Integrand size = 16, antiderivative size = 42 \[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=(2-2 i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-e^{i x}\right )-\frac {e^x \sin (x)}{1+\cos (x)} \]
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Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4551, 4548, 4527, 2225, 2283, 2326} \[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=-4 i e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-e^{i x}\right )+2 i e^x+\frac {e^x \sin (x)}{\cos (x)+1} \]
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Rule 2225
Rule 2283
Rule 2326
Rule 4527
Rule 4548
Rule 4551
Rubi steps \begin{align*} \text {integral}& = -\left (2 \int \frac {e^x \sin (x)}{1+\cos (x)} \, dx\right )+\int \frac {e^x (1+\sin (x))}{1+\cos (x)} \, dx \\ & = \frac {e^x \sin (x)}{1+\cos (x)}-2 \int e^x \tan \left (\frac {x}{2}\right ) \, dx \\ & = \frac {e^x \sin (x)}{1+\cos (x)}-2 i \int \left (-e^x+\frac {2 e^x}{1+e^{i x}}\right ) \, dx \\ & = \frac {e^x \sin (x)}{1+\cos (x)}+2 i \int e^x \, dx-4 i \int \frac {e^x}{1+e^{i x}} \, dx \\ & = 2 i e^x-4 i e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-e^{i x}\right )+\frac {e^x \sin (x)}{1+\cos (x)} \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(42)=84\).
Time = 0.43 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.07 \[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=-\frac {2 e^x \cos \left (\frac {x}{2}\right ) \left (2 i \cos \left (\frac {x}{2}\right ) \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-e^{i x}\right )-(1+i) e^{i x} \cos \left (\frac {x}{2}\right ) \operatorname {Hypergeometric2F1}\left (1,1-i,2-i,-e^{i x}\right )-\sin \left (\frac {x}{2}\right )\right )}{1+\cos (x)} \]
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\[\int \frac {{\mathrm e}^{x} \left (-\sin \left (x \right )+1\right )}{\cos \left (x \right )+1}d x\]
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\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) - 1\right )} e^{x}}{\cos \left (x\right ) + 1} \,d x } \]
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\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=- \int \left (- \frac {e^{x}}{\cos {\left (x \right )} + 1}\right )\, dx - \int \frac {e^{x} \sin {\left (x \right )}}{\cos {\left (x \right )} + 1}\, dx \]
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\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) - 1\right )} e^{x}}{\cos \left (x\right ) + 1} \,d x } \]
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\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) - 1\right )} e^{x}}{\cos \left (x\right ) + 1} \,d x } \]
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Timed out. \[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (\sin \left (x\right )-1\right )}{\cos \left (x\right )+1} \,d x \]
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