+\sin (x)} \, dx+\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 = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.70 \[ \int \frac {e^x (1+\cos (x))}{1+\sin (x)} \, dx=\frac {1}{2} (1+\cos (x)) \sec ^2\left (\frac {x}{2}\right ) \left (-4 i \operatorname {Hypergeometric2F1}(-i,1,1-i,i \cos (x)-\sin (x)) (\cosh (x)+\sinh (x))+\frac {e^x \left ((-1+2 i)+(1+2 i) \tan \left (\frac {x}{2}\right )\right )}{1+\tan \left (\frac {x}{2}\right )}\right ) \]
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\[\int \frac {{\mathrm e}^{x} \left (\cos \left (x \right )+1\right )}{\sin \left (x \right )+1}d x\]
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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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