Integrand size = 12, antiderivative size = 30 \[ \int \frac {e^x}{1-\sin (x)} \, dx=(1+i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-i e^{i x}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4541, 4535} \[ \int \frac {e^x}{1-\sin (x)} \, dx=(1+i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-i e^{i x}\right ) \]
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Rule 4535
Rule 4541
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int e^x \sec ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx \\ & = (1+i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-i e^{i x}\right ) \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(61\) vs. \(2(30)=60\).
Time = 0.62 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03 \[ \int \frac {e^x}{1-\sin (x)} \, dx=\frac {2 e^x \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )}+(1+i) (1-(1+i) \operatorname {Hypergeometric2F1}(-i,1,1-i,-i \cos (x)+\sin (x))) (\cosh (x)+\sinh (x)) \]
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\[\int \frac {{\mathrm e}^{x}}{-\sin \left (x \right )+1}d x\]
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\[ \int \frac {e^x}{1-\sin (x)} \, dx=\int { -\frac {e^{x}}{\sin \left (x\right ) - 1} \,d x } \]
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\[ \int \frac {e^x}{1-\sin (x)} \, dx=- \int \frac {e^{x}}{\sin {\left (x \right )} - 1}\, dx \]
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\[ \int \frac {e^x}{1-\sin (x)} \, dx=\int { -\frac {e^{x}}{\sin \left (x\right ) - 1} \,d x } \]
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\[ \int \frac {e^x}{1-\sin (x)} \, dx=\int { -\frac {e^{x}}{\sin \left (x\right ) - 1} \,d x } \]
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Timed out. \[ \int \frac {e^x}{1-\sin (x)} \, dx=-\int \frac {{\mathrm {e}}^x}{\sin \left (x\right )-1} \,d x \]
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